- A scale factor compares new length with original length. - Since this is a reduction, should the factor be above or below \(1\)?

1. Divide the new length by the original length: \(k=\frac{10}{15}=\frac{2}{3}\).

\(k=\frac{2}{3}\)

- Write enlarged width as scale factor times original width. - Which operation undoes the enlargement?

1. The enlarged width equals \(2.5\) times the original. 2. Original width \(=20\div2.5=8\) inches.

\(8\) inches

- Which side in the second triangle corresponds to \(AB\)? - Divide its length by the matching length in the first triangle.

1. The matching sides \(AB\) and \(DE\) are labeled \(4\) and \(8\) inches. 2. Their ratio gives \(k=\frac{8}{4}=2\).

\(k=2\)

- Does a scaled copy add a fixed amount or multiply by a fixed number? - Compare more than one pair of corresponding sides.

1. A scaled copy must multiply every corresponding side by the same positive number. 2. The shortest-side ratio is \(\frac{6}{3}=2\), while the middle-side ratio is \(\frac{8}{5}=1.6\). 3. These differ, so the triangles are not scaled copies, even though each side increased by \(3\).

No. Adding \(3\) to each side does not give a common scale factor.

- Think of the perimeter as four side lengths added together. - What happens to each term in that sum when the side length is enlarged?

1. A perimeter is a sum of lengths, so it changes by the same scale factor. 2. New perimeter \(=12\cdot4=48\) inches.

\(48\) inches

- A square's area is its side length times itself. - What happens to that product if both factors are doubled?

1. Doubling both side lengths multiplies area by \(2^2=4\). 2. New area \(=9\cdot4=36\) square feet.

\(36\) square feet

- Look at the two smaller triangles on either side of the angle bisector. - Which whole-triangle sides meet the two parts of \(BC\)?

1. The Angle Bisector Theorem gives \(\frac{BD}{DC}=\frac{AB}{AC}\). 2. Substitute \(\frac{6}{9}=\frac{2}{3}\), so \(BD:DC=2:3\).

\(BD:DC=2:3\) because \(BD\) matches \(AB\) and \(DC\) matches \(AC\).

- What does the triangle midsegment theorem say about a segment joining two side midpoints? - How does each new side length compare with the original side length? - Use the perimeter formula for an equilateral triangle.

1. A segment joining two side midpoints of a triangle is parallel to the third side and half its length. Each new segment therefore has length \(12\,\text{cm} \div 2 = 6\,\text{cm}\). 2. Every smaller triangle has three sides of length \(6\,\text{cm}\), so all four are equilateral. 3. The perimeter of one smaller triangle is \(3 \cdot 6\,\text{cm} = 18\,\text{cm}\). 4. The original perimeter is \(3 \cdot 12\,\text{cm} = 36\,\text{cm}\). Thus, \(\frac{18}{36} = \frac{1}{2}\).

a) All four smaller triangles are equilateral. b) The perimeter of one smaller triangle is \(18\,\text{cm}\). c) Its perimeter is \(\frac{1}{2}\) of the original perimeter.

- How is an area factor related to a linear scale factor? - Is a height a linear measurement or an area measurement? - Which operation reverses squaring?

1. For similar figures, the area factor is the square of the linear scale factor: \(k^2=2.25\). 2. Therefore, \(k=\sqrt{2.25}=1.5\). 3. Corresponding heights are linear measurements, so they scale by \(k\). Thus, \(\frac{h_{a,1}}{h_a}=1.5=\frac{3}{2}\).

The ratio of \(h_{a,1}\) to \(h_a\) is \(3\) to \(2\).

- How do the perimeters of similar triangles compare? - What scale factor is determined by the two perimeters? - How does the scale factor affect every corresponding side length?

1. The perimeter of the first triangle is \(P_1=4.5\,\text{cm}+6\,\text{cm}+7.5\,\text{cm}=18\,\text{cm}\). 2. The scale factor is the perimeter ratio: \(k=\frac{P_2}{P_1}=\frac{54}{18}=3\). 3. Multiply each side length by \(3\): \(4.5\,\text{cm}\cdot3=13.5\,\text{cm}\), \(6\,\text{cm}\cdot3=18\,\text{cm}\), and \(7.5\,\text{cm}\cdot3=22.5\,\text{cm}\).

The side lengths of the second triangle are \(13.5\,\text{cm}\), \(18\,\text{cm}\), and \(22.5\,\text{cm}\).

- How are the area ratio and the side-length ratio related for similar figures? - What happens to area when every side length is doubled or tripled? - How can you use the area ratio to find the scale factor for lengths?

1. The ratio of the areas is \(\frac{A_{\text{large}}}{A_{\text{small}}}=\frac{162\,\text{cm}^2}{18\,\text{cm}^2}=9\). 2. For similar figures, the area ratio equals the square of the scale factor, so \(k^2=9\). 3. Since a scale factor is positive, \(k=\sqrt{9}=3\). 4. Therefore, every side length of the larger pentagon is three times the corresponding side length of the smaller pentagon.

The scale factor is \(k=3\). The ratio of a side length of the larger pentagon to the corresponding side length of the smaller pentagon is \(3\) to \(1\).

- Which side of the first triangle corresponds to the given side of the second triangle? - How does a dilation change every side length? - What ratio gives the scale factor?

1. The longest side of the first triangle is \(9\,\text{cm}\), so it corresponds to the \(13.5\,\text{cm}\) side. 2. The scale factor is \(k=\frac{13.5}{9}=1.5\). 3. The side corresponding to \(4.5\,\text{cm}\) is \(4.5\,\text{cm}\cdot1.5=6.75\,\text{cm}\). 4. The side corresponding to \(6\,\text{cm}\) is \(6\,\text{cm}\cdot1.5=9\,\text{cm}\).

The other two side lengths are \(6.75\,\text{cm}\) and \(9\,\text{cm}\).

- First find \(ZB\) and \(ZD\). - Compare the fractions of the full ray lengths cut off by the inner segment. - What does the converse of the Triangle Proportionality Theorem imply when those ratios are equal?

1. Find the full distances from \(Z\): \(ZB=3.5+5.25=8.75\,\text{cm}\) and \(ZD=4.2+6.3=10.5\,\text{cm}\). 2. Compare the ratios: \(\frac{ZA}{ZB}=\frac{3.5}{8.75}=0.4\) and \(\frac{ZC}{ZD}=\frac{4.2}{10.5}=0.4\). 3. Because the two rays are divided proportionally, the converse of the Triangle Proportionality Theorem shows that \(AC\parallel BD\).

Yes. Since \(\frac{ZA}{ZB}=\frac{ZC}{ZD}=0.4\), the converse of the Triangle Proportionality Theorem gives \(AC\parallel BD\).

- Use the diagram to identify the two triangles on opposite sides of \(S\). - Which sides of those triangles are parallel? - Match each parallel side with its corresponding distance from \(S\).

1. Because \(AB\parallel CD\), triangles \(SAB\) and \(SCD\) are similar by AA. 2. Corresponding sides are proportional: \(\frac{AB}{CD}=\frac{SA}{SC}\). 3. Substitute the measurements: \(\frac{x}{6}=\frac{24}{8}=3\). 4. Therefore, \(x=3\cdot6=18\,\text{m}\).

The river is \(18\,\text{m}\) wide.

- Convert the percent increase in length to a linear scale factor. - Square the factor for area and cube it for volume. - Subtract \(100\%\) to find the percent increase.

1. A \(40\%\) increase in length gives a linear scale factor of \(k=1.4\). 2. For part a), the area factor is \(1.4^2=1.96\), so the area increase is \(96\%\). 3. For part b), the volume factor is \(1.4^3=2.744\), so the volume increase is \(174.4\%\).

a) The area increases by \(96\%\). b) The volume increases by \(174.4\%\).

- Match the distances from \(S\) with the corresponding parallel segments. - Use the given ratio to find \(SA\). - Subtract \(SA\) from \(SB\) to find \(AB\).

1. The triangles formed by the parallel segments are similar, so \(\frac{SA}{SB}=\frac{AC}{BD}=\frac{2}{5}\). 2. Therefore, \(SA=15\cdot\frac{2}{5}=6\,\text{cm}\). 3. Since \(A\) lies between \(S\) and \(B\), \(AB=SB-SA=15-6=9\,\text{cm}\).

\(SA=6\,\text{cm}\) and \(AB=9\,\text{cm}\)

- Imagine the solid fitted inside a box. How many independent length directions change when it is enlarged? - How would the box's volume change if the same multiplier applied to every edge? - Use the fraction form of the scale factor to simplify the calculation.

1. Volume scales by the cube of the linear factor: \(V_{\text{new}}=250\left(\frac{6}{5}\right)^3\). 2. Since \(\left(\frac{6}{5}\right)^3=\frac{216}{125}\), \(V_{\text{new}}=250\cdot\frac{216}{125}=2\cdot216=432\,\text{cm}^3\).

\(432\,\text{cm}^3\)

- What is the ratio of the two corresponding altitudes? - Does the same ratio apply to the corresponding bases? - Write a proportion with the unknown base as the only variable.

1. Corresponding lengths in similar triangles have the same ratio: \(\frac{c_2}{c}=\frac{h_{c,2}}{h_c}\). 2. Substitute the values: \(\frac{c_2}{12.5}=\frac{6}{8}=0.75\). 3. Solve for \(c_2\): \(c_2=0.75\cdot12.5\,\text{cm}=9.375\,\text{cm}\).

\(c_2=9.375\,\text{cm}\)

- How does area change when every length is multiplied by a scale factor? - First find the scale factor from the corresponding bases. - Square the length scale factor to obtain the area scale factor.

1. The length scale factor is \(k=\frac{g_2}{g_1}=\frac{12}{9}=\frac{4}{3}\). 2. The area ratio is \(k^2=\left(\frac{4}{3}\right)^2=\frac{16}{9}\). 3. Therefore, \(A_2=A_1\cdot\frac{16}{9}=54\cdot\frac{16}{9}=96\,\text{cm}^2\).

\(A_2=96\,\text{cm}^2\)

- First find the full distance \(SB\). - Match the parallel segment lengths with the distances measured from \(S\). - Set up a proportion using corresponding lengths.

1. The full distance to the farther parallel line is \(SB=SA+AB=4.0+6.0=10.0\,\text{cm}\). 2. The two triangles are similar, so \(\frac{a}{b}=\frac{SA}{SB}\). 3. Substitute the values: \(\frac{3.0}{b}=\frac{4.0}{10.0}\). 4. Solve for \(b\): \(b=\frac{3.0\cdot10.0}{4.0}=7.5\,\text{cm}\).

\(b=7.5\,\text{cm}\)

- Identify which labeled segments correspond across the two rays. - Compare distances measured from \(Z\) when using the parallel segment lengths. - Equivalent proportions can produce more than one correct expression.

1. Corresponding distances from \(Z\) and the parallel segment lengths are proportional: \(\frac{a}{a+b}=\frac{c}{c+d}=\frac{e}{f}\). Therefore, in part a), either \(x=c\), \(y=c+d\), or \(x=e\), \(y=f\). 2. The segments between the two parallel lines are proportional to the nearer distances from \(Z\): \(\frac{c}{d}=\frac{a}{b}\). Therefore, in part b), \(x=a\) and \(y=b\). 3. From \(\frac{e}{f}=\frac{a}{a+b}=\frac{c}{c+d}\), part c) has two valid forms.

a) \(x=c\), \(y=c+d\), or \(x=e\), \(y=f\) b) \(x=a\), \(y=b\) c) \(x=a\), \(y=a+b\), or \(x=c\), \(y=c+d\)

- Identify the two vertical segments in the diagram. - Compare each vertical height with its horizontal distance from \(S\). - Use the common line of sight to justify the same acute angle in both triangles.

1. The pole and tower are parallel vertical segments, so the two right triangles with vertex \(S\) are similar by AA. 2. Read the displayed measurements and compare corresponding heights and horizontal distances: \(\frac{2}{3}=\frac{h}{24}\). 3. Solve: \(h=24\cdot\frac{2}{3}=16\,\text{m}\).

The observation tower is \(16\,\text{m}\) tall.

- First combine the two adjacent labeled segments on the lower ray. - Compare distances from \(Z\) to the two parallel lines along each ray. - After finding the full upper-ray distance to \(h\), subtract the nearer segment.

1. Read the displayed lengths and find the full distance on the lower ray: \(ZC=ZA+AC=4+6=10\,\text{cm}\). 2. Because \(g\parallel h\), corresponding distances from \(Z\) are proportional: \(\frac{ZA}{ZC}=\frac{ZB}{ZD}\). 3. Substitute: \(\frac{4}{10}=\frac{5}{ZD}\), so \(ZD=\frac{5\cdot10}{4}=12.5\,\text{cm}\). 4. Therefore, \(BD=ZD-ZB=12.5-5=7.5\,\text{cm}\).

\(BD=7.5\,\text{cm}\)

- Identify the intersection point \(S\). - Find the ratio of the known distances from \(S\). - Apply the same ratio to the other line and to the parallel segments.

1. Corresponding distances from \(S\) are proportional: \(\frac{SA}{SB}=\frac{SC}{SD}\). 2. From the first line, \(\frac{SA}{SB}=\frac{5}{10}=0.5\). 3. Therefore, \(\frac{x}{6}=0.5\), so \(x=3\). 4. The parallel segment lengths have the same ratio: \(\frac{AC}{BD}=\frac{SA}{SB}\). 5. Thus, \(\frac{4}{y}=0.5\), so \(y=8\).

\(x=3\) and \(y=8\)

- Combine the two labeled pieces on the lower ray to find the full distance from \(Z\) to \(A\). - Compare distances from \(Z\) with the two parallel cross-segments. - Solve the resulting proportion for the unknown parallel segment.

1. Read the two adjacent lengths on the lower ray and find the full distance: \(ZA=ZC+CA=4+2=6\,\text{cm}\). 2. Because \(CD\parallel AB\), corresponding lengths are proportional: \(\frac{ZC}{ZA}=\frac{CD}{AB}\). 3. Substitute the displayed values: \(\frac{4}{6}=\frac{3}{x}\). 4. Cross-multiply: \(4x=18\), so \(x=4.5\,\text{cm}\).

\(x=4.5\,\text{cm}\)

- Use the converse of the Triangle Proportionality Theorem. - Compare the distances from \(Z\) to each line on both rays. - Equal corresponding ratios indicate parallel lines.

1. For lines \(g\) and \(h\), compare the distances from \(Z\) on the two rays: \(\frac{3.0}{2.0}=1.5\) and \(\frac{6.0}{4.0}=1.5\). Because the ratios are equal, the converse of the Triangle Proportionality Theorem gives \(g\parallel h\). 2. For line \(k\), the distances from \(Z\) are \(9.0\) and \(5.5\), and \(\frac{9.0}{5.5}\ne1.5\). 3. Therefore, \(k\) is not parallel to either \(g\) or \(h\).

\(g\parallel h\). Line \(k\) is not parallel to either of the other two lines.

- Match the corresponding intervals on the two rays. - Find the total distance \(SB\). - Compare the parallel segment lengths using distances measured from \(S\).

1. Corresponding segments between the parallel lines are proportional to the nearer distances from \(S\): \(\frac{y}{5}=\frac{6}{4}\). 2. Therefore, \(y=5\cdot\frac{6}{4}=7.5\). 3. The total distance on the lower ray is \(SB=4+6=10\). 4. The parallel segment lengths are proportional to the corresponding distances from \(S\): \(\frac{x}{3}=\frac{10}{4}\). 5. Therefore, \(x=3\cdot\frac{10}{4}=7.5\).

\(x=7.5\) and \(y=7.5\)

- Match the corresponding segments on the two rays. - Set up a proportion using the intervals before and between the parallel lines. - Solve the proportion for \(x\).

1. The given lengths are \(ZA=3\,\text{cm}\), \(AB=4.5\,\text{cm}\), and \(ZC=4\,\text{cm}\). 2. Corresponding segments on the two rays are proportional: \(\frac{ZA}{AB}=\frac{ZC}{CD}\). 3. Substitute the values: \(\frac{3}{4.5}=\frac{4}{x}\). 4. Solve for \(x\): \(x=\frac{4\cdot4.5}{3}=6\,\text{cm}\).

\(x=6\,\text{cm}\)

- Find the ratio of the two parallel connecting segments. - Use full distances measured from \(Z\). - Express each full distance as the known first segment plus the unknown segment.

1. The nearer and farther connecting segments have lengths \(3\,\text{cm}\) and \(9\,\text{cm}\), so their ratio is \(\frac{1}{3}\). 2. On the lower ray, \(\frac{4}{4+x}=\frac{1}{3}\). Thus, \(12=4+x\), so \(x=8\,\text{cm}\). 3. On the upper ray, \(\frac{5}{5+y}=\frac{1}{3}\). Thus, \(15=5+y\), so \(y=10\,\text{cm}\).

\(x=8\,\text{cm}\) and \(y=10\,\text{cm}\)

- Match corresponding distances on the two intersecting lines. - Relate the parallel segment lengths to the distances from \(Z\). - Use cross-multiplication to test equivalent proportions.

1. Corresponding distances from \(Z\) are proportional, so \(\frac{p}{q}=\frac{r}{s}\). Statement A is correct. 2. The parallel segment lengths have the same ratio as corresponding distances from \(Z\), so \(\frac{u}{v}=\frac{p}{q}\). Statement B is correct. 3. From \(\frac{p}{q}=\frac{r}{s}\), cross-multiplication gives \(ps=qr\), which is equivalent to \(\frac{p}{r}=\frac{q}{s}\). Statement C is correct. 4. Statement D reverses the ratio on the right and is not generally true.

A, B, and C

- Identify corresponding distances on the crossed lines. - Decide whether each part compares ray distances or parallel segment lengths. - Solve the appropriate proportion for the missing value.

1. For a), corresponding distances from \(Z\) satisfy \(\frac{a}{a_1}=\frac{b}{b_1}\). 2. Substitute the values: \(\frac{3}{4.5}=\frac{2}{b_1}\), so \(b_1=\frac{2\cdot4.5}{3}=3\,\text{cm}\). 3. For b), the parallel segment lengths satisfy \(\frac{x}{y}=\frac{a}{a_1}\). 4. Substitute the values: \(\frac{3.2}{y}=\frac{4}{10}\), so \(y=\frac{3.2\cdot10}{4}=8\,\text{cm}\).

a) \(b_1=3\,\text{cm}\) b) \(y=8\,\text{cm}\)

- Identify the two triangles meeting at the pinhole. - Read the object height and the two horizontal distances from the diagram. - Compare height ratios with distance ratios.

1. The object and image form similar triangles with the pinhole as their common vertex. 2. Read the measurements from the diagram. The ratio of image height to object height equals the ratio of image distance to object distance: \(\frac{h}{12}=\frac{15}{20}\). 3. Solve: \(h=12\cdot\frac{15}{20}=12\cdot0.75=9\,\text{cm}\).

The image height is \(9\,\text{cm}\).

- Identify the similar triangles on opposite sides of \(Z\). - Match corresponding distances from \(Z\). - Use the same scale factor for the parallel segment lengths.

1. Since \(AB\parallel CD\), \(\triangle ZAB\sim\triangle ZCD\). 2. Corresponding distances satisfy \(\frac{ZB}{ZD}=\frac{ZA}{ZC}=\frac{5}{12.5}=0.4\). Therefore, \(ZB=15\cdot0.4=6\,\text{cm}\). 3. The parallel segment lengths satisfy \(\frac{CD}{AB}=\frac{ZC}{ZA}=\frac{12.5}{5}=2.5\). Therefore, \(CD=4\cdot2.5=10\,\text{cm}\).

\(ZB=6\,\text{cm}\) and \(CD=10\,\text{cm}\)

- Identify the two similar triangles with vertex \(Z\). - Find the scale factor using the two horizontal distances. - Apply the same scale factor to corresponding segments.

1. The similar triangles have scale factor \(\frac{ZB}{ZA} = \frac{4 + 6}{4} = 2.5\). 2. Corresponding parallel segments satisfy \(\frac{BD}{AC} = 2.5\). Thus, \(b = 2.5(5) = 12.5\,\text{cm}\). 3. Corresponding distances from \(Z\) satisfy \(\frac{ZD}{ZC} = 2.5\). Thus, \(ZD = 2.5(3.2) = 8\,\text{cm}\).

a) \(b = 12.5\,\text{cm}\) b) \(ZD = 8\,\text{cm}\)

- How does each side of the medial triangle compare with the corresponding side of the original triangle? - If every side length is halved, what happens to the perimeter?

1. Each side of the medial triangle is a midsegment of triangle \(ABC\). 2. A triangle midsegment is parallel to the third side and half its length. 3. Therefore, the medial triangle has scale factor \(\frac{1}{2}\) relative to the original triangle. 4. Its perimeter is \(\frac{1}{2}\cdot42\,\text{cm}=21\,\text{cm}\).

\(21\,\text{cm}\)

- Read the two pieces of the hypotenuse from the diagram. - How do those two pieces combine to make the full hypotenuse? - The altitude is the geometric mean of the two hypotenuse segments.

1. Read the two hypotenuse segments from the diagram. The hypotenuse is their sum: \(c=p+q=9+16=25\,\text{cm}\). 2. The similar triangles formed by the altitude give \(h^2=pq\). 3. Substitute: \(h^2=9\cdot16=144\), so \(h=12\,\text{cm}\).

\(h=12\,\text{cm}\) and \(c=25\,\text{cm}\)

- Which leg is labeled in the diagram, and which part of the hypotenuse lies next to that leg? - Use the leg-projection relationship produced by the similar triangles. - Check that the projection you find is shorter than the full hypotenuse.

1. Read \(BC=12\,\text{cm}\) from the diagram. 2. The similar triangles formed by the altitude give \(BC^2=AB\cdot BD\). 3. Substitute: \(12^2=20\cdot BD\). 4. Therefore, \(BD=\frac{144}{20}=7.2\,\text{cm}\).

\(BD=7.2\,\text{cm}\)

- Use the named vertices to match corresponding sides in the two triangles. - The given ratio is the scale factor from the smaller displayed triangle to the larger one. - Apply the same factor to each labeled side in the diagram.

1. The scale factor from \(A_1B_1C_1\) to \(ABC\) is \(k=1.5\). 2. Read the three side lengths of \(A_1B_1C_1\) from the diagram and multiply each corresponding length by \(1.5\): \(x=1.5\cdot6=9\), \(y=1.5\cdot8=12\), and \(z=1.5\cdot10=15\).

\(x=9\), \(y=12\), and \(z=15\)

- Match the displayed side of length \(16\) to its position in the \(3:4:5\) ratio. - Find the length represented by one ratio part. - Use the other two ratio numbers for the unknown sides.

1. Because the triangles are similar, the side lengths of \(A_1B_1C_1\) are also in the ratio \(3:4:5\). The diagram shows that the side corresponding to \(4\) ratio parts has length \(16\). 2. One ratio part has length \(16\div4=4\). 3. Therefore, \(x=a_1=3\cdot4=12\) and \(y=c_1=5\cdot4=20\).

\(x=12\) and \(y=20\)

- Read the full hypotenuse and one of its two parts from the diagram. - Find the other hypotenuse segment by subtraction. - The altitude is the geometric mean of the two hypotenuse segments.

1. Read \(AB=13\,\text{cm}\) and \(AD=4\,\text{cm}\) from the diagram. 2. The other hypotenuse segment is \(DB=13-4=9\,\text{cm}\). 3. The similar triangles formed by the altitude give \(h^2=AD\cdot DB\). 4. Thus, \(h=\sqrt{4\cdot9}=6\,\text{cm}\).

\(h=6\,\text{cm}\)

- Read the pair of corresponding side lengths from the diagram. - How does the area of a similar figure change when each length is multiplied by \(k\)? - Apply the area scale factor to the known area.

1. From the diagram, the linear scale factor from \(\triangle ABC\) to \(\triangle DEF\) is \(k=\frac{10}{5}=2\). 2. Areas of similar figures scale by the square of the linear scale factor, so the area factor is \(k^2=2^2=4\). 3. Therefore, \(x=15\cdot4=60\,\text{cm}^2\).

\(x=60\,\text{cm}^2\)

- Read the two collinear lengths from \(S\) to \(A\) and from \(A\) to \(B\), then find \(SB\). - Which two triangles are similar because the cross-segments are parallel? - Match each parallel segment with its distance from \(S\).

1. From the diagram, \(SA=8\) and \(AB=12\), so \(SB=8+12=20\). 2. Because \(AC\parallel BD\), the triangles with vertex \(S\) are similar, so \(\frac{BD}{AC}=\frac{SB}{SA}\). 3. The diagram shows \(AC=6\), so \(\frac{x}{6}=\frac{20}{8}=2.5\), and \(x=15\).

\(x=15\)

- Read the three side lengths of the medial triangle from the diagram. - How does each triangle midsegment compare with the parallel side of the original triangle? - Once you know the scale factor, apply it to the perimeter as a linear measure.

1. The diagram shows that the medial triangle \(DEF\) has side lengths \(7\,\text{cm}\), \(8\,\text{cm}\), and \(9\,\text{cm}\). 2. Each side of a medial triangle is half the length of the parallel side of the original triangle, so the side lengths of \(\triangle ABC\) are \(14\,\text{cm}\), \(16\,\text{cm}\), and \(18\,\text{cm}\). 3. Therefore, \(x=14+16+18=48\,\text{cm}\).

\(x=48\,\text{cm}\)

- What is the linear scale factor from half a side length to the full side length? - How does area change when all lengths are doubled?

1. Because \(D\) and \(E\) are midpoints, the linear scale factor from \(\triangle ADE\) to \(\triangle ABC\) is \(2\). 2. The area scale factor is \(2^2=4\). 3. Therefore, \(x=10\cdot4=40\,\text{cm}^2\).

\(x=40\,\text{cm}^2\)

- Which two triangles are formed by the crossing diagonals and the parallel bases? - Match the diagonal segments with their corresponding bases. - Read the three given lengths from the diagram before solving the proportion.

1. Since \(AB\parallel CD\), \(\triangle ABS\sim\triangle CDS\). 2. Corresponding sides satisfy \(\frac{AS}{SC}=\frac{AB}{CD}\). 3. Read the values from the diagram: \(\frac{6}{SC}=\frac{9}{6}=\frac{3}{2}\). 4. Therefore, \(SC=4\,\text{cm}\).

\(SC=4\,\text{cm}\)

- How does a perimeter ratio compare with the ratio of corresponding side lengths? - Read \(AB\) from panel a). - Use the order of the two perimeters to determine the scale factor from panel a) to panel b).

1. For similar figures, the ratio of the perimeters equals the ratio of corresponding side lengths. 2. Therefore, \(\frac{DE}{AB}=\frac{5}{2}\). 3. From panel a), \(AB=6\,\text{cm}\), so \(DE=6\cdot\frac{5}{2}=15\,\text{cm}\).

\(DE=15\,\text{cm}\)

- Apply the triangle midsegment theorem to each segment through \(M\). - How long is a midsegment compared with the side parallel to it? - Add the four side lengths of \(AFME\).

1. Since \(M\) is the midpoint of \(BC\) and \(ME\parallel AB\), the triangle midsegment theorem gives \(ME=\frac{1}{2}AB=8\,\text{m}\) and \(AE=\frac{1}{2}AC=10\,\text{m}\). 2. Since \(MF\parallel AC\), the same theorem gives \(MF=\frac{1}{2}AC=10\,\text{m}\) and \(AF=\frac{1}{2}AB=8\,\text{m}\). 3. Therefore, the perimeter is \(8+10+8+10=36\,\text{m}\).

\(36\,\text{m}\)

- Read the two base lengths from the diagram and compare them. - Parallel cross-sections change by equal amounts when the legs are divided into equal parts. - Divide the total change in base length into three equal changes.

1. From the diagram, the base lengths differ by \(18-9=9\,\text{cm}\). 2. Because the legs are divided into three equal parts and the cross-sections are parallel to the bases, the cross-section length increases by equal amounts. The increase per step is \(9\div3=3\,\text{cm}\). 3. One step from \(CD\), \(s_1=9+3=12\,\text{cm}\). 4. Two steps from \(CD\), \(s_2=9+2\cdot3=15\,\text{cm}\).

\(s_1=12\,\text{cm}\) and \(s_2=15\,\text{cm}\)

- What do two equal parts of a side tell you about the point between them? - Which triangle theorem connects the line joining two such points to the third side?

1. The matching tick marks show that \(AD=DB\) and \(AE=EC\), so \(D\) and \(E\) are midpoints. 2. Segment \(DE\) joins the midpoints of two sides of a triangle. By the triangle midsegment theorem, \(DE\parallel BC\) and \(DE=\frac{1}{2}BC\).

The matching marks make \(D\) and \(E\) midpoints. By the midsegment theorem, \(DE\parallel BC\) and \(\frac{DE}{BC}=\frac{1}{2}\).

- Read the four side-part lengths from the diagram. - Compare how \(D\) and \(E\) divide their respective sides rather than comparing raw lengths across different sides. - Which converse theorem turns proportional side divisions into a parallel conclusion?

1. From the diagram, \(\frac{AD}{DB}=\frac{4}{6}=\frac{2}{3}\). 2. Also, \(\frac{AE}{EC}=\frac{6}{9}=\frac{2}{3}\). 3. Thus, points \(D\) and \(E\) divide the two sides proportionally. 4. By the converse of the triangle proportionality theorem, the line through \(D\) and \(E\) is parallel to the third side. 5. Therefore, \(DE\parallel BC\).

Since \(\frac{AD}{DB}=\frac{4}{6}=\frac{2}{3}\) and \(\frac{AE}{EC}=\frac{6}{9}=\frac{2}{3}\), the two sides are divided proportionally. The converse of the triangle proportionality theorem gives \(DE\parallel BC\).

- Relate the two pieces of \(\overline{BC}\) to the two sides adjacent to \(\angle A\). - The two pieces must have a fixed ratio and also add to the whole side. - Check that your two lengths add to \(12\,\text{cm}\).

1. By the Angle Bisector Theorem, \(\frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{6}=\frac{5}{3}\). 2. Let \(BD=5k\) and \(DC=3k\). 3. Since \(BD+DC=BC=12\,\text{cm}\), \(8k=12\), so \(k=1.5\). 4. Therefore, \(BD=7.5\,\text{cm}\) and \(DC=4.5\,\text{cm}\).

\(BD=7.5\,\text{cm}\) and \(DC=4.5\,\text{cm}\).

- Which two triangles have the same angles because the two bases are parallel? - Compare a small-triangle side with the full matching side, not only with its leftover segment.

1. The parallel line makes \(\triangle ADE\sim\triangle ABC\) by AA. 2. The whole side \(AB=4+8=12\,\text{cm}\), so the small-to-large factor is \(\frac{AD}{AB}=\frac{4}{12}=\frac{1}{3}\). 3. Hence \(DE=15\cdot\frac{1}{3}=5\,\text{cm}\).

\(DE=5\,\text{cm}\)

- Match each original length with the enlarged length directly below it. - What multiplier works for both given pairs?

1. The first two pairs show a common factor \(\frac{6}{2}=\frac{15}{5}=3\). 2. The third enlarged length is \(x=3\cdot7=21\).

\(x=21\)

- How does enlarging each side affect their sum? - Use the perimeters to recover the scale factor.

1. Every side length changes by the same factor \(k\). 2. Adding the scaled side lengths gives a perimeter \(k\) times as large. 3. Thus \(k=\frac{36}{15}=\frac{12}{5}=2.4\).

\(k=2.4\)

- Does area grow by \(k\) or \(k^2\)? - What positive number has square equal to the area ratio?

1. The area scale factor is \(\frac{98}{50}=1.96\). 2. If the length factor is \(k\), the area factor is \(k^2\). Therefore \(k=\sqrt{1.96}=1.4\).

\(k=1.4\)

- Is \(DB\) the whole matching side or only a remaining part? - Compare the small triangle with the whole triangle.

1. The whole side is \(AB=AD+DB=8\,\text{cm}\). 2. Similar triangles give \(\frac{DE}{BC}=\frac{AD}{AB}=\frac{3}{8}\). 3. Thus \(BC=6\cdot\frac{8}{3}=16\,\text{cm}\).

\(BC=16\,\text{cm}\)

- Which angle pairs make the two smaller triangles similar? - Use their corresponding sides to write an equation involving \(h\).

1. Corresponding sides of the two smaller triangles give \(\frac{h}{2}=\frac{8}{h}\). 2. Thus \(h^2=16\), so the altitude is \(h=4\,\text{cm}\).

\(\frac{h}{2}=\frac{8}{h}\), so \(h=4\,\text{cm}\).

- Pair height with shadow length in each right triangle. - What proportion is the same when the Sun's elevation is the same?

1. The Sun's rays create similar right triangles. 2. If the tree's height is \(h\), then \(\frac{h}{12}=\frac{5}{3}\). 3. Therefore \(h=20\) feet.

\(20\) feet

- This time you know the two pieces of the base rather than the angle. - Which ratios would need to agree for \(AD\) to be an angle bisector?

1. Compare the sides beside angle \(A\): \(\frac{AB}{AC}=\frac{8}{12}=\frac{2}{3}\). 2. Compare the two parts of the opposite side: \(\frac{BD}{DC}=\frac{6}{9}=\frac{2}{3}\). 3. Because these ratios agree and \(D\) lies on \(BC\), the converse of the Angle Bisector Theorem shows that \(AD\) bisects angle \(A\).

Yes. \(\frac{AB}{AC}=\frac{BD}{DC}=\frac{2}{3}\), so \(AD\) bisects angle \(A\).

- Each smaller triangle uses one of the original sides as its base. - What length serves as the perpendicular height for all three? - Add their areas rather than using a new triangle-area formula.

1. The perpendicular distance from \(I\) to each side is \(4\,\text{cm}\), so the three triangles with bases \(13\), \(14\), and \(15\) each have height \(4\,\text{cm}\). 2. Their areas add to \(A=\frac{1}{2}(13)(4)+\frac{1}{2}(14)(4)+\frac{1}{2}(15)(4)\). 3. Thus \(A=\frac{1}{2}(13+14+15)(4)=84\,\text{cm}^2\).

\(A=\frac{1}{2}(13+14+15)\cdot4=84\,\text{cm}^2\). Each smaller triangle has height \(4\,\text{cm}\), the perpendicular distance from \(I\) to its base.

- What is the base of each of the three smaller triangles? - How is the incenter's perpendicular distance related to each triangle's height? - Their three areas add to the given total area.

1. Each smaller triangle has height \(r\), with one of the original sides as its base. 2. Their areas add to \(36=\frac{1}{2}(9r+10r+17r)=18r\). 3. Therefore \(r=2\,\text{cm}\).

\(36=\frac{1}{2}(9+10+17)r\), so \(r=2\,\text{cm}\).

- Find the vertex angle of the first triangle. - Use the vertex angle of the second triangle to find its two base angles. - How is the area scale factor related to the length scale factor for similar figures?

1. In the first triangle, both base angles are \(70^\circ\), so the vertex angle is \(180^\circ-70^\circ-70^\circ=40^\circ\). 2. In the second triangle, the two base angles share the remaining \(180^\circ-40^\circ=140^\circ\), so each base angle is \(\frac{140^\circ}{2}=70^\circ\). 3. Both triangles have angle measures \(70^\circ,70^\circ,40^\circ\), so they are similar by AA. 4. The area scale factor is \(k^2=2.25\), so the length scale factor is \(k=\sqrt{2.25}=1.5\). 5. The base of the larger triangle is \(1.5\cdot5\,\text{cm}=7.5\,\text{cm}\).

a) Yes. Both triangles have angle measures \(70^\circ,70^\circ,40^\circ\). b) The base of the larger triangle is \(7.5\,\text{cm}\).

- First determine whether the triangle with side lengths \(3\), \(4\), and \(5\) is a right triangle. - Then find the ratio of the two areas. - How do you obtain the length scale factor from an area ratio?

1. Triangle \(D_1\) is a right triangle because \(3^2+4^2=9+16=25=5^2\). 2. Its area is \(A_1=\frac{1}{2}\cdot3\,\text{cm}\cdot4\,\text{cm}=6\,\text{cm}^2\). 3. The area ratio is \(\frac{A_2}{A_1}=\frac{54}{6}=9\). 4. Since \(\frac{A_2}{A_1}=k^2\), the positive scale factor is \(k=\sqrt{9}=3\).

The area of \(D_1\) is \(6\,\text{cm}^2\). The scale factor from \(D_1\) to \(D_2\) is \(k=3\).

- Convert both edge lengths to the same unit first. - Surface area scales with the square of the linear factor. - Volume scales with the cube of the linear factor. - Recall the relationship between liters and cubic meters.

1. Convert to the same unit: \(12\,\text{m}=1200\,\text{cm}\). The scale factor from the actual tower to the model is \(k=\frac{30}{1200}=\frac{1}{40}=0.025\). 2. Surface area scales by the square of the linear factor. From the model to the actual tower, the linear factor is \(40\), so the surface-area factor is \(40^2=1600\). 3. Volume scales by the cube of the linear factor, so the volume factor is \(40^3=64{,}000\). 4. The actual capacity is \(27\cdot64{,}000=1{,}728{,}000\,\text{L}\). Since \(1000\,\text{L}=1\,\text{m}^3\), this is \(1728\,\text{m}^3\).

a) \(k=\frac{1}{40}=0.025\) b) The actual surface area is \(1600\) times the model’s surface area. c) \(1728\,\text{m}^3\)

- How do the perimeters of similar figures compare with their corresponding side lengths? - What happens to the perimeter when every side length is multiplied by the same factor? - How is the area scale factor related to the length scale factor?

1. The perimeter of \(R_1\) is \(P_1=2\cdot(4\,\text{cm}+6\,\text{cm})=20\,\text{cm}\). 2. Perimeters of similar figures scale by the same factor as corresponding lengths, so \(k=\frac{P_2}{P_1}=\frac{40}{20}=2\). 3. The side lengths of \(R_2\) are \(a_2=2\cdot4\,\text{cm}=8\,\text{cm}\) and \(b_2=2\cdot6\,\text{cm}=12\,\text{cm}\). 4. Areas scale by the square of the scale factor. Therefore, \(\frac{A_2}{A_1}=k^2=2^2=4\).

The side lengths of \(R_2\) are \(8\,\text{cm}\) and \(12\,\text{cm}\). The ratio of the area of \(R_2\) to the area of \(R_1\) is \(4\) to \(1\).

- Use the ratio between the two known sides of the second triangle to identify which pair of sides in the first triangle corresponds to them. - Once the correspondence is fixed, determine the scale factor between the triangles. - Apply that same scale factor to the remaining side.

1. Compare the ratio of the two known sides in the second triangle: \(\frac{18.75}{10}=1.875\). 2. In the first triangle, \(\frac{15}{8}=1.875\), so \(10\,\text{cm}\) corresponds to \(8\,\text{cm}\) and \(18.75\,\text{cm}\) corresponds to \(15\,\text{cm}\). 3. The scale factor from the first triangle to the second is \(\frac{10}{8}=1.25\). 4. The remaining side corresponds to \(12\,\text{cm}\), so its length is \(12\cdot1.25=15\,\text{cm}\).

The third side of the second triangle is \(15\,\text{cm}\).

- How does area change when every side length is multiplied by the same factor? - First find the area of the original rectangle. - How is the area ratio related to the length scale factor for similar figures?

1. The area of the first rectangle is \(A=ab=4\,\text{cm}\cdot9\,\text{cm}=36\,\text{cm}^2\). 2. The ratio of the areas is \(\frac{A_1}{A}=\frac{144\,\text{cm}^2}{36\,\text{cm}^2}=4\). 3. For similar figures, the area ratio equals \(k^2\), where \(k\) is the length scale factor. Thus, \(k^2=4\), so \(k=\sqrt{4}=2\). 4. The side lengths of the second rectangle are \(a_1=4\,\text{cm}\cdot2=8\,\text{cm}\) and \(b_1=9\,\text{cm}\cdot2=18\,\text{cm}\).

The side lengths of the second rectangle are \(a_1=8\,\text{cm}\) and \(b_1=18\,\text{cm}\).

- How do you find the area of a right triangle from its leg lengths? - How is the area ratio related to the length scale factor? - Multiply corresponding side lengths by the same scale factor. - Use the Pythagorean theorem to find the hypotenuse.

1. The area of \(D_1\) is \(A_1=\frac{1}{2}\cdot5\,\text{cm}\cdot12\,\text{cm}=30\,\text{cm}^2\). 2. The area ratio is \(\frac{A_2}{A_1}=\frac{120}{30}=4\). 3. Since the area ratio equals \(k^2\), the positive scale factor is \(k=\sqrt{4}=2\). 4. The legs of \(D_2\) are \(a_2=2\cdot5\,\text{cm}=10\,\text{cm}\) and \(b_2=2\cdot12\,\text{cm}=24\,\text{cm}\). 5. The hypotenuse is \(c_2=\sqrt{10^2+24^2}\,\text{cm}=\sqrt{676}\,\text{cm}=26\,\text{cm}\).

The scale factor is \(k=2\). Triangle \(D_2\) has legs of \(10\,\text{cm}\) and \(24\,\text{cm}\), and a hypotenuse of \(26\,\text{cm}\).

- The scale ratio describes lengths. How many directions of length are involved in a solid's volume? - Think about how changing every dimension affects the number of unit cubes the model can hold. - Once you have the smaller volume, make sure its unit matches the unit requested.

1. The linear scale factor from the aquarium to the model is \(k=\frac{1}{20}\). 2. The volume scale factor is \(k^3=\left(\frac{1}{20}\right)^3=\frac{1}{8000}\). 3. The model’s volume is \(12\div8000=0.0015\,\text{m}^3\). 4. Since \(1\,\text{m}^3=1{,}000{,}000\,\text{cm}^3\), the model’s volume is \(0.0015\cdot1{,}000{,}000=1500\,\text{cm}^3\).

\(1500\,\text{cm}^3\)

- Convert both volumes to the same unit. - Take a cube root to move from the volume ratio to the linear ratio. - Square the linear factor to obtain the surface-area factor. - Convert square decimeters to square meters.

1. Convert the actual volume: \(270\,\text{m}^3=270{,}000\,\text{dm}^3\). 2. The actual-to-model volume ratio is \(\frac{270000}{10}=27{,}000\). 3. If the actual-to-model linear factor is \(n\), then \(n^3=27{,}000\). Thus, \(n=30\), so the model scale is \(1:30\). 4. Surface area scales by the square of the linear factor. The actual surface area is \(25\cdot30^2=22{,}500\,\text{dm}^2\). 5. Since \(100\,\text{dm}^2=1\,\text{m}^2\), this is \(225\,\text{m}^2\).

a) \(1:30\) b) \(225\,\text{m}^2\)

- Volume scales with the cube of the linear factor. - Surface area scales with the square of the linear factor. - Use a cube root when the desired volume factor is known.

1. For part a), the linear factor is \(2\), so the volume factor is \(2^3=8\). The new capacity is \(80\cdot8=640\,\text{L}\). 2. For part b), surface area scales by the square of the linear factor. Since \(2^2=4\), the material needed is multiplied by \(4\). 3. For part c), let the linear factor be \(k\). Doubling the volume requires \(k^3=2\), so \(k=\sqrt[3]{2}\approx1.26\).

a) \(640\,\text{L}\) b) The material needed is multiplied by \(4\). c) The edge lengths should be multiplied by approximately \(1.26\).

- Which side of the first triangle corresponds to the hypotenuse of the second triangle? - Find the missing side of the first triangle before calculating the scale factor. - How is the area ratio related to the length scale factor?

1. By the Pythagorean theorem, \(c=\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{cm}\). 2. Since \(c_2=8\,\text{cm}\), the scale factor from \(ABC\) to \(A_2B_2C_2\) is \(k=\frac{8}{10}=0.8\). 3. The area ratio is \(k^2=(0.8)^2=0.64=\frac{16}{25}\). 4. The new legs are \(a_2=0.8\cdot6\,\text{cm}=4.8\,\text{cm}\) and \(b_2=0.8\cdot8\,\text{cm}=6.4\,\text{cm}\).

a) \(c=10\,\text{cm}\) b) The area ratio of \(A_2B_2C_2\) to \(ABC\) is \(16\) to \(25\). c) \(a_2=4.8\,\text{cm}\) and \(b_2=6.4\,\text{cm}\)

- Which measurement is given for both triangles? - Imagine enlarging the triangle together with its circumcircle. - How does changing a circle's radius affect its area?

1. Corresponding linear measures of similar figures have the same scale factor, so \(k=\frac{R_2}{R}=\frac{6.0}{4.5}=\frac{4}{3}\). 2. The corresponding side length is \(c_2=kc=\frac{4}{3}\cdot7.2\,\text{cm}=9.6\,\text{cm}\). 3. Circle areas scale by the square of the radius scale factor. Therefore, \(\frac{A_{\text{circle},2}}{A_{\text{circle},1}}=k^2=\left(\frac{4}{3}\right)^2=\frac{16}{9}\).

a) \(k=\frac{4}{3}\) b) \(c_2=9.6\,\text{cm}\) c) The circumcircle area ratio is \(16\) to \(9\).

- Find the linear scale factor from the two heights. - Cube the linear factor to obtain the volume factor. - Recall the relationship between cubic decimeters and liters.

1. The linear scale factor from the smaller model to the larger model is \(k=\frac{40}{25}=1.6\). 2. Volume scales by the cube of the linear factor: \(V_2=1.2\cdot1.6^3=1.2\cdot4.096=4.9152\,\text{dm}^3\). 3. Since \(1\,\text{dm}^3=1\,\text{L}\), the volume is \(4.9152\,\text{L}\).

\(4.9152\,\text{L}\)

- Compare all three pairs of corresponding side lengths. - How do perimeters scale for similar figures? - How do areas scale for similar figures? - Check whether the first triangle is a right triangle to simplify the area calculation.

1. The corresponding side-length ratios are \(\frac{7.5}{3}=2.5\), \(\frac{10}{4}=2.5\), and \(\frac{12.5}{5}=2.5\). Therefore, the triangles are similar by SSS with \(k=2.5\). 2. The perimeters are \(P_1=3+4+5=12\,\text{cm}\) and \(P_2=7.5+10+12.5=30\,\text{cm}\). 3. Thus, \(\frac{P_2}{P_1}=\frac{30}{12}=2.5=k\). 4. Because \(3^2+4^2=5^2\), the first triangle is right. Its area is \(A_1=\frac{1}{2}\cdot3\cdot4=6\,\text{cm}^2\). The second triangle is also right, with area \(A_2=\frac{1}{2}\cdot7.5\cdot10=37.5\,\text{cm}^2\). 5. Therefore, \(\frac{A_2}{A_1}=\frac{37.5}{6}=6.25=(2.5)^2=k^2\).

a) The triangles are similar by SSS, and \(k=2.5\). b) \(\frac{P_2}{P_1}=2.5=k\), and \(\frac{A_2}{A_1}=6.25=k^2\).

- Which two right triangles are formed by the horizontal and vertical segments in the diagram? - What angle relationship occurs where the two straight paths cross at \(B\)? - Match the horizontal sides and the perpendicular sides before writing a proportion.

1. Consider \(\triangle BED\) and \(\triangle BAC\). 2. The right-angle marks show that \(\angle BED\cong\angle BAC\). Angles \(\angle EBD\) and \(\angle ABC\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle BED\sim\triangle BAC\) by AA. 4. Corresponding sides are proportional: \(\frac{AC}{ED}=\frac{AB}{BE}\). 5. Substitute the displayed measurements: \(\frac{AC}{4}=\frac{20}{5}=4\). 6. Therefore, \(AC=16\,\text{m}\).

The river is \(16\,\text{m}\) wide. The triangles are similar by AA because each has a right angle and the angles at \(B\) are vertical angles.

- How is the area ratio related to \(k\)? - How do perimeters and side lengths change once \(k\) is known? - Does a dilation scale only the sides, or every length in the figure? - Is a median a length or an area?

1. The area ratio is \(\frac{A_2}{A_1}=\frac{150}{24}=6.25\). 2. Since \(k^2=6.25\), the positive scale factor is \(k=\sqrt{6.25}=2.5\). 3. Perimeters scale by \(k\), so \(P_2=2.5\cdot24\,\text{cm}=60\,\text{cm}\). 4. Corresponding side lengths scale by \(k\), so \(a_2=2.5\cdot6\,\text{cm}=15\,\text{cm}\). 5. A dilation scales every length, including medians, by \(k\). Therefore, \(\frac{m_2}{m_1}=2.5\).

a) \(k=2.5\) b) \(P_2=60\,\text{cm}\) c) \(a_2=15\,\text{cm}\) d) \(\frac{m_2}{m_1}=2.5\)

- The object and its image form a pair of similar triangles. - Compare each height with its corresponding distance from the pinhole. - Use consistent units in each proportion. - Write image height as a function of camera depth to analyze part c.

1. The object and image form similar triangles, so \(\frac{h}{b}=\frac{G}{g}\), where \(h\) is image height, \(b\) is camera depth, \(G\) is object height, and \(g\) is object distance. 2. For part a, use centimeters: \(G=1200\,\text{cm}\), \(h=15\,\text{cm}\), and \(b=25\,\text{cm}\). Then \(\frac{15}{25}=\frac{1200}{g}\). 3. Since \(\frac{15}{25}=0.6\), \(g=1200\div0.6=2000\,\text{cm}=20\,\text{m}\). 4. For part b, the new distance is \(40\,\text{m}\). Keeping \(h\) and \(b\) unchanged gives the same scale factor, so \(\frac{G_2}{40}=0.6\). Thus, \(G_2=0.6\cdot40=24\,\text{m}\). 5. For part c, \(h=\frac{Gb}{g}\). With \(G\) and \(g\) fixed, decreasing \(b\) decreases \(h\) proportionally.

a) \(20\,\text{m}\) b) \(24\,\text{m}\) c) The image height decreases in direct proportion to the camera depth.

- Use the line of sight from the observer's eye through the top of the fence. - Compare vertical changes measured from eye level, not from the ground. - What is the observer's total horizontal distance from the wall?

1. The observer is \(3+12=15\,\text{m}\) from the wall. 2. Measure vertical changes from eye level. At the fence, the vertical change is \(2-1.60=0.40\,\text{m}\) over a horizontal distance of \(3\,\text{m}\). 3. Similar triangles give \(\frac{0.40}{3}=\frac{h-1.60}{15}\), where \(h\) is the blocked height on the wall. 4. Since \(15\div3=5\), \(h-1.60=0.40\cdot5=2\). 5. Therefore, \(h=3.60\,\text{m}\).

The fence blocks the wall up to a height of \(3.60\,\text{m}\).

- Convert all lengths to the same unit before writing a proportion. - For equal apparent sizes, compare each object's length with its distance from the lens. - In part b, decide whether the model must move closer to or farther from the lens when the real car moves farther away.

1. For part a, convert the real car's length to centimeters: \(4.80\,\text{m}=480\,\text{cm}\). 2. Objects have the same apparent size when their lengths are proportional to their distances from the lens: \(\frac{12}{80}=\frac{480}{d}\). 3. Solve: \(12d=80\cdot480\), so \(d=3200\,\text{cm}=32\,\text{m}\). 4. For part b, the real car's new distance is \(32+10=42\,\text{m}=4200\,\text{cm}\). 5. Let \(m\) be the model's new distance. Then \(\frac{12}{m}=\frac{480}{4200}\). 6. Solve: \(480m=12\cdot4200\), so \(m=105\,\text{cm}\). 7. The model must move \(105-80=25\,\text{cm}\) farther from the lens.

a) \(32\,\text{m}\) b) Move the model \(25\,\text{cm}\) farther from the lens.

- Use the diagram to identify the smaller triangle ending at the ruler and the larger triangle ending at the wall. - Find the total horizontal distance from the light source to the wall in terms of the unknown distance. - Compare vertical height with horizontal distance in the two triangles. - Identify the two angle relationships that establish AA similarity.

1. Convert the ruler height to meters: \(20\,\text{cm}=0.20\,\text{m}\). 2. Let \(x\) be the distance from the light source to the ruler. The distance from the light source to the wall is \(x+2\). 3. The ruler and wall form corresponding vertical sides of similar triangles, so \(\frac{0.20}{x}=\frac{1}{x+2}\). 4. Cross-multiply: \(0.20(x+2)=x\), so \(0.20x+0.40=x\). 5. Therefore, \(0.40=0.80x\), and \(x=0.50\,\text{m}\). 6. The triangles are similar by AA: both contain a right angle because the ruler and wall are perpendicular to the tabletop, and they share the angle formed by the tabletop and the light ray from the source to the top of the shadow.

a) \(0.50\,\text{m}\), or \(50\,\text{cm}\) b) The triangles are similar by AA because the ruler and wall are parallel vertical segments and the light rays begin at the same point.

- Look for the two triangles that meet at the mirror. - Think about what happens to a ray of light when it reflects off a flat mirror. - How do the two heights relate to their distances from the mirror? - Which feature of the ground makes the right-angle marks valid?

1. Triangles JEM and TMH are right triangles, as shown by the angle marks at J and T. 2. Light reflects from a flat mirror at equal angles to the normal. This makes the angles at M in the two triangles equal, so the triangles are similar by AA. 3. Corresponding heights and ground distances give \(\frac{TH}{45}=\frac{1.70}{2.50}\). 4. Therefore, \(TH=45\cdot\frac{1.70}{2.50}=30.6\,\text{m}\). 5. The ground must be level between Jordan and the tower, so the ground distances form one horizontal line and the two heights are perpendicular to it.

a) \(30.6\,\text{m}\) b) The angle of incidence equals the angle of reflection. c) The ground between Jordan and the tower must be level.

- Use the diagram to identify the large light-pole triangle and the smaller person-shadow triangle. - The base of the large triangle contains both the pole-to-person distance and the shadow length. - Think about how that geometry changes when the person moves toward the pole.

1. The large triangle has height \(5.00\,\text{m}\) and base \(4.25+s\). The smaller similar triangle has height \(1.60\,\text{m}\) and base \(s\). 2. Write a proportion: \(\frac{5.00}{4.25+s}=\frac{1.60}{s}\). 3. Cross-multiply: \(5.00s=1.60(4.25+s)\). 4. Simplify: \(5.00s=6.80+1.60s\), so \(3.40s=6.80\). 5. Therefore, \(s=2.00\,\text{m}\). 6. If the person moves toward the streetlight, the horizontal distance from the pole decreases. The corresponding similar triangle becomes narrower, so the shadow becomes shorter.

a) \(s=2.00\,\text{m}\) b) The shadow becomes shorter.

- Express the areas of \(D_2\) and \(D_3\) in terms of the area of \(D_1\). - How do you obtain a length scale factor from an area ratio? - First determine the area ratio from \(D_1\) to \(D_2\).

1. The area ratio from \(D_1\) to \(D_2\) is \(\frac{A_2}{A_1}=(1.5)^2=2.25\). 2. Since \(A_3=4A_2\), \(A_3=4(2.25A_1)=9A_1\). 3. The length scale factor from \(D_1\) to \(D_3\) is \(k_{31}=\sqrt{\frac{A_3}{A_1}}=\sqrt{9}=3\).

The ratio of corresponding side lengths from \(D_1\) to \(D_3\) is \(3\) to \(1\).

- Find the entire length \(AC\). - Use corresponding side lengths measured from the shared vertex \(C\). - Square the linear scale factor to get the area factor. - Subtract the smaller triangle's area from the larger triangle's area.

1. The full side length is \(AC=CD+DA=6+9=15\,\text{cm}\). 2. Since \(DE\parallel AB\), the triangles are similar. The scale factor from \(\triangle ABC\) to \(\triangle DEC\) is \(k=\frac{CD}{CA}=\frac{6}{15}=0.4\). 3. The area factor is \(k^2=0.4^2=0.16=\frac{4}{25}\). 4. The area of \(\triangle DEC\) is \(75\cdot0.16=12\,\text{cm}^2\). 5. Therefore, the area of trapezoid \(ABED\) is \(75-12=63\,\text{cm}^2\).

a) \(k=0.4\) b) \(4\) to \(25\) c) \(63\,\text{cm}^2\)

- Use the diagram to compare the right angles and the angle shared at \(B\). - Match the hypotenuse and the leg adjacent to \(\angle B\) in each triangle. - Write one proportion using corresponding leg-to-hypotenuse ratios.

1. Triangles \(BDC\) and \(ABC\) are both right triangles because \(\angle BDC=90^\circ\) and \(\angle BCA=90^\circ\). They also share \(\angle B\). Therefore, they are similar by AA. 2. In the large triangle, the ratio of the leg adjacent to \(\angle B\) to the hypotenuse is \(\frac{a}{c}\). In the smaller triangle, the corresponding ratio is \(\frac{q}{a}\). 3. Set the ratios equal: \(\frac{q}{a}=\frac{a}{c}\), so \(q=\frac{a^2}{c}\). 4. Substitute the values: \(q=\frac{6^2}{10}\,\text{cm}=3.6\,\text{cm}\).

a) The triangles are similar by AA because they share \(\angle B\) and both have a right angle. b) \(q=3.6\,\text{cm}\)

- Similar solids use the same linear factor for all corresponding lengths. - Surface area scales with \(k^2\), and volume scales with \(k^3\). - Take a cube root when a volume factor is given. - Distinguish the new total percent from the percent increase.

1. Since the cylinders are similar, doubling the radius gives a linear scale factor of \(k=2\). 2. For part a), the volume factor is \(2^3=8\), so \(8\) containers the size of \(Z_1\) would fill \(Z_2\). 3. For part b), lateral surface area scales by \(2^2=4\), which is an increase of \(300\%\). 4. For part c), \(k^3=27\), so \(k=3\). Total surface area scales by \(3^2=9\), which is an increase of \(800\%\).

a) \(8\) containers b) \(300\%\) c) \(k=3\), and the total surface area increases by \(800\%\).

- Find the entire length \(AB\). - Use corresponding sides of the similar triangles to find the scale factor. - How are the areas of similar figures related to the linear scale factor?

1. The full side length is \(AB=AD+DB=4+6=10\,\text{cm}\). 2. Since \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). The scale factor from the larger triangle to the smaller triangle is \(k=\frac{AD}{AB}=\frac{4}{10}=0.4\). 3. Corresponding lengths scale by \(k\), so \(DE=15\cdot0.4=6\,\text{cm}\). 4. Areas scale by \(k^2\), so the area ratio is \(0.4^2=0.16=\frac{4}{25}\).

a) \(DE=6\,\text{cm}\) b) \(4\) to \(25\)

- Convert the two volumes to the same unit. - A scale describes a linear ratio, not a volume ratio. - Take the cube root of the volume factor.

1. Convert the actual volume: \(54\,\text{m}^3=54{,}000\,\text{dm}^3\). 2. Let \(k\) be the linear factor from the model to the actual building. Then \(k^3=\frac{54000}{16}=3375\). 3. Therefore, \(k=\sqrt[3]{3375}=15\), so the model scale is \(1:15\).

\(1:15\)

- After a \(125\%\) increase, how large is the new surface compared with the original? - How is the change in area related to the change in each length? - Once you know the length factor, how many length directions make up a volume?

1. A \(125\%\) increase means the new surface area is \(225\%\) of the original, so the area factor is \(2.25\). 2. If the length factor is \(k\), then \(k^2=2.25\), so \(k=1.5\). 3. The volume factor is \(k^3=1.5^3=3.375\). 4. The new volume is \(337.5\%\) of the original, an increase of \(237.5\%\).

The volume increases by \(237.5\%\).

- How can you recover the linear scale factor from an area factor? - How is the volume factor related to the linear scale factor? - Find \(k\) before calculating the new volume.

1. The surface-area factor is the square of the linear scale factor, so \(k^2=2.25\). 2. Therefore, \(k=\sqrt{2.25}=1.5\). 3. The volume factor is \(k^3=1.5^3=3.375\). 4. The enlarged capacity is \(400\cdot3.375=1350\,\text{mL}\).

a) \(k=1.5\) b) \(1350\,\text{mL}\)

- Use the diagram to identify the two similar triangles on opposite sides of \(S\). - Compare the perpendicular distances from \(S\) to the two parallel bases. - How do those two partial heights combine to form the full trapezoid height?

1. Corresponding altitudes of similar triangles have the same ratio as corresponding bases: \(\frac{h_1}{h_2}=\frac{AB}{CD}\). 2. Substitute the values: \(\frac{h_1}{4}=\frac{12}{8}=1.5\), so \(h_1=6\,\text{cm}\). 3. Point \(S\) lies between the parallel bases, so the trapezoid height is the sum of the two perpendicular distances: \(H=h_1+h_2=6\,\text{cm}+4\,\text{cm}=10\,\text{cm}\).

\(H=10\,\text{cm}\)

- How is the perimeter of a polygon calculated? - What does similarity imply about corresponding side-length ratios? - How can each side of one polygon be written using a scale factor and the corresponding side of the other polygon? - What common factor can be factored from the perimeter sum?

1. Because \(V\) and \(V'\) are similar, all corresponding side-length ratios are equal: \(\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}=\cdots=k\). 2. Therefore, \(a=ka'\), \(b=kb'\), \(c=kc'\), and so on. 3. The perimeter of \(V\) is \(P=a+b+c+\cdots\). 4. Substitute the expressions from Step 2: \(P=ka'+kb'+kc'+\cdots\). 5. Factor out \(k\): \(P=k(a'+b'+c'+\cdots)=kP'\). 6. Thus, \(\frac{P}{P'}=k\). Since \(\frac{a}{a'}=k\), it follows that \(\frac{P}{P'}=\frac{a}{a'}\).

Since each side length of \(V\) is \(k\) times its corresponding side length in \(V'\), \(P=\sum s_i=\sum(ks_i')=k\sum s_i'=kP'\). Therefore, \(\frac{P}{P'}=k=\frac{a}{a'}\).

- Which side of the first triangle corresponds to the given side of the second triangle? - Use the Pythagorean theorem to find the first hypotenuse. - Apply the same scale factor to both legs.

1. The hypotenuse of the first triangle is \(c=\sqrt{6^2+8^2}=10\,\text{cm}\). 2. The scale factor is \(k=\frac{c_2}{c}=\frac{25}{10}=2.5\). 3. The corresponding legs are \(a_2=2.5\cdot6\,\text{cm}=15\,\text{cm}\) and \(b_2=2.5\cdot8\,\text{cm}=20\,\text{cm}\).

\(a_2=15\,\text{cm}\) and \(b_2=20\,\text{cm}\)

- Use the diagram to identify the large right triangle, the smaller similar triangle, and the trapezoid between them. - How is an area ratio related to the corresponding length scale factor? - After finding the small triangle's sides, determine which remaining segments form the trapezoid perimeter.

1. The hypotenuse of the large triangle is \(AB=\sqrt{6^2+8^2}=10\,\text{cm}\). 2. The area ratio is \(0.25\), so the positive length scale factor from \(ABC\) to \(DEC\) is \(k=\sqrt{0.25}=0.5\). 3. The small triangle has side lengths \(DC=0.5\cdot6\,\text{cm}=3\,\text{cm}\), \(EC=0.5\cdot8\,\text{cm}=4\,\text{cm}\), and \(DE=0.5\cdot10\,\text{cm}=5\,\text{cm}\). 4. The remaining side segments are \(AD=6\,\text{cm}-3\,\text{cm}=3\,\text{cm}\) and \(BE=8\,\text{cm}-4\,\text{cm}=4\,\text{cm}\). 5. The trapezoid perimeter is \(AB+BE+ED+DA=10+4+5+3=22\,\text{cm}\).

\(DE=5\,\text{cm}\), and the perimeter of trapezoid \(ABED\) is \(22\,\text{cm}\).

- Express the area of a square in terms of its side length. - Determine how a ninefold area change affects the side length. - Relate the diagonal to the side length using the Pythagorean theorem. - Use variables to show that the result holds for any square.

1. Let the original side length be \(s\). Its area is \(s^2\), and its diagonal is \(d=s\sqrt{2}\). 2. If the new area is nine times the original area, then \(s_{\text{new}}^2=9s^2\). Because lengths are positive, \(s_{\text{new}}=3s\), so the side length is multiplied by \(3\). 3. The new diagonal is \(d_{\text{new}}=s_{\text{new}}\sqrt{2}=3s\sqrt{2}=3d\). Therefore, the diagonal is also multiplied by \(3\).

Both the side length and the diagonal length are multiplied by \(3\).

- Recall the area formula for a triangle. - In an isosceles right triangle, the two legs have equal lengths. - Write a Pythagorean equation before and after the legs are doubled. - Consider what happens to a squared length when the length is doubled.

1. Let each original leg have length \(a\). Then \(c^2=a^2+a^2=2a^2\), so \(c=a\sqrt{2}\). 2. After both legs are doubled, the new hypotenuse is \(c_{\text{new}}=\sqrt{(2a)^2+(2a)^2}=\sqrt{8a^2}=2a\sqrt{2}=2c\). 3. The original area is \(A=\frac{1}{2}a^2\). The new area is \(A_{\text{new}}=\frac{1}{2}(2a)(2a)=2a^2=4A\).

The hypotenuse length doubles, and the area becomes four times as large.

- Find the total distance from \(S\) to \(h\) on the first ray. - Match corresponding distances on the two rays. - After finding \(b_1\), subtract it from the total length on the second ray.

1. On the first ray, the total distance from \(S\) to \(h\) is \(12+18=30\,\text{cm}\). 2. Corresponding distances on the rays are proportional, so \(\frac{b_1}{45}=\frac{12}{30}\). 3. Thus, \(b_1=45\cdot\frac{12}{30}=18\,\text{cm}\). 4. The remaining distance is \(b_2=45-18=27\,\text{cm}\).

a) \(b_1=18\,\text{cm}\) b) \(b_2=27\,\text{cm}\)

- Identify the intersection point \(S\). - Match the distances from \(S\) to the two parallel lines. - The ratio of the parallel segments matches the ratio of corresponding distances from \(S\).

1. Corresponding distances from \(S\) to the parallel lines are proportional: \(\frac{u}{v}=\frac{w}{z}\). Therefore, in part a), \(p=w\) and \(q=z\). 2. The parallel segment lengths have the same ratio as either pair of corresponding distances from \(S\): \(\frac{m}{n}=\frac{u}{v}=\frac{w}{z}\). Therefore, part b) has two valid answers. 3. Rearranging \(\frac{u}{v}=\frac{w}{z}\) gives \(\frac{w}{u}=\frac{z}{v}\). Therefore, in part c), \(p=u\) and \(q=v\).

a) \(p=w\), \(q=z\) b) \(p=u\), \(q=v\), or \(p=w\), \(q=z\) c) \(p=u\), \(q=v\)

- Identify the two triangles with common vertex \(Z\). - Match segments that lie on the same diagonal brace. - Then match the two parallel beam segments.

1. Since \(AB\parallel CD\), triangles \(ZAB\) and \(ZCD\) are similar by AA. 2. Corresponding brace segments satisfy \(\frac{ZA}{ZC}=\frac{ZB}{ZD}\). Using the displayed measurements, \(\frac{1.2}{2.4}=\frac{1.0}{ZD}\). 3. Since \(\frac{1.2}{2.4}=\frac{1}{2}\), \(ZD=2.0\,\text{m}\). 4. Corresponding parallel beams satisfy \(\frac{AB}{CD}=\frac{ZA}{ZC}\). Thus, \(\frac{0.8}{CD}=\frac{1.2}{2.4}=\frac{1}{2}\). 5. Therefore, \(CD=1.6\,\text{m}\).

1. \(ZD=2.0\,\text{m}\) 2. \(CD=1.6\,\text{m}\)

- Distinguish the total ray lengths from the segments between the parallel lines. - Compare the parallel segment lengths using distances measured from \(S\). - Express the distance from \(S\) to \(h\) on the lower ray in terms of \(f\) and \(e\).

1. The parallel segment lengths are proportional to the corresponding distances from \(S\): \(\frac{c}{d}=\frac{b}{a}\). 2. Substitute the values: \(\frac{6}{d}=\frac{8}{12}\), so \(d=\frac{6\cdot12}{8}=9\). 3. On the lower ray, the distance from \(S\) to \(h\) is \(f-e\). Therefore, \(\frac{f-e}{f}=\frac{b}{a}\). 4. Substitute the values: \(\frac{15-e}{15}=\frac{8}{12}=\frac{2}{3}\). 5. Thus, \(15-e=10\), so \(e=5\).

\(d=9\) and \(e=5\)

- Use the diagram to identify the two right triangles with the same shadow tip. - Decide which full ground distance corresponds to the flagpole's height. - Match each vertical height with its corresponding horizontal shadow length.

1. Read the ground distances in the diagram. The flagpole's full shadow length is \(2+8=10\,\text{m}\). 2. Sanele and the flagpole form similar right triangles because the sun's rays are parallel. 3. Corresponding heights and shadow lengths are proportional: \(\frac{h}{10}=\frac{1.60}{2}\). 4. Solve: \(h=10\cdot\frac{1.60}{2}=8\,\text{m}\).

The flagpole is \(8\,\text{m}\) tall.

- Read the ratio of the two parallel segments from the diagram. - Use that ratio together with \(y-x=2\). - Apply the same similarity ratio to the second intersecting line.

1. Read the parallel segment lengths from the diagram: \(\frac{a}{b}=\frac{6}{9}=\frac{2}{3}\). 2. Corresponding distances from \(S\) have the same ratio, so \(\frac{x}{y}=\frac{2}{3}\). Thus, \(y=1.5x\). 3. Use the condition \(y-x=2\): \(1.5x-x=2\), so \(0.5x=2\) and \(x=4\). 4. Then \(y=x+2=6\). 5. On the second line, the diagram gives \(\frac{5}{v}=\frac{2}{3}\), so \(v=7.5\).

a) \(x=4\) and \(y=6\) b) \(v=7.5\)

- Separate the segments on the rays from the segments on the parallel lines. - Match corresponding intervals on the two rays. - When comparing parallel segment lengths, use distances measured from \(Z\).

1. Corresponding segments between the parallel lines are proportional, so \(\frac{p}{q}=\frac{s}{t}\). 2. The parallel segment lengths are proportional to the corresponding distances from \(Z\), so \(\frac{a}{b}=\frac{p}{p+q}\). 3. Comparing the second and third parallel segments gives \(\frac{b}{c}=\frac{s+t}{s+t+u}\). 4. Corresponding segments between the parallel lines are proportional, so \(\frac{q}{r}=\frac{t}{u}\).

a) \(t\) b) \(p+q\) c) \(b\) d) \(t\)

- Match corresponding distances from the intersection point \(Z\). - Relate the parallel segment lengths to those distances. - Equivalent proportions can be formed by cross-multiplication or by adding \(1\) to both sides.

1. Corresponding distances from \(Z\) satisfy \(\frac{x}{y}=\frac{w}{z}\), so the missing label in part a) is \(z\). 2. The parallel segment lengths have the same ratio as corresponding distances from \(Z\): \(\frac{m}{n}=\frac{x}{y}\). The missing label in part b) is \(x\). 3. From \(\frac{m}{n}=\frac{w}{z}\), cross-multiplication gives \(mz=nw\), so \(\frac{z}{n}=\frac{w}{m}\). The missing label in part c) is \(m\). 4. Adding \(1\) to both sides of \(\frac{x}{y}=\frac{w}{z}\) gives \(\frac{x+y}{y}=\frac{w+z}{z}\). The missing label in part d) is \(z\).

a) \(z\) b) \(x\) c) \(m\) d) \(z\)

- Express each full distance from \(S\) as a sum of labeled segments. - Set up a proportion that contains only \(x\). - Use the same scale factor to compare the parallel segment lengths.

1. Corresponding distances from \(S\) are proportional: \(\frac{x+9}{x}=\frac{4+6}{4}\). 2. Thus, \(\frac{x+9}{x}=\frac{10}{4}=2.5\). Solving gives \(x+9=2.5x\), so \(x=6\). 3. The parallel segment lengths have the same ratio as the distances from \(S\): \(\frac{10}{y}=\frac{10}{4}\). 4. Therefore, \(y=4\).

\(x=6\) and \(y=4\)

- In the crossed-line figure, match segments on opposite sides of \(S\). - First find the ratio of the parallel segment lengths. - Write a separate proportion for each unknown.

1. The ratio of the parallel segment lengths is \(\frac{16.8}{7}=2.4\). 2. Corresponding distances from \(S\) have the same ratio, so \(\frac{x}{5}=2.4\). Therefore, \(x=12\). 3. On the other intersecting line, \(\frac{12}{y}=2.4\). Therefore, \(y=5\).

\(x=12\) and \(y=5\)

- Identify the full distances measured from \(Z\). - Match corresponding intervals on the two rays. - Compare the parallel segment lengths using distances from \(Z\).

1. Corresponding distances from \(Z\) are proportional: \(\frac{a}{a+b}=\frac{d}{d+e}\). Therefore, \(x=d+e\). 2. The parallel segment lengths are proportional to the corresponding distances from \(Z\): \(\frac{u}{w}=\frac{a}{a+b+c}\). Therefore, \(y=a\). 3. Similarly, \(\frac{v}{u}=\frac{a+b}{a}\). Therefore, \(z=a+b\).

a) \(x=d+e\) b) \(y=a\) c) \(z=a+b\)

- Read the ratio of the two parallel connecting segments from the diagram. - Match distances from \(Z\) along the first intersecting line. - Apply the same ratio along the second intersecting line.

1. Read the two parallel-segment lengths from the diagram. Their ratio is \(\frac{6}{15}=0.4\). 2. Corresponding distances from \(Z\) have the same ratio, so \(\frac{4}{x}=0.4\). Therefore, \(x=10\,\text{cm}\). 3. On the other intersecting line, \(\frac{y}{12.5}=0.4\). Therefore, \(y=5\,\text{cm}\).

\(x=10\,\text{cm}\) and \(y=5\,\text{cm}\)

- Read the two adjacent segments on the first ray and find the full distance from \(Z\) to \(B\). - Use corresponding distances from \(Z\) to find \(x\). - Use the same scale factor for the two parallel segments to find \(y\).

1. Read the two adjacent lengths on the first ray and find the full distance: \(ZB=ZA+AB=4+6=10\,\text{cm}\). 2. Corresponding distances from \(Z\) are proportional: \(\frac{ZA}{ZB}=\frac{ZC}{ZD}\). Thus, \(\frac{4}{10}=\frac{6}{6+x}\). 3. Solving gives \(4(6+x)=60\), so \(x=9\,\text{cm}\). 4. The parallel segment lengths have the same ratio: \(\frac{ZA}{ZB}=\frac{AC}{BD}\). Thus, \(\frac{4}{10}=\frac{5}{y}\). 5. Solving gives \(y=12.5\,\text{cm}\).

\(x=9\,\text{cm}\) and \(y=12.5\,\text{cm}\)

- Identify the two similar triangles on opposite sides of \(S\). - Match the two distances measured from \(S\) along each intersecting line. - Use the same scale factor for the parallel segments.

1. Since \(AC\parallel BD\), \(\triangle ASC\sim\triangle BSD\). 2. Corresponding distances satisfy \(\frac{SA}{SB}=\frac{SC}{SD}\). Using the displayed values, \(\frac{3}{4.5}=\frac{2}{u}\), so \(u=3\,\text{cm}\). 3. The parallel segment lengths satisfy \(\frac{SA}{SB}=\frac{AC}{BD}\). Thus, \(\frac{3}{4.5}=\frac{2.4}{v}\), so \(v=3.6\,\text{cm}\).

\(u=3\,\text{cm}\) and \(v=3.6\,\text{cm}\)

- Find the scale factor from the two parallel segment lengths. - Express \(SD\) as \(e+f\). - Use \(SA+SC=16\,\text{cm}\) together with the scale factor to form an equation for \(c\).

1. The scale factor from the nearer parallel segment to the farther one is \(\frac{b}{a}=\frac{7.5}{4.5}=\frac{5}{3}\). 2. On the upper ray, \(SD=e+f\), and \(\frac{SD}{SB}=\frac{5}{3}\). Thus, \(\frac{6+f}{6}=\frac{5}{3}\), so \(f=4\,\text{cm}\). 3. On the lower ray, \(SA=c\), and the condition gives \(SA+SC=c+SC=16\). 4. Since \(\frac{SC}{SA}=\frac{5}{3}\), \(SC=\frac{5}{3}c\). Therefore, \(c+\frac{5}{3}c=16\). 5. Solving \(\frac{8}{3}c=16\) gives \(c=6\,\text{cm}\).

\(c=6\,\text{cm}\) and \(f=4\,\text{cm}\)

- Read the two parallel base lengths from the diagram and use the diagonal similarity they create. - Determine how point \(S\) divides diagonal \(DB\). - Relate the perpendicular distance from \(S\) to \(AD\) to base \(AB\) with another pair of similar triangles.

1. Read the base lengths from the diagram. Because \(AB\parallel CD\), \(\triangle ABS\sim\triangle CDS\), and \(\frac{CD}{AB}=\frac{4}{12}=\frac{1}{3}\). 2. Therefore, the diagonal is divided so that \(SD\) to \(SB\) has a ratio of \(1\) to \(3\). Thus, \(\frac{SD}{DB}=\frac{1}{4}\). 3. In triangle \(DAB\), the segment through \(S\) perpendicular to \(AD\) is parallel to \(AB\). Similar triangles give \(\frac{x}{AB}=\frac{SD}{DB}=\frac{1}{4}\). 4. Therefore, \(x=12\cdot\frac{1}{4}=3\,\text{cm}\).

\(3\,\text{cm}\)

- Identify the parallelogram formed by the two pairs of parallel sides. - Find the full length \(AC\). - Use the similar triangles \(DCE\) and \(ACB\). - Express \(AB\) in terms of \(DE\) before solving.

1. Since \(D\) lies on \(AC\), \(AD\parallel BF\). Together with \(AB\parallel DF\), this makes \(ABFD\) a parallelogram. Therefore, \(AB=DF=DE+EF=DE+5\). 2. The full side length is \(AC=AD+DC=4+6=10\,\text{cm}\). 3. Since \(DE\parallel AB\), \(\triangle DCE\sim\triangle ACB\). Thus, \(\frac{DE}{AB}=\frac{DC}{AC}=\frac{6}{10}=0.6\). 4. Substitute \(AB=DE+5\): \(DE=0.6(DE+5)\). 5. Solving gives \(0.4DE=3\), so \(DE=7.5\,\text{cm}\). 6. Therefore, \(AB=7.5+5=12.5\,\text{cm}\).

\(DE=7.5\,\text{cm}\) and \(AB=12.5\,\text{cm}\)

- Identify the two similar triangles formed by the parallel segments. - Find the scale factor using \(BD\) and \(BA\). - Apply that factor to the corresponding sides \(BC\) and \(AC\).

1. Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BAC\). 2. The scale factor from \(\triangle BAC\) to \(\triangle BDE\) is \(\frac{BD}{BA}=\frac{3}{6}=0.5\). 3. Therefore, \(BE=0.5\cdot BC=0.5\cdot12=6\,\text{cm}\). 4. Also, \(DE=0.5\cdot AC=0.5\cdot12=6\,\text{cm}\).

\(BE=6\,\text{cm}\) and \(DE=6\,\text{cm}\)

- Use the shared angle at \(A\) and the angle relationships created by the parallel segments. - Use the two labeled pieces on \(AB\) to find the whole side. - Match corresponding sides of the similar triangles before setting up a proportion.

1. The triangles share \(\angle A\). Since \(DE\parallel BC\), a second pair of corresponding angles is congruent, so \(\triangle ADE\sim\triangle ABC\) by AA. 2. From the diagram, \(AB=AD+DB=4+2=6\,\text{cm}\). 3. Corresponding sides satisfy \(\frac{DE}{BC}=\frac{AD}{AB}\), so \(\frac{DE}{7.5}=\frac{4}{6}\). 4. Therefore, \(DE=7.5\cdot\frac{4}{6}=5\,\text{cm}\).

a) \(\triangle ADE\sim\triangle ABC\) by AA. b) \(DE=5\,\text{cm}\)

- Identify the two similar triangles. - Use corresponding sides to find the linear scale factor. - Square the linear scale factor to obtain the area factor.

1. Triangles \(SAB\) and \(SCD\) are similar. Their scale factor is \(k = \frac{CD}{AB} = \frac{6}{2.4} = 2.5\). 2. Since \(\frac{SC}{SA} = 2.5\), \(y = 2(2.5) = 5\,\text{cm}\). 3. Since \(\frac{SD}{SB} = 2.5\), \(x = \frac{7.5}{2.5} = 3\,\text{cm}\). 4. Areas scale by \(k^2\), so the area factor is \(2.5^2 = 6.25\).

a) \(x = 3\,\text{cm}\); \(y = 5\,\text{cm}\) b) \(6.25\)

- Start with the nearer pair of parallel segments and read their displayed data from the diagram. - Apply the same scale factor to the corresponding cross-segments. - Then use the farther parallel segment together with the given \(ZE\).

1. Read the first three displayed measurements. From \(AB\parallel CD\), \(\frac{ZA}{ZD}=\frac{ZB}{ZC}\). Thus, \(\frac{3}{ZD}=\frac{4}{6}\), so \(ZD=4.5\,\text{cm}\). 2. The parallel segment lengths use the same ratio: \(\frac{AB}{CD}=\frac{ZA}{ZD}\). Thus, \(\frac{2}{CD}=\frac{3}{4.5}\), so \(CD=3\,\text{cm}\). 3. From \(CD\parallel EF\), \(\frac{ZD}{ZF}=\frac{ZC}{ZE}\). Thus, \(\frac{4.5}{ZF}=\frac{6}{9}\), so \(ZF=6.75\,\text{cm}\). 4. The parallel segment lengths satisfy \(\frac{CD}{EF}=\frac{ZC}{ZE}\). Thus, \(\frac{3}{EF}=\frac{6}{9}\), so \(EF=4.5\,\text{cm}\).

\(ZD=4.5\,\text{cm}\), \(CD=3\,\text{cm}\), \(ZF=6.75\,\text{cm}\), and \(EF=4.5\,\text{cm}\)

- Use the altitude-to-hypotenuse similarity relationships shown by the diagram. - Start with the given leg and its adjacent hypotenuse segment to find the full hypotenuse. - After finding the other hypotenuse segment, use the corresponding leg and altitude relationships.

1. The similar triangles formed by the altitude give \(b^2=cp\). Reading \(b=12\,\text{cm}\) and \(p=9.6\,\text{cm}\) from the diagram, \(c=\frac{b^2}{p}=\frac{12^2}{9.6}=15\,\text{cm}\). 2. Since \(c=p+q\), \(q=15-9.6=5.4\,\text{cm}\). 3. Use \(a^2=cq\): \(a^2=15\cdot5.4=81\), so \(a=9\,\text{cm}\). 4. Use \(h_c^2=pq\): \(h_c^2=9.6\cdot5.4=51.84\), so \(h_c=7.2\,\text{cm}\).

a) \(c=15\,\text{cm}\) b) \(q=5.4\,\text{cm}\); \(a=9\,\text{cm}\) c) \(h_c=7.2\,\text{cm}\)

- Read the given leg and hypotenuse from the diagram. - Relate that leg to its adjacent hypotenuse segment and the full hypotenuse. - After finding both hypotenuse segments, use the altitude geometric-mean relationship.

1. Read \(c=7.5\,\text{cm}\) and \(a=4.5\,\text{cm}\) from the diagram. The similar triangles formed by the altitude give \(a^2=cp\), so \(p=\frac{4.5^2}{7.5}=2.7\,\text{cm}\). 2. Since \(p+q=c\), \(q=7.5-2.7=4.8\,\text{cm}\). 3. The altitude relationship is \(h^2=pq\). Therefore, \(h=\sqrt{2.7\cdot4.8}=\sqrt{12.96}=3.6\,\text{cm}\).

\(p=2.7\,\text{cm}\), \(q=4.8\,\text{cm}\), and \(h=3.6\,\text{cm}\)

- Read the two leg lengths and first find the hypotenuse. - Match each leg with its adjacent segment of the hypotenuse. - Use the two projection relationships before finding the altitude.

1. Read the leg lengths from the diagram and find the hypotenuse: \(c=\sqrt{5^2+12^2}=13\,\text{cm}\). 2. The similar triangles give \(a^2=cp\), so \(p=\frac{25}{13}\,\text{cm}\approx1.92\,\text{cm}\). 3. Similarly, \(b^2=cq\), so \(q=\frac{144}{13}\,\text{cm}\approx11.08\,\text{cm}\). 4. The altitude satisfies \(h^2=pq\). Using exact values, \(h=\sqrt{\frac{25}{13}\cdot\frac{144}{13}}=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

\(p\approx1.92\,\text{cm}\), \(q\approx11.08\,\text{cm}\), and \(h\approx4.62\,\text{cm}\)

- Assume for a moment that \(KP\) is parallel to \(AC\). - What does the midpoint-and-parallel relationship force about where \(P\) lies on \(BC\)? - Compare that consequence with the given statement about \(P\).

1. Suppose \(KP\parallel AC\). 2. A line through the midpoint of one side of a triangle that is parallel to a second side meets the third side at its midpoint. 3. Since \(K\) is the midpoint of \(AB\), point \(P\) would have to be the midpoint of \(BC\). 4. This contradicts the given information. Therefore, \(KP\not\parallel AC\).

No. If \(KP\parallel AC\), then \(P\) would have to be the midpoint of \(BC\), contrary to the given condition.

- Why are the two triangles similar? - How is the area ratio related to the ratio of corresponding side lengths?

1. Because \(AB\parallel CD\), alternate interior angles show that \(\triangle CDS\sim\triangle ABS\) by AA. 2. The length scale factor from triangle \(ABS\) to triangle \(CDS\) is \(k=\frac{CD}{AB}=\frac{5}{20}=\frac{1}{4}\). 3. Areas of similar figures scale by the square of the length scale factor, so \(\frac{A_{CDS}}{A_{ABS}}=k^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}\).

The area ratio of triangle \(CDS\) to triangle \(ABS\) is \(1\) to \(16\).

- Match the two corresponding intervals between the same pair of parallel lines. - The same unknown appears on two different ray segments. - After cross-multiplying, keep only a value that can represent a length.

1. Read the four segment labels from the diagram. Corresponding intervals on the two rays are proportional: \(\frac{AB}{ZA}=\frac{CD}{ZC}\). 2. Substitute: \(\frac{x}{3}=\frac{12}{x}\). 3. Cross-multiplication gives \(x^2=36\). 4. Since a length is positive, \(x=6\).

\(x=6\)

- Read the two corresponding intervals on the first ray from the diagram. - Use them to relate \(x\) and \(y\). - Combine that proportion with the given total \(ZD\).

1. Read the two labeled intervals on the first ray. Corresponding intervals on the two rays are proportional: \(\frac{x}{5}=\frac{y}{3}\), so \(x=\frac{5}{3}y\). 2. The two unknown segments make the given total length: \(x+y=16\). 3. Substitute: \(\frac{5}{3}y+y=16\), so \(\frac{8}{3}y=16\) and \(y=6\). 4. Therefore, \(x=16-6=10\).

\(x=10\) and \(y=6\)

- Find the pair of corresponding sides in the diagram whose two lengths are both known. - Use that pair to determine the scale factor. - Apply the factor in the correct direction for each unknown.

1. Read the known corresponding pair \(AB=12\) and \(A_1B_1=4\) from the diagram. The scale factor from \(A_1B_1C_1\) to \(ABC\) is \(k=\frac{12}{4}=3\). 2. Since \(BC\) corresponds to \(B_1C_1\), \(x=3\cdot5=15\). 3. Since \(AC\) corresponds to \(A_1C_1\), \(18=3y\), so \(y=6\).

\(x=15\) and \(y=6\)

- Read all three side lengths of the smaller triangle from the diagram. - Compare its perimeter with the given perimeter of the similar triangle. - Use the perimeter ratio as the linear scale factor.

1. Read the side lengths of \(\triangle ABC\) from the diagram. Its perimeter is \(6+8+10=24\). 2. The ratio of the perimeters equals the scale factor: \(k=\frac{48}{24}=2\). 3. Multiply each side length of \(\triangle ABC\) by \(2\): \(x=2\cdot6=12\), \(y=2\cdot8=16\), and \(z=2\cdot10=20\).

\(x=12\), \(y=16\), and \(z=20\)

- Use the two perimeters to determine the linear scale factor. - Read the corresponding side \(c\) from the diagram and scale it first. - Split the remaining part of the smaller perimeter in the stated \(2:3\) ratio.

1. The scale factor from \(\triangle ABC\) to \(\triangle A_1B_1C_1\) is the ratio of the perimeters: \(k=\frac{26}{39}=\frac{2}{3}\). 2. Read \(c=15\,\text{cm}\) from the diagram. The corresponding side is \(c_1=\frac{2}{3}\cdot15=10\,\text{cm}\). 3. The other two sides have a combined length of \(26-10=16\,\text{cm}\). 4. Let \(a_1=2t\) and \(b_1=3t\). Then \(5t=16\), so \(t=3.2\). 5. Thus, \(a_1=2\cdot3.2=6.4\,\text{cm}\) and \(b_1=3\cdot3.2=9.6\,\text{cm}\).

\(a_1=6.4\,\text{cm}\), \(b_1=9.6\,\text{cm}\), and \(c_1=10\,\text{cm}\)

- Read the given leg and full hypotenuse from the diagram. - Use the leg-projection relationship to find the adjacent part of the hypotenuse. - Find the other part, then use the altitude geometric-mean relationship.

1. Read \(BC=8\,\text{cm}\) and \(AB=10\,\text{cm}\) from the diagram. 2. The similar triangles give \(BC^2=AB\cdot BD\), so \(BD=\frac{8^2}{10}=6.4\,\text{cm}\). 3. The other hypotenuse segment is \(AD=10-6.4=3.6\,\text{cm}\). 4. The altitude satisfies \(h^2=AD\cdot BD\), so \(h=\sqrt{3.6\cdot6.4}=4.8\,\text{cm}\).

\(BD=6.4\,\text{cm}\) and \(h=4.8\,\text{cm}\)

- Why are \(\triangle DEC\) and \(\triangle ABC\) similar? - How is the area ratio of similar figures related to their linear scale factor? - Use a square root to move from the area ratio to the side-length ratio. - Is area proportional to length or to the square of length?

1. Because \(DE\parallel AB\), \(\triangle DEC\sim\triangle ABC\) by AA. 2. The area ratio equals the square of the linear scale factor: \(\frac{81}{144}=k^2\). 3. Therefore, \(k=\sqrt{\frac{81}{144}}=\frac{9}{12}=\frac{3}{4}\). 4. Since \(CD\) corresponds to \(CA\), the ratio \(CD\) to \(CA\) is \(3\) to \(4\).

The ratio \(CD\) to \(CA\) is \(3\) to \(4\).

- Simplify the ratio of the two areas first. - How is the length scale factor related to the area ratio of similar figures? - Read the known corresponding side length from the diagram and apply the length scale factor.

1. The area ratio is \(k^2=\frac{75}{27}=\frac{25}{9}\). 2. Since the length scale factor is positive, \(k=\sqrt{\frac{25}{9}}=\frac{5}{3}\). 3. The diagram shows that the corresponding side of the smaller figure is \(9\,\text{cm}\), so \(x=9\cdot\frac{5}{3}=15\,\text{cm}\).

\(x=15\,\text{cm}\)

- Square the side-length ratio to obtain the ratio of the two areas. - Express the larger area in terms of the smaller one. - Use the given difference between the areas to solve for one area, then recover the other.

1. The linear scale factor from \(T_1\) to \(T_2\) is \(\frac{4}{3}\), so \(\frac{S_2}{S_1}=\left(\frac{4}{3}\right)^2=\frac{16}{9}\). Thus, \(S_2=\frac{16}{9}S_1\). 2. Use the area difference: \(\frac{16}{9}S_1-S_1=28\). 3. Then \(\frac{7}{9}S_1=28\), so \(S_1=36\,\text{cm}^2\). 4. Therefore, \(S_2=36+28=64\,\text{cm}^2\).

\(x=36\,\text{cm}^2\) and \(y=64\,\text{cm}^2\)

- Combine adjacent labeled segments in the diagram to express the two full sides of the large triangle. - Which smaller triangle is similar to \(\triangle ABC\) because \(DE\parallel BC\)? - Write a proportion using corresponding sides that contains \(x\).

1. From the diagram, \(AD=x\) and \(DB=2\), so \(AB=x+2\). Also, \(AE=6\) and \(EC=4\), so \(AC=10\). 2. Because \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\), so \(\frac{AD}{AB}=\frac{AE}{AC}\). 3. Substitute the diagram values: \(\frac{x}{x+2}=\frac{6}{10}\). 4. Cross-multiply: \(10x=6(x+2)\), so \(4x=12\) and \(x=3\).

\(x=3\)

- Use the two labeled pieces on side \(AB\) to determine \(AD:AB\). - Convert the linear scale factor between the similar triangles into an area scale factor. - The requested quadrilateral is the part of the large triangle outside the smaller triangle.

1. From the diagram, \(AD=1\) and \(DB=2\), so \(AB=3\) and \(\frac{AD}{AB}=\frac{1}{3}\). 2. Because \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). The linear scale factor from \(\triangle ADE\) to \(\triangle ABC\) is \(3\). 3. The area scale factor is \(3^2=9\), so the area of \(\triangle ABC\) is \(5\cdot9=45\,\text{cm}^2\). 4. Therefore, \(x=45-5=40\,\text{cm}^2\).

\(x=40\,\text{cm}^2\)

- Start with the ratio of the two areas. - Take the positive square root of the area ratio to obtain the linear scale factor. - Perimeters scale by the same factor as corresponding side lengths; combine that fact with the given perimeter sum.

1. The area ratio is \(\frac{27}{12}=\frac{9}{4}\). 2. The linear scale factor is \(k=\sqrt{\frac{9}{4}}=\frac{3}{2}\). 3. Perimeters scale by the same linear factor, so \(y=\frac{3}{2}x\). 4. Use the perimeter sum: \(x+\frac{3}{2}x=45\), so \(\frac{5}{2}x=45\) and \(x=18\,\text{cm}\). 5. Then \(y=45-18=27\,\text{cm}\).

\(x=18\,\text{cm}\) and \(y=27\,\text{cm}\)

- Find the ratio of \(BD\) to the entire side \(BA\). - Similar-figure areas scale by the square of the side-length scale factor.

1. Since \(BD:DA = 1:2\), the ratio \(BD:BA\) is \(1:3\). 2. Because \(DE \parallel AC\), \(\triangle BDE \sim \triangle BAC\) with linear scale factor \(\frac{1}{3}\). 3. The area scale factor is \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\), so the small triangle has area \(\frac{x}{9}\). 4. The quadrilateral has area \(x - \frac{x}{9} = \frac{8x}{9}\). 5. Solve \(\frac{8x}{9} = 40\): \(x = 40 \cdot \frac{9}{8} = 45\).

\(x = 45\) square units

- Write a proportion using corresponding sides of the similar triangles. - Use the order in the similarity statement to match the vertices.

1. From \(\triangle ABC\sim\triangle CAD\), corresponding sides give \(\frac{AB}{AC}=\frac{AC}{CD}\). 2. Substitute the known values: \(\frac{9}{x}=\frac{x}{4}\). 3. Then \(x^2=9\cdot4=36\). Since a length is positive, \(x=6\).

\(x=6\)

- Read the three given side lengths from the diagram. - Use the order of the vertices in the similarity statement to match corresponding sides. - One proportion gives \(QS\); a second proportion can then give \(PS\).

1. Corresponding sides give \(\frac{PQ}{QS}=\frac{QS}{RS}\). 2. Substitute the values from the diagram: \(\frac{8}{x}=\frac{x}{18}\). Then \(x^2=144\), so \(x=12\). 3. Another pair of corresponding sides gives \(\frac{PS}{QR}=\frac{PQ}{QS}\). 4. Thus, \(\frac{y}{15}=\frac{8}{12}=\frac{2}{3}\), so \(y=10\).

\(x=12\) and \(y=10\)

- Which two triangles share the angle at \(P\)? - What additional angle congruence is marked in the diagram? - After establishing similarity, match the side containing \(x\) with its corresponding side.

1. Triangles \(PRS\) and \(PQR\) share the angle at \(P\), and the marked angles show \(\angle PRS\cong\angle PQR\). Therefore, \(\triangle PRS\sim\triangle PQR\) by AA. 2. Corresponding sides give \(\frac{PQ}{PR}=\frac{PR}{PS}\). 3. Read \(PR=12\) and \(PS=9\) from the diagram: \(\frac{x}{12}=\frac{12}{9}\). Then \(9x=144\), so \(x=16\). 4. Therefore, \(y=SQ=PQ-PS=16-9=7\).

\(x=16\) and \(y=7\)

- Compare the two right triangles formed by the altitude. - Use the matching acute-angle marks to establish AA similarity. - Match the two base segments with the altitude in a proportion.

1. The marked right angle shows that triangles \(QPS\) and \(PSR\) are right triangles. Since \(S\) lies on \(QR\), \(\angle PRQ=\angle PRS\). 2. The marked acute angles give \(\angle QPS\cong\angle PRS\), so \(\triangle QPS\sim\triangle PRS\) by AA. 3. Corresponding sides give \(\frac{QS}{PS}=\frac{PS}{SR}\). 4. Read \(QS=4\) and \(SR=9\) from the diagram: \(\frac{4}{x}=\frac{x}{9}\). Then \(x^2=36\), so \(x=6\).

\(x=6\)

- First find the hypotenuse of the large right triangle. - Which smaller triangle shares an angle with \(\triangle ABC\) and also has a right angle? - Write a proportion using corresponding sides.

1. Use the Pythagorean theorem to find the hypotenuse: \(AB=\sqrt{8^2+6^2}=10\,\text{cm}\). 2. Triangles \(ADE\) and \(ABC\) share \(\angle A\), and \(\angle AED\) and \(\angle ACB\) are right angles. Therefore, \(\triangle ADE\sim\triangle ABC\) by AA. 3. Corresponding sides give \(\frac{DE}{BC}=\frac{AD}{AB}\). 4. Thus, \(\frac{DE}{6}=\frac{5}{10}\), so \(DE=3\,\text{cm}\).

\(DE=3\,\text{cm}\)

- Identify the two similar triangles formed by the transversals and parallel lines. - Read the two base lengths from the diagram. - The two perpendicular distances from \(P\) to the parallel lines add to \(10\,\text{cm}\).

1. The two triangles with vertex \(P\) and bases on the parallel lines are similar. 2. Let \(d_g\) be the distance from \(P\) to \(g\), and let \(d_h\) be the distance from \(P\) to \(h\). Then \(d_g+d_h=10\). 3. From the diagram, the corresponding bases are \(12\,\text{cm}\) and \(8\,\text{cm}\), so \(\frac{d_g}{d_h}=\frac{12}{8}=\frac{3}{2}\). 4. Thus, \(d_g=\frac{3}{2}d_h\). Substitute into the sum: \(\frac{3}{2}d_h+d_h=10\). 5. Therefore, \(\frac{5}{2}d_h=10\), so \(d_h=4\,\text{cm}\).

\(4\,\text{cm}\)

- Read the two parts of \(BC\) from the diagram and find the full length. - Which marked angles establish AA similarity? - Match the side \(AB\) in the two similar triangles carefully.

1. From the diagram, \(BC=BD+DC=3+9=12\,\text{cm}\). 2. Triangles \(ABD\) and \(CBA\) share the angle at \(B\), and the marked angles show \(\angle BAD\cong\angle BCA\). Therefore, \(\triangle ABD\sim\triangle CBA\) by AA. 3. Corresponding sides give \(\frac{AB}{BC}=\frac{BD}{AB}\). 4. Thus, \(\frac{AB}{12}=\frac{3}{AB}\), so \(AB^2=36\). Since a length is positive, \(AB=6\,\text{cm}\).

\(AB=6\,\text{cm}\)

- Match vertices that have the same angle-mark pattern. - Use the known corresponding sides to determine the scale factor between panels. - Apply that same factor to each unknown side expression.

1. The angle marks give the correspondence \(M\leftrightarrow C\), \(N\leftrightarrow A\), and \(K\leftrightarrow B\). Thus, \(\triangle KMN\sim\triangle BCA\). 2. Using corresponding sides \(KN\) and \(BA\), the scale factor from panel b) to panel a) is \(\frac{5}{10}=\frac{1}{2}\). 3. Since \(MN\) corresponds to \(CA\), \(\frac{y}{12}=\frac{1}{2}\), so \(y=6\). 4. Since \(KM\) corresponds to \(BC\), \(\frac{x}{x+3}=\frac{1}{2}\). Solving gives \(x=3\).

\(x=3\) and \(y=6\)

- Identify the similar triangles on opposite sides of \(O\). - Read the two corresponding transversal segments from the diagram to form the scale ratio. - Combine that ratio with \(x+y=20\), then reuse it for \(QO\) and \(ON\).

1. Since \(MQ\parallel LN\), \(\triangle MOQ\sim\triangle LON\). 2. From the diagram, \(\frac{x}{y}=\frac{MO}{OL}=\frac{12}{8}=\frac{3}{2}\), so \(x=\frac{3}{2}y\). 3. Use \(x+y=20\): \(\frac{3}{2}y+y=20\), so \(\frac{5}{2}y=20\) and \(y=8\). 4. Therefore, \(x=12\). 5. Also, \(\frac{QO}{ON}=\frac{MO}{OL}=\frac{3}{2}\). Thus, \(\frac{15}{ON}=\frac{3}{2}\), so \(ON=10\).

\(x=12\), \(y=8\), and \(ON=10\) units

- Use the Pythagorean Theorem in the smaller right triangle. - Find the full base of the larger triangle. - Use similarity to compare corresponding legs.

1. Because \(EF\parallel BC\), \(\triangle AEF\sim\triangle ABC\). 2. In right triangle \(AEF\), apply the Pythagorean Theorem: \(y=\sqrt{15^2-12^2}=\sqrt{81}=9\) units. 3. The full base of the larger triangle is \(AC=6+9=15\) units. 4. Corresponding legs satisfy \(\frac{x}{12}=\frac{15}{9}\). 5. Therefore, \(x=12\cdot\frac{15}{9}=20\) units.

\(x=20\) units and \(y=9\) units

- Which two triangles share the angle at \(M\)? - What second angle pair is indicated by the matching marks? - Match corresponding sides carefully before writing the two proportions.

1. Triangles \(MNL\) and \(MKN\) share the angle at \(M\), and the matching angle marks show \(\angle MNL\cong\angle MKN\). Therefore, \(\triangle MNL\sim\triangle MKN\) by AA. 2. Since \(MK=9+7=16\), corresponding sides give \(\frac{MN}{ML}=\frac{MK}{MN}\), or \(\frac{x}{9}=\frac{16}{x}\). 3. Thus, \(x^2=144\), so \(x=12\). 4. Also, \(\frac{NL}{KN}=\frac{ML}{MN}\), so \(\frac{y}{20}=\frac{9}{12}=\frac{3}{4}\). Therefore, \(y=15\).

\(x=12\) and \(y=15\)

- Identify the similar triangles formed by the diagonals and parallel bases. - Find the scale factor using the known diagonal segments. - Apply that factor to the corresponding base and diagonal segment.

1. Since \(TF\parallel SE\), \(\triangle FOT\sim\triangle SOE\). 2. The scale factor from the larger triangle to the smaller triangle is \(k=\frac{OF}{SO}=\frac{10}{16}=0.625\). 3. Therefore, \(x=TF=40\cdot0.625=25\) units. 4. Also, \(y=TO=24\cdot0.625=15\) units.

\(x=25\) units and \(y=15\) units

- Use the two labeled parts of \(AD\) to determine the small-to-large scale factor. - Apply that factor to the full side \(AC\). - Use the same factor for the parallel sides \(MN\) and \(DC\).

1. Since \(MN\parallel DC\), \(\triangle AMN\sim\triangle ADC\). 2. The scale factor from the larger triangle to the smaller triangle is \(\frac{AM}{AD}=\frac{10}{10+5}=\frac{2}{3}\). 3. Thus, \(\frac{x}{18}=\frac{2}{3}\), so \(x=12\) units. 4. Therefore, \(y=18-12=6\) units. 5. Also, \(\frac{14}{z}=\frac{2}{3}\), so \(z=21\) units.

\(x=12\) units, \(y=6\) units, and \(z=21\) units

- Which two triangles share the angle at \(R\)? - Use the matching angle marks to establish AA similarity. - Read the side lengths from the diagram and match corresponding sides before writing each proportion.

1. Triangles \(RKE\) and \(RTK\) share the angle at \(R\), and the matching angle marks show \(\angle RKE\cong\angle RTK\). Therefore, \(\triangle RKE\sim\triangle RTK\) by AA. 2. Corresponding sides give \(\frac{RE}{RK}=\frac{RK}{RT}\). Thus, \(\frac{x}{15}=\frac{15}{25}\), so \(x=9\). 3. Also, \(\frac{KE}{KT}=\frac{RK}{RT}\). Thus, \(\frac{y}{20}=\frac{15}{25}\), so \(y=12\).

\(x=9\) and \(y=12\)

- Use the ratio of the two parallel sides to find the similarity scale factor. - Express each full side as the sum of its two labeled parts. - Use one proportion for \(x\) and a second for \(y\).

1. Similar triangles give \(\frac{DE}{AC}=\frac{BD}{BA}\). 2. Substitute the labeled lengths: \(\frac{15}{25}=\frac{x}{x+8}\). 3. Solve: \(\frac{3}{5}(x+8)=x\), so \(3x+24=5x\) and \(x=12\) units. 4. The same scale factor gives \(\frac{BE}{BC}=\frac{3}{5}\). Thus, \(\frac{y}{y+10}=\frac{3}{5}\). 5. Solving gives \(5y=3y+30\), so \(y=15\) units.

\(x=12\) units and \(y=15\) units

- Which angle is shared by the two triangles? - What do the two right-angle marks tell you? - Read the leg lengths from the diagram, find the hypotenuse, and then match corresponding sides.

1. For part a), the right-angle marks show \(\angle ADC\cong\angle ACB\), and the triangles share \(\angle A\). Therefore, \(\triangle ACD\sim\triangle ABC\) by AA. 2. From the diagram, \(AC=6\,\text{cm}\) and \(BC=8\,\text{cm}\). Use the Pythagorean theorem: \(AB=\sqrt{6^2+8^2}=10\,\text{cm}\). 3. Corresponding sides give \(\frac{CD}{BC}=\frac{AC}{AB}\). 4. Therefore, \(CD=\frac{6\cdot8}{10}=4.8\,\text{cm}\).

a) \(\triangle ACD\sim\triangle ABC\) by AA because they share \(\angle A\) and the diagram marks a right angle in each triangle. b) \(CD=4.8\,\text{cm}\)

- Look for two triangles that both include vertex \(C\). - What do the right-angle marks show about those triangles? - Read the three given lengths from the diagram and match corresponding sides.

1. For part a), the right-angle marks show that triangles \(ADC\) and \(BEC\) are right triangles, and they share the angle at \(C\). Therefore, \(\triangle ADC\sim\triangle BEC\) by AA. 2. Corresponding sides give \(\frac{AD}{AC}=\frac{BE}{BC}\). 3. Read \(AC=10\,\text{cm}\), \(BC=15\,\text{cm}\), and \(BE=6\,\text{cm}\) from the diagram. Then \(AD=\frac{6\cdot10}{15}=4\,\text{cm}\).

a) \(\triangle ADC\sim\triangle BEC\) by AA because both are right triangles and share the angle at \(C\). b) \(AD=4\,\text{cm}\)

- Which side of the rectangle is parallel to \(AB\)? - Find the height of the triangle above the rectangle. - Use corresponding heights and bases of the similar triangles.

1. Since opposite sides of a rectangle are parallel, \(SR\parallel AB\). Therefore, \(\triangle SRC\sim\triangle ABC\) by AA. 2. The height of \(\triangle SRC\) is \(10-4=6\,\text{cm}\). 3. Corresponding bases and heights are proportional: \(\frac{SR}{AB}=\frac{6}{10}\). 4. Thus, \(SR=15\cdot\frac{6}{10}=9\,\text{cm}\).

a) \(\triangle SRC\) b) \(SR=9\,\text{cm}\)

- View the picture as two nested right triangles. - Identify the shared angle and the right angles. - Use the ratio of the pole heights to find the total distance \(SC\). - Subtract \(SA\) to find only the distance between the poles.

1. Both triangles are right triangles because the poles are perpendicular to the ground. They also share the angle at \(S\). Therefore, \(\triangle SAB\sim\triangle SCD\) by AA. 2. Corresponding heights and ground distances are proportional: \(\frac{CD}{AB}=\frac{SC}{SA}\). 3. Substitute: \(\frac{5}{2}=\frac{SC}{3}\), so \(SC=3\cdot\frac{5}{2}=7.5\,\text{m}\). 4. The distance between the poles is \(AC=SC-SA=7.5-3=4.5\,\text{m}\).

a) The triangles are similar by AA because each has a right angle and they share the angle at \(S\). b) \(AC=4.5\,\text{m}\)

- Why are the small and large triangles similar? - What fraction of \(BC\) is \(MC\)? - Square the linear scale factor to obtain the area ratio.

1. Because \(MN\parallel AB\), \(\triangle MNC\sim\triangle BAC\) by AA. 2. Since \(M\) is the midpoint of \(BC\), the linear scale factor from the large triangle to the small triangle is \(\frac{MC}{BC}=\frac{1}{2}\). 3. The area ratio is the square of the linear scale factor: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\). 4. Therefore, the ratio of the small area to the large area is \(1\) to \(4\).

\(1\) to \(4\)

- Read \(h\) and \(q\) from the diagram. - Use the geometric-mean relationship between the altitude and the two hypotenuse segments. - After finding \(p\), apply the Pythagorean theorem in each smaller right triangle.

1. The similar triangles give \(h^2=pq\), so from the diagram \(p=\frac{2.4^2}{1.8}=3.2\,\text{cm}\). 2. In right triangle \(BCD\), \(a=\sqrt{2.4^2+1.8^2}=\sqrt{9}=3\,\text{cm}\). 3. In right triangle \(ACD\), \(b=\sqrt{2.4^2+3.2^2}=\sqrt{16}=4\,\text{cm}\).

a) \(p=3.2\,\text{cm}\) b) \(a=3\,\text{cm}\); \(b=4\,\text{cm}\)

- Identify the complete distances from \(Z\) to the farther parallel segment. - Use a proportion involving distances measured from \(Z\) to find \(x\). - Then compare the lengths of the parallel segments to their distances from \(Z\).

1. The distances from \(Z\) to the two parallel segments are proportional on both rays: \(\frac{ZA}{ZC}=\frac{ZB}{ZD}\). 2. Substitute \(ZA=5\), \(ZC=5+3=8\), \(ZB=4\), and \(ZD=4+x\): \(\frac{5}{8}=\frac{4}{4+x}\). 3. Cross-multiply: \(5(4+x)=32\), so \(20+5x=32\), \(5x=12\), and \(x=2.4\,\text{cm}\). 4. The parallel segments are proportional to their distances from \(Z\): \(\frac{AB}{CD}=\frac{ZA}{ZC}\). 5. Substitute: \(\frac{6}{y}=\frac{5}{8}\). Cross-multiplying gives \(5y=48\), so \(y=9.6\,\text{cm}\).

\(x=2.4\,\text{cm}\) and \(y=9.6\,\text{cm}\)

- Which two triangles share angle \(A\)? - How can parallel lines establish equal angles in those triangles? - Compare the two distances along the extra ray before using corresponding sides.

1. Triangles \(APQ\) and \(ARB\) share the angle at \(A\). Their other corresponding angles are equal because \(PQ\parallel RB\). Thus the triangles are similar by AA. 2. Two equal steps reach \(P\) and five reach \(R\), so \(\frac{AP}{AR}=\frac{2}{5}=\frac{AQ}{AB}\). 3. Since \(AB=15\,\text{cm}\), \(AQ=\frac{2}{5}\cdot15=6\,\text{cm}\). The rest is \(QB=15-6=9\,\text{cm}\).

Triangles \(APQ\) and \(ARB\) are similar by AA: they share angle \(A\), and \(PQ\parallel RB\) gives another equal angle pair. Therefore \(AQ=6\,\text{cm}\) and \(QB=9\,\text{cm}\).

- Use a median through the centroid to identify a known ratio. - Determine which triangles become similar after the parallel is constructed. - Match the centroid ratio to corresponding sides of the similar triangles.

1. Let \(M\) be the midpoint of \(BC\). Since \(G\) is the centroid, \(G\) lies on median \(AM\) and \(AG=\frac{2}{3}AM\). 2. Because \(EF\parallel BC\), triangles \(AEF\) and \(ABC\) are similar. 3. Segment \(AG\) corresponds to \(AM\), so the similarity scale factor from \(ABC\) to \(AEF\) is \(\frac{AG}{AM}=\frac{2}{3}\). 4. Therefore, \(\frac{AE}{AB}=\frac{AF}{AC}=\frac{EF}{BC}=\frac{2}{3}\). 5. Hence \(AE=\frac{2}{3}AB\), \(AF=\frac{2}{3}AC\), and \(EF=\frac{2}{3}BC\).

The constructed parallel creates similar triangles with scale factor \(\frac{AG}{AM}=\frac{2}{3}\). Thus \(AE=\frac{2}{3}AB\), \(AF=\frac{2}{3}AC\), and \(EF=\frac{2}{3}BC\).

- Apply the Triangle Midsegment Theorem in the triangles formed by the diagonals of \(ABCD\). - Pair the two midpoint segments associated with the same diagonal. - Establish both pairs of opposite sides of \(EFGH\) as parallel.

1. In triangle \(ABC\), segment \(\overline{EF}\) joins two side midpoints, so \(EF\parallel AC\) and \(EF=\frac{1}{2}AC\). 2. In triangle \(CDA\), segment \(\overline{GH}\) joins two side midpoints, so \(GH\parallel AC\) and \(GH=\frac{1}{2}AC\). 3. Therefore, \(EF\parallel GH\) and \(EF=GH\). 4. Similarly, the Triangle Midsegment Theorem in triangles \(BCD\) and \(DAB\) gives \(FG\parallel BD\), \(HE\parallel BD\), and \(FG=HE=\frac{1}{2}BD\). 5. Both pairs of opposite sides of \(EFGH\) are parallel, so \(EFGH\) is a parallelogram.

\(EFGH\) is a parallelogram. Its sides \(\overline{EF}\) and \(\overline{GH}\) are parallel to diagonal \(\overline{AC}\), while \(\overline{FG}\) and \(\overline{HE}\) are parallel to diagonal \(\overline{BD}\).

- Consider adding one diagonal and using its midpoint. - Look for two triangles in which pairs of side midpoints create segments parallel to the bases. - Relate the two smaller segment lengths to the base lengths shown in the diagram.

1. Let \(M\) be the midpoint of diagonal \(\overline{AC}\). 2. In triangle \(ADC\), points \(E\) and \(M\) are midpoints, so \(\overline{EM}\parallel\overline{DC}\) and \(EM=\frac{1}{2}CD=4.5\,\text{cm}\). 3. In triangle \(ABC\), points \(M\) and \(F\) are midpoints, so \(\overline{MF}\parallel\overline{AB}\) and \(MF=\frac{1}{2}AB=7.5\,\text{cm}\). 4. Since \(\overline{AB}\parallel\overline{CD}\), segments \(\overline{EM}\) and \(\overline{MF}\) lie on one line. Therefore, \(\overline{EF}\) is parallel to both bases. 5. The length is \(EF=EM+MF=4.5\,\text{cm}+7.5\,\text{cm}=12\,\text{cm}\).

\(\overline{EF}\parallel\overline{AB}\parallel\overline{CD}\), and \(EF=12\,\text{cm}\).

- Use the midpoint of a diagonal in two different triangles. - Apply the converse of the Triangle Midsegment Theorem with the constructed parallel. - Combine the two half-lengths along \(\overline{EF}\).

1. The diagonals of a parallelogram bisect each other, so \(O\) is the midpoint of \(\overline{AC}\). 2. In triangle \(ADC\), line \(EO\) is parallel to \(DC\). Since \(O\) is the midpoint of \(AC\), the converse of the Triangle Midsegment Theorem gives that \(E\) is the midpoint of \(AD\), and \(EO=\frac{1}{2}DC\). 3. In triangle \(ABC\), line \(OF\) is parallel to \(AB\). Since \(O\) is the midpoint of \(AC\), point \(F\) is the midpoint of \(BC\), and \(OF=\frac{1}{2}AB\). 4. Opposite sides of a parallelogram are congruent, so \(DC=AB\). Thus, \(EO=OF=\frac{1}{2}AB\). 5. Therefore, \(EF=EO+OF=AB\).

The parallel through the diagonal midpoint makes \(E\) and \(F\) the midpoints of \(\overline{AD}\) and \(\overline{BC}\). Also, \(EO=OF=\frac{1}{2}AB\), so \(EF=AB\).

- Look at the two smaller triangles made by the altitude. - What angle pairs show that those triangles are similar? - Match the altitude to its corresponding side in a proportion.

1. Because \(DB\perp AC\), triangles \(ABD\) and \(DBC\) both have a right angle at \(B\). Their other angles match those of the original right triangle, so \(\triangle ABD\sim\triangle DBC\) by AA. 2. Corresponding sides give \(\frac{AB}{BD}=\frac{BD}{BC}\), so \(BD^2=AB\cdot BC\). 3. Substituting, \(BD^2=4\cdot9=36\). Since a length is positive, \(BD=6\,\text{cm}\). 4. Therefore, \(BD=\sqrt{AB\cdot BC}\), the geometric mean of the two hypotenuse segments.

\(BD=6\,\text{cm}\). The smaller right triangles are similar, giving \(BD^2=AB\cdot BC\) and \(BD=\sqrt{4\cdot9}\).

- Use the two pairs of parallel sides to classify the inner quadrilateral first. - Compare \(\triangle BDE\) with the original triangle. - Use the midpoint condition to determine the similarity scale factor.

1. Since \(AD\) lies on \(AB\) and \(EF\parallel AB\), \(AD\parallel EF\). Since \(AF\) lies on \(AC\) and \(DE\parallel AC\), \(AF\parallel DE\). Thus \(ADEF\) is a parallelogram. 2. Because \(AB\perp AC\), adjacent sides \(AD\) and \(AF\) are perpendicular. Therefore, \(ADEF\) is a rectangle. 3. Point \(D\) is the midpoint of \(AB\), so \(AD=DB=5\,\text{cm}\). 4. In \(\triangle BDE\), \(DE\parallel AC\), so \(\triangle BDE\sim\triangle BAC\). The scale factor is \(\frac{BD}{BA}=\frac12\), giving \(DE=\frac12 AC=5\,\text{cm}\). 5. The rectangle has adjacent sides \(AD=DE=5\,\text{cm}\), so it is a square. Its area is \(25\,\text{cm}^2\).

\(ADEF\) is a square with side length \(5\,\text{cm}\) and area \(25\,\text{cm}^2\).

- Use the two areas to find how lengths change. - Once you have the length factor, apply it to the perimeter, not its square.

1. The area ratio is \(\frac{75}{12}=\frac{25}{4}\). 2. The positive length factor is \(k=\sqrt{\frac{25}{4}}=\frac{5}{2}\). 3. The new perimeter is \(21\cdot\frac{5}{2}=52.5\) inches.

\(52.5\) inches

- What are the area and all three side lengths of the large triangle? - How can the three triangles around the incenter help you find its perpendicular distance from the sides? - After finding that distance, compare the two triangles with parallel bases.

1. The right triangle has legs \(5\) and \(12\,\text{cm}\), so its area is \(\frac{1}{2}\cdot5\cdot12=30\,\text{cm}^2\). Its hypotenuse is \(\sqrt{5^2+12^2}=13\,\text{cm}\). 2. The three triangles formed by joining \(I\) to the vertices have the same height \(r\). Thus \(30=\frac{1}{2}(5+12+13)r\), so \(r=2\,\text{cm}\). 3. Since \(EF\parallel BC\), \(\triangle AEF\sim\triangle ABC\). Their corresponding heights are \(5-r=3\) and \(5\,\text{cm}\). 4. Therefore \(\frac{EF}{12}=\frac{3}{5}\), giving \(EF=7.2\,\text{cm}\).

\(r=2\,\text{cm}\) and \(EF=7.2\,\text{cm}\). The smaller and larger triangles have height ratio \(\frac{3}{5}\).

- What does the perimeter ratio tell you about the length scale factor? - How is the area ratio related to the length scale factor? - Write an equation using the \(100\,\text{cm}^2\) difference between the areas.

1. The length scale factor from the smaller triangle to the larger triangle is \(k=\frac{5}{3}\). 2. Therefore, \(\frac{A_2}{A_1}=k^2=\left(\frac{5}{3}\right)^2=\frac{25}{9}\). 3. The area difference gives \(A_2-A_1=100\). 4. Substitute \(A_2=\frac{25}{9}A_1\): \(\frac{25}{9}A_1-A_1=100\), so \(\frac{16}{9}A_1=100\). 5. Thus, \(A_1=\frac{100\cdot9}{16}=56.25\,\text{cm}^2\). 6. The larger area is \(A_2=56.25\,\text{cm}^2+100\,\text{cm}^2=156.25\,\text{cm}^2\).

\(A_1=56.25\,\text{cm}^2\) and \(A_2=156.25\,\text{cm}^2\)

- Use the parallel bases and the vertical angles at \(S\) to compare the triangles. - The diagonal segments have the same ratio as the corresponding parallel sides. - Use the corresponding side-length ratio to split the full diagonal into \(AS\) and \(CS\). - How do the areas of similar figures compare to their corresponding side lengths?

1. Since \(AB\parallel CD\), alternate interior angles give two pairs of congruent angles. The vertical angles at \(S\) are also congruent, so \(\triangle ABS\sim\triangle CDS\) by AA. 2. Corresponding sides are proportional: \(\frac{AS}{CS}=\frac{AB}{CD}=1.5=\frac{3}{2}\). 3. Let \(AS=3x\) and \(CS=2x\). Since \(AS+CS=15\), \(5x=15\), so \(x=3\). 4. Thus, \(AS=9\,\text{cm}\) and \(CS=6\,\text{cm}\). 5. The area ratio of similar triangles is the square of the side-length ratio: \(\left(\frac{3}{2}\right)^2=\frac{9}{4}\).

a) \(\triangle ABS\sim\triangle CDS\) by AA because \(AB\parallel CD\) and the angles at \(S\) are vertical angles. b) \(AS=9\,\text{cm}\) and \(CS=6\,\text{cm}\) c) The area ratio is \(9\) to \(4\).

- What equation describes equal side-length ratios for similar rectangles? - What are the side lengths of each smaller rectangle after the longer side is divided into three equal parts? - Which side of a smaller rectangle corresponds to the longer side of the original rectangle? - Set the two longer-side-to-shorter-side ratios equal.

1. The original rectangle has side-length ratio \(\frac{b}{a}\). Dividing the longer side \(b\) into three equal parts produces smaller rectangles with side lengths \(a\) and \(\frac{b}{3}\). 2. Each smaller rectangle is similar to the original after a \(90^\circ\) rotation. Therefore, the ratio of the longer side to the shorter side satisfies \(\frac{b}{a}=\frac{a}{b/3}\). 3. Rewrite the equation as \(\frac{b}{a}=\frac{3a}{b}\). Multiplying by \(ab\) gives \(b^2=3a^2\). 4. Divide by \(a^2\) to obtain \(\frac{b^2}{a^2}=3\). Since side lengths are positive, \(\frac{b}{a}=\sqrt{3}\).

The ratio of the longer side to the shorter side is \(\sqrt{3}\) to \(1\).

- For part a, represent the longer side as a multiple of the shorter side and use the known area. - For part b, track the multiplicative change in area through four halvings. - For part c, write the two side lengths after exactly one halving before comparing ratios. - Rotation changes which side is called longer, but it does not change either side length.

1. Let the shorter A0 side be \(x\,\text{cm}\). The longer side is \(\sqrt{2}x\,\text{cm}\), and \(1\,\text{m}^2=10{,}000\,\text{cm}^2\). 2. Thus, \(x(\sqrt{2}x)=10{,}000\), so \(x=\sqrt{\frac{10000}{\sqrt{2}}}\approx84.1\,\text{cm}\). The longer side is \(\sqrt{2}x\approx118.9\,\text{cm}\). 3. Four halvings divide the area by \(2^4=16\), so the A4 area is \(\frac{10000}{16}=625\,\text{cm}^2\). 4. Write the original side lengths as \(s\) and \(\sqrt{2}s\). After cutting the longer side in half, the new rectangle has side lengths \(s\) and \(\frac{\sqrt{2}}{2}s=\frac{s}{\sqrt{2}}\). 5. After rotating the smaller rectangle, its longer-to-shorter side ratio is \(\frac{s}{s/\sqrt{2}}=\sqrt{2}\), the same as the original rectangle. Since both are rectangles, equal corresponding angles and proportional side lengths make them similar. 6. The same ratio calculation applies to every later halving, so each successive A-series rectangle is similar to the previous one and therefore to A0.

a) Approximately \(84.1\,\text{cm}\) by \(118.9\,\text{cm}\) b) \(625\,\text{cm}^2\) c) If the original sides are \(s\) and \(\sqrt{2}s\), one half has sides \(s\) and \(s/\sqrt{2}\). After rotation its side ratio is \(\sqrt{2}:1\), so it is similar to the original; the same argument repeats at every halving.

- Compare each rectangle by dividing its longer side by its shorter side. - Which side changes when the rectangle is cut as described? - Let the unknown longer side be a variable and write a proportion between the original rectangle and a rotated half.

1. For part a, \(\frac{18}{12}=1.5\) and \(\frac{24}{16}=1.5\). The corresponding side-length ratios are equal, and both figures are rectangles, so \(R_1\) and \(R_2\) are similar. 2. Cutting \(R_1\) as described produces rectangles with side lengths \(12\,\text{cm}\) and \(9\,\text{cm}\). Their longer-side-to-shorter-side ratio is \(\frac{12}{9}=\frac{4}{3}\approx1.33\), which is not equal to the original ratio \(1.5\). The halves are not similar to \(R_1\). 3. Let \(x\) be the required longer side. A half has side lengths \(12\) and \(\frac{x}{2}\), and it must correspond to the original after rotation. Set \(\frac{x}{12}=\frac{12}{x/2}\). Then \(\frac{x}{12}=\frac{24}{x}\), so \(x^2=288\). Since a length is positive, \(x=\sqrt{288}=12\sqrt{2}\approx16.97\,\text{cm}\).

a) Yes. Both rectangles have a longer-side-to-shorter-side ratio of \(1.5\). b) No. Each half has ratio \(\frac{4}{3}\approx1.33\), while the original has ratio \(1.5\). c) The longer side must be \(12\sqrt{2}\,\text{cm}\approx16.97\,\text{cm}\).

- Use the surface-area ratio to find \(k^2\), then take a square root. - For objects made from the same material, mass scales with volume. - Move from a volume factor to a linear factor with a cube root. - Square the linear factor to obtain the surface-area factor.

1. For part a), the surface-area factor is \(\frac{1.8}{0.8}=2.25\), so \(k=\sqrt{2.25}=1.5\). 2. For part b), mass is proportional to volume when the material is the same. The volume factor is \(1.5^3=3.375\), so the larger mass is \(15\cdot3.375=50.625\,\text{kg}\). 3. For part c), doubling the mass gives a volume factor of \(2\), so the linear factor is \(\sqrt[3]{2}\). 4. The surface-area factor is \(\left(\sqrt[3]{2}\right)^2=\sqrt[3]{4}\approx1.5874\). The increase is approximately \(58.7\%\).

a) \(k=1.5\) b) \(50.625\,\text{kg}\) c) The surface area is approximately \(58.7\%\) greater.

- Identify the parallelogram and use its opposite-side relationship. - Use the similar triangles \(RST\) and \(RPQ\), reading \(ST\) from the diagram. - Find each full side before subtracting the known segment.

1. Because \(PQ\parallel SU\) and \(PR\parallel QU\), quadrilateral \(PQUS\) is a parallelogram. Therefore, \(QU=PS\). 2. Since \(ST\parallel PQ\), \(\triangle RST\sim\triangle RPQ\). The displayed value of \(ST\) gives \(\frac{ST}{PQ}=\frac{RT}{RQ}=\frac{10}{15}=\frac{2}{3}\). 3. Solve \(\frac{8}{RQ}=\frac{2}{3}\) to get \(RQ=12\,\text{cm}\). Therefore, \(TQ=RQ-RT=12-8=4\,\text{cm}\). 4. Also, \(\frac{RS}{RP}=\frac{2}{3}\). Since \(RS=12\,\text{cm}\), \(RP=18\,\text{cm}\). 5. Then \(PS=RP-RS=18-12=6\,\text{cm}\), so \(QU=6\,\text{cm}\).

\(TQ=4\,\text{cm}\) and \(QU=6\,\text{cm}\)

- Read the two post heights from the diagram and locate the crossing point. - Use the smaller similar triangles formed by each cable and the vertical through the crossing point. - Express the crossing height once from each cable, then combine the relationships.

1. Read the two post heights from the diagram. Let \(d\) be the distance between the posts, \(x\) the horizontal distance from the left post to the crossing point, and \(h\) the crossing height. 2. From the similar triangles along the cable descending from the \(4\,\text{m}\) post, \(\frac{h}{4}=\frac{d-x}{d}\). 3. From the similar triangles along the cable rising to the \(12\,\text{m}\) post, \(\frac{h}{12}=\frac{x}{d}\). 4. Add the equations: \(\frac{h}{4}+\frac{h}{12}=1\). 5. Thus, \(h\left(\frac{1}{4}+\frac{1}{12}\right)=1\), so \(\frac{h}{3}=1\) and \(h=3\,\text{m}\).

The cables cross \(3\,\text{m}\) above the ground.

- What does the circle tell you about \(CD_1\) and \(CD_2\)? - Write the proportion that follows from \(AB\parallel CD_1\). - Compare the directions of \(AB\) and \(CD_2\).

1. Since \(AB\parallel CD_1\), proportionality gives \(\frac{ZA}{ZC}=\frac{AB}{CD_1}\). 2. Both \(D_1\) and \(D_2\) lie on the circle centered at \(C\), so \(CD_1=CD_2\). 3. Replacing \(CD_1\) with the equal length \(CD_2\) gives \(\frac{ZA}{ZC}=\frac{AB}{CD_2}\). 4. However, the diagram shows that \(CD_2\) has a different direction from \(AB\), so the segments are not parallel. Equal ratios of this form are therefore not sufficient to prove parallelism.

Because \(CD_1=CD_2\), the proportion remains true when \(CD_1\) is replaced by \(CD_2\). However, \(CD_2\) is not parallel to \(AB\), so this proportion alone is not a valid converse condition for parallel lines.

- Use the parallel top and bottom edges to establish two matching angles. - Compare the corresponding bases \(AB\) and \(DF\) to get the similarity scale factor. - Corresponding altitudes of similar triangles have the same linear scale factor. - The two altitudes together span the distance between the parallel sides of the squares.

1. Because \(AB\parallel DF\), \(\angle SAB\cong\angle SDF\). Also, \(\angle SBA\cong\angle SFD\). Therefore, \(\triangle ABS\sim\triangle DFS\) by AA. 2. Since the two adjacent squares each have side length \(a\), \(AB=a\) and \(DF=2a\). Thus, the scale factor from \(\triangle ABS\) to \(\triangle DFS\) is \(2\). 3. Let the perpendicular height from \(S\) to \(AB\) be \(h\). Corresponding altitudes of similar triangles scale by the same factor, so the perpendicular height from \(S\) to \(DF\) is \(2h\). 4. The parallel lines containing \(AB\) and \(DF\) are \(a\) units apart, so \(h+2h=a\). Hence, \(h=\frac{a}{3}\). 5. Therefore, \([ABS]=\frac{1}{2}\cdot a\cdot\frac{a}{3}=\frac{1}{6}a^2\).

a) \(\triangle ABS\sim\triangle DFS\) by AA. b) \([ABS]=\frac{1}{6}a^2\)

- Why is each shaded triangle similar to the large triangle? - How does an area ratio relate to the corresponding linear scale factor? - How could all three scale factors describe one large triangle? - What happens to area when all lengths are multiplied by a scale factor?

1. Each shaded triangle is similar to \(\triangle ABC\) because its sides are parallel to the corresponding sides of \(\triangle ABC\). 2. Let the area of \(\triangle ABC\) be \(S\). For similar figures, each linear scale factor equals the square root of the corresponding area ratio. 3. The corresponding side segments of the three shaded triangles can be translated to partition one side of \(\triangle ABC\), so their linear scale factors add to \(1\). Therefore, \(\frac{\sqrt{4}}{\sqrt{S}}+\frac{\sqrt{9}}{\sqrt{S}}+\frac{\sqrt{25}}{\sqrt{S}}=1\). 4. Thus, \(\sqrt{S}=2+3+5=10\), so \(S=10^2=100\,\text{cm}^2\).

\(100\,\text{cm}^2\)

- Interpret the three matching angle marks in the diagram before writing any proportions. - Which triangle becomes isosceles because two of its angles are congruent? - After establishing the larger similarity relationship, use \(AC=CD+DA\).

1. The matching angle marks at \(B\) show that \(BD\) bisects \(\angle B\). The marked angle at \(A\) is congruent to those half-angles, so \(\triangle ABD\) is isosceles and \(BD=AD=5\,\text{cm}\). 2. Triangles \(BCD\) and \(ACB\) share the angle at \(C\), and the matching angle marks show \(\angle CBD\cong\angle A\). Therefore, \(\triangle BCD\sim\triangle ACB\) by AA. 3. Since \(AC=4+5=9\,\text{cm}\), corresponding sides give \(\frac{BC}{AC}=\frac{CD}{BC}\). Thus, \(\frac{x}{9}=\frac{4}{x}\), so \(x=6\,\text{cm}\). 4. Also, \(\frac{BD}{AB}=\frac{CD}{BC}\). Hence, \(\frac{5}{y}=\frac{4}{6}\), so \(y=7.5\,\text{cm}\).

\(x=6\,\text{cm}\) and \(y=7.5\,\text{cm}\)

- Identify the similar triangles formed by the diagonals and parallel bases. - Take the square root of the area ratio to find the linear ratio. - Combine the linear ratio with \(x+y=20\,\text{cm}\).

1. Since \(BC\parallel AD\), \(\triangle BOC\sim\triangle AOD\). 2. The area ratio is the square of the linear ratio. Since the area ratio from the smaller triangle to the larger triangle is \(1\) to \(9\), the linear scale factor is \(\sqrt{9}=3\). 3. Therefore, \(y=3x\). Using \(x+y=20\), \(x+3x=20\), so \(x=5\,\text{cm}\) and \(y=15\,\text{cm}\). 4. The corresponding diagonal segments use the same factor: \(OD=3\cdot BO=3\cdot4=12\,\text{cm}\).

\(x=5\,\text{cm}\), \(y=15\,\text{cm}\), and \(OD=12\,\text{cm}\)

- Interpret all three right-angle marks before comparing the triangles. - Determine which acute angles must be congruent. - After using similarity to find \(x\), use the Pythagorean theorem for \(y\).

1. The marked right angles at \(R\), \(L\), and \(O\) imply that \(\angle RKO\cong\angle MOL\). Therefore, \(\triangle KRO\sim\triangle OLM\) by AA. 2. Corresponding legs give \(\frac{KR}{RO}=\frac{OL}{LM}\), so \(\frac{x}{15}=\frac{8}{6}\). Thus, \(x=20\). 3. In right triangle \(KRO\), \(y=KO=\sqrt{20^2+15^2}=25\).

\(x=20\) and \(y=25\)

- First compare the two triangles formed by the intersecting diagonals and the parallel bases. - Read the two base lengths from the diagram and convert their ratio into a fraction of each whole diagonal. - Use the line through \(O\) in one triangle on each side of \(O\).

1. Triangles \(AOB\) and \(COD\) are similar because their angles at \(O\) are vertical and the bases \(AB\) and \(CD\) are parallel. 2. From the diagram, \(\frac{AO}{OC}=\frac{BO}{OD}=\frac{AB}{CD}=\frac{12}{8}=\frac{3}{2}\). Therefore, \(\frac{AO}{AC}=\frac{BO}{BD}=\frac{3}{5}\). 3. In triangle \(ADC\), \(EO\parallel DC\), so similarity gives \(\frac{EO}{DC}=\frac{AO}{AC}=\frac{3}{5}\). Hence \(EO=\frac{3}{5}\cdot8=\frac{24}{5}\,\text{cm}\). 4. In triangle \(BCD\), \(OF\parallel CD\), so \(\frac{OF}{CD}=\frac{BO}{BD}=\frac{3}{5}\). Hence \(OF=\frac{24}{5}\,\text{cm}\). 5. Therefore, \(EO=OF\), and \(EF=EO+OF=\frac{48}{5}=9.6\,\text{cm}\).

\(EO=OF=\frac{24}{5}\,\text{cm}\), so \(EF=\frac{48}{5}\,\text{cm}=9.6\,\text{cm}\).

- Which triangle-midpoint fact could help prove parallelism? - The diagonal midpoints belong to different triangles. What might connect their midpoint relationships? - Once you prove parallelism, how can you compare two lengths along a line?

1. Let \(E\) be the midpoint of leg \(AD\). 2. In \(\triangle ADC\), points \(E\) and \(M\) are midpoints, so \(EM\parallel DC\) and \(EM=\frac{1}{2}CD=4\,\text{cm}\). 3. In \(\triangle ADB\), points \(E\) and \(N\) are midpoints, so \(EN\parallel AB\) and \(EN=\frac{1}{2}AB=7\,\text{cm}\). 4. Because \(AB\parallel CD\), segments \(EM\) and \(EN\) lie on the same line through \(E\). Thus, \(MN\) is parallel to both bases. 5. Both \(M\) and \(N\) lie on the same side of \(E\), with \(EM=4\,\text{cm}\) and \(EN=7\,\text{cm}\). Therefore, \(MN=7-4=3\,\text{cm}\).

\(MN\parallel AB\parallel CD\), and \(MN=3\,\text{cm}\).

- Relate the small triangle to the original triangle by similarity. - Compare their corresponding altitudes using the inradius. - Express the area of the original triangle in two different ways.

1. Because \(EF\parallel BC\), triangles \(AEF\) and \(ABC\) are similar. Let their scale factor be \(k=\frac{AE}{AB}=\frac{AF}{AC}=\frac{EF}{BC}\). 2. Let \(h\) be the altitude from \(A\) to \(BC\), and let \(r\) be the inradius. Since \(I\) lies on \(EF\) and its perpendicular distance to \(BC\) is \(r\), the altitude from \(A\) to \(EF\) is \(h-r\). 3. Therefore, \(k=\frac{h-r}{h}=1-\frac{r}{h}\). 4. If \(s=\frac{a+b+c}{2}\), the triangle area satisfies \(\frac{1}{2}ah=rs\). Hence \(\frac{r}{h}=\frac{a}{a+b+c}\). 5. Thus, \(k=1-\frac{a}{a+b+c}=\frac{b+c}{a+b+c}\). 6. The perimeter of \(\triangle AEF\) is \(k(a+b+c)=b+c\).

Let \(h\) be the altitude to \(BC\), \(r\) the inradius, and \(s=\frac{a+b+c}{2}\). Similarity gives scale factor \(k=\frac{h-r}{h}=1-\frac{r}{h}\). Since \(\frac{1}{2}ah=rs\), \(\frac{r}{h}=\frac{a}{a+b+c}\), so \(k=\frac{b+c}{a+b+c}\). Therefore, the perimeter of \(\triangle AEF\) is \(k(a+b+c)=b+c=CA+AB\).

- Do not match vertices by the order in which the side lengths are listed. - First determine which three side pairs share one scale factor. - A median corresponds according to its vertex and the opposite side, so use the vertex correspondence you established.

1. Match the side lengths by a common factor: \(QR=3AB\), \(RP=3BC\), and \(PQ=3CA\). 2. Therefore, all three corresponding side pairs are proportional, so the triangles are similar by SSS. 3. Since \(AB\leftrightarrow QR\), \(BC\leftrightarrow RP\), and \(CA\leftrightarrow PQ\), the vertex correspondence is \(A\leftrightarrow Q\), \(B\leftrightarrow R\), \(C\leftrightarrow P\). Thus, \(\triangle ABC\sim\triangle QRP\). 4. The scale factor from \(\triangle ABC\) to \(\triangle QRP\) is \(3\). A median from \(A\) to \(BC\) therefore corresponds to the median from \(Q\) to \(RP\). 5. Its length is \(3\cdot5.5=16.5\).

a) \(\triangle ABC\sim\triangle QRP\) by SSS. b) The corresponding median is the median from \(Q\) to \(RP\), and its length is \(16.5\).

- First combine the two pieces of the hypotenuse. - Use the parallel segment to relate the small right triangle to the whole triangle. - Pay attention to the matching tick marks after expressing \(EF\) and \(NF\) in terms of the large triangle's legs.

1. The hypotenuse is \(KM=40+30=70\). 2. Because \(EF\parallel KN\), \(\triangle EFM\sim\triangle KNM\), with scale factor \(\frac{EM}{KM}=\frac{30}{70}=\frac{3}{7}\). 3. Therefore, \(EF=\frac{3}{7}KN\) and \(FM=\frac{3}{7}NM\), so \(NF=NM-FM=\frac{4}{7}NM\). 4. The matching tick marks show \(EF=NF\). Thus, \(\frac{3}{7}KN=\frac{4}{7}NM\), so \(KN=\frac{4}{3}NM\). 5. Apply the Pythagorean Theorem: \(NM^2+\left(\frac{4}{3}NM\right)^2=70^2\). 6. This gives \(\frac{25}{9}NM^2=4900\), so \(NM=42\) and \(KN=56\). 7. The perimeter is \(42+56+70=168\).

\(x=168\) units