Free Math Worksheets

- Use the geometric-mean relationship connecting a leg, the hypotenuse, and the adjacent hypotenuse segment.

1. The similar triangles formed by the altitude give \(b^2=cq\). 2. Substitute: \(12^2=20q\). 3. Solve: \(q=\frac{144}{20}=7.2\,\text{cm}\).

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

- 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 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 \(1\).

- 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 \(AC\parallel BD\).

- Identify the two intersecting lines through \(S\). - Identify the two parallel segments and the two similar triangles. - Compare the parallel sides with their corresponding distances from \(S\).

1. Because \(AB\parallel CD\), triangles \(\triangle SAB\) and \(\triangle SCD\) 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 \(AB=9\,\text{cm}\)

- Volume scales with the cube of the linear factor. - Cube the numerator and denominator of the fraction. - Simplify before multiplying.

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 \(c_2\): \(c_2=0.75\cdot12.5\,\text{cm}=9.375\,\text{cm}\).

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

- 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, parallel segments. - Compare each height with its distance from \(S\). - Use a proportion based on the two similar triangles.

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

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

- Find the full distance \(ZC\). - Compare corresponding distances from \(Z\) to the two parallel lines. - After finding \(ZD\), subtract \(ZB\).

1. The full distance on the lower ray is \(ZC=ZA+AC=4+6=10\,\text{cm}\). 2. Because \(g\parallel h\), corresponding distances from \(Z\) are proportional: \(\frac{ZA}{ZC}=\frac{ZB}{ZD}\). 3. Substitute the values: \(\frac{4}{10}=\frac{5}{ZD}\), 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 \(y=8\)

- Find the full distance \(ZA\). - Compare the two parallel segment lengths with the corresponding distances from \(Z\). - Solve the resulting proportion for \(x\).

1. The full distance is \(ZA=ZC+CA=4+2=6\,\text{cm}\). 2. Because \(CD\parallel AB\), the corresponding lengths are proportional: \(\frac{ZC}{ZA}=\frac{CD}{AB}\). 3. Substitute the 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 rays from the top and bottom of the object through the pinhole to the screen. - Identify the two similar triangles with vertex at the pinhole. - Compare image height with object height and image distance with object distance.

1. The object and image form similar triangles with the pinhole as their common vertex. 2. 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 \(\frac{1}{2}\cdot42\,\text{cm}=21\,\text{cm}\).

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

- The altitude is the geometric mean of the two hypotenuse segments. - Add the two segments to find the entire hypotenuse.

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

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

- Match corresponding sides in the two triangles. - The ratio of one pair of corresponding sides is the same for every pair.

1. The scale factor from \(A_1B_1C_1\) to \(ABC\) is \(k=1.5\). 2. Multiply each corresponding side 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 \(z=15\)

- Which ratio number corresponds to the side of length \(16\)? - Find the length represented by one ratio part. - Multiply that value by the other two ratio numbers.

1. Because the triangles are similar, the side lengths of \(A_1B_1C_1\) are also in the ratio \(3\) to \(4\) to \(5\). The side \(b_1=16\) corresponds to \(4\) equal ratio parts. 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 \(y=20\)

- First find the linear scale factor from the corresponding side lengths. - 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. The linear scale factor from \(\triangle ABC\) to \(\triangle DEF\) is \(k=\frac{DE}{AB}=\frac{10}{5}=2\). 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, \(x=15\cdot4=60\,\text{cm}^2\).

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

- Find the full distance \(SB\). - Compare the parallel segments with the corresponding distances from \(S\).

1. The full distance is \(SB=SA+AB=8+12=20\). 2. The parallel segment lengths are proportional to the distances from \(S\): \(\frac{BD}{AC}=\frac{SB}{SA}\). 3. Thus, \(\frac{x}{6}=\frac{20}{8}=2.5\), so \(x=15\).

\(x=15\)

- How does the length of a triangle midsegment compare with the parallel side? - How does doubling every side length affect the perimeter?

1. Each side of a medial triangle is half the length of the parallel side of the original triangle. 2. Therefore, the side lengths of \(\triangle ABC\) are \(2\cdot7=14\,\text{cm}\), \(2\cdot8=16\,\text{cm}\), and \(2\cdot9=18\,\text{cm}\). 3. The perimeter is \(x=14+16+18=48\,\text{cm}\).

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

- Identify the similar triangles formed by the diagonals and parallel bases. - Match the diagonal segments with the corresponding bases. - Solve the resulting proportion.

1. Since \(AB\parallel CD\), \(\triangle ABS\sim\triangle CDS\). 2. Corresponding sides satisfy \(\frac{AS}{SC}=\frac{AB}{CD}\). 3. Substitute the values: \(\frac{6}{SC}=\frac{9}{6}=1.5\). 4. Therefore, \(SC=6\div1.5=4\,\text{cm}\).

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

- How does a perimeter ratio compare with the ratio of corresponding side lengths? - Use the order of the two perimeters to determine the correct scale factor.

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. Thus, \(DE=6\cdot\frac{5}{2}=15\,\text{cm}\).

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

- 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 \(8+10+8+10=36\,\text{m}\).

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

- Compare the lengths of the two bases. - Parallel cross-sections change by equal amounts when the legs are divided into equal parts. - Divide the difference between the base lengths into three equal changes.

1. The difference between the base lengths is \(18-9=9\,\text{cm}\). 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 \(s_2=15\,\text{cm}\)

- 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 \(k=3\).

- 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 \(2\).

- 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 \(1\).

- Corresponding side lengths of similar figures have a constant ratio. - Test which sides correspond by comparing the given lengths. - Ordering the side lengths from least to greatest can help identify corresponding sides.

1. Match the shortest known side of the second triangle to the shortest side of the first triangle: \(k=\frac{10}{8}=1.25\). 2. Check the longest sides: \(15\,\text{cm}\cdot1.25=18.75\,\text{cm}\). The correspondence is consistent. 3. The unknown side corresponds to the \(12\,\text{cm}\) side of the first triangle. 4. Its length is \(12\,\text{cm}\cdot1.25=15\,\text{cm}\).

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 \(b_1=18\,\text{cm}\).

- 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 \(9\,\text{cm}\).

- 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 \(26\,\text{cm}\).

- Cube the linear scale factor to obtain the volume scale factor. - Convert cubic meters to cubic centimeters after finding the model volume.

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 \(b_2=6.4\,\text{cm}\)

- Do circumradii scale in the same way as corresponding side lengths? - How can two corresponding radii be used to find the scale factor? - How are area ratios related to length scale factors?

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 \(\frac{A_2}{A_1}=6.25=k^2\).

- Identify the two right triangles formed by the measurements and the line of sight. - Identify the pair of vertical angles at \(B\). - Match the corresponding sides of the similar triangles. - Use a proportion to find \(AC\).

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

- 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 side-view geometry of the observer, fence, and wall. - Use the observer’s eye as the common vertex of the similar triangles. - Compare height differences measured from eye level, not from the ground. - Find the observer’s total 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 \(3.60\,\text{m}\).

- Use the top-view geometry with the smaller pillar between the observer and the larger pillar. - A sightline tangent to a circle is perpendicular to the radius at the point of tangency. - Compare each radius with the distance from the observer to the corresponding center.

1. The pillar radii are \(r_1=0.60\,\text{m}\) and \(r_2=0.90\,\text{m}\). 2. Let \(x\) be the distance from the observer to the center of the smaller pillar. The distance to the center of the larger pillar is \(x+6\). 3. When the pillars have the same apparent width, the right triangles formed by a sightline and a radius to each point of tangency are similar. Therefore, \(\frac{r_1}{x}=\frac{r_2}{x+6}\). 4. Substitute: \(\frac{0.60}{x}=\frac{0.90}{x+6}\). 5. Cross-multiply: \(0.60(x+6)=0.90x\), so \(0.60x+3.60=0.90x\). 6. Thus, \(3.60=0.30x\), and \(x=12\,\text{m}\).

The observer must stand \(12\,\text{m}\) from the center of the smaller pillar.

- 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.

- Identify the light source, ruler, wall, and the ray reaching the top of the shadow. - Find the total distance from the light source to the wall. - Compare the ruler height and shadow height with their distances from the light source. - Identify the equal angles 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.

- Identify the two right triangles that meet at the mirror. - Recall the relationship between the angle of incidence and the angle of reflection. - Match each vertical height with its horizontal distance from the mirror. - Consider how sloped or uneven ground would affect the angles and measured distances.

1. The person-mirror triangle and the tower-mirror triangle are right triangles. 2. By the law of reflection, the angle of incidence equals the angle of reflection. Together with the right angles made with level ground, this establishes AA similarity. 3. Let \(h\) be the tower height. Corresponding sides give \(\frac{h}{45}=\frac{1.70}{2.50}\). 4. Solve: \(h=45\cdot\frac{1.70}{2.50}=45\cdot0.68=30.6\,\text{m}\). 5. The calculation assumes the ground is level, so the measured ground distances are horizontal and both vertical heights are perpendicular to the same line.

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

- First consider where the shadow would end if the wall were not present. - Use the geometry of the post, wall, ground, and the ray that forms the shadow’s edge. - Identify the large and small similar triangles. - Use the remaining horizontal distance from the wall to the tip of the full shadow.

1. Without the wall, the shadow would reach the ground \(9.0\,\text{m}\) from the post. 2. The wall is \(9.0-6.0=3.0\,\text{m}\) from the tip of the full ground shadow. 3. The small triangle beside the wall is similar to the triangle formed by the post and its full shadow. Therefore, \(\frac{h}{3.0}=\frac{6.0}{9.0}\). 4. Solve: \(h=3.0\cdot\frac{6.0}{9.0}=2.0\,\text{m}\).

The edge of the shadow ends \(2.0\,\text{m}\) above the ground on the wall.

- Identify the streetlight, person, shadow, and the ray from the light to the tip of the shadow. - The large triangle’s base includes both the pole-to-person distance and the shadow length. - Write a proportion in which \(s\) appears in two places. - Think about how the base of the large triangle 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.

- Draw altitudes to form right triangles at the ends of each trapezoid. - Similar figures must have proportional corresponding side lengths. - Compare all types of corresponding sides.

1. Dropping altitudes creates right triangles whose horizontal legs have length \(\frac{a-c}{2}\). Thus, \(\cos(\alpha)=\frac{a-c}{2b}\). 2. For \(T_1\), \(\cos(\alpha_1)=\frac{10-6}{2\cdot 4}=\frac{1}{2}\), so \(\alpha_1=60^\circ\). 3. For \(T_2\), \(\cos(\alpha_2)=\frac{15-11}{2\cdot 4}=\frac{1}{2}\), so \(\alpha_2=60^\circ\). Both trapezoids have angles \(60^\circ,60^\circ,120^\circ,120^\circ\). 4. Compare corresponding side ratios: \(\frac{15}{10}=1.5\), but \(\frac{11}{6}\approx 1.83\) and \(\frac{4}{4}=1\). 5. Because the corresponding side ratios are not equal, the trapezoids are not similar.

a) Both longer-base angles are \(60^\circ\). Each trapezoid has the same four angle measures. b) No. The corresponding side ratios are not constant: \(1.5\), approximately \(1.83\), and \(1\).

- 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) \(63\,\text{cm}^2\)

- 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 \(1:15\).

\(1:15\)

- 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) \(1350\,\text{mL}\)

- 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, \(A_2=A_1\cdot\frac{16}{9}=54\cdot\frac{16}{9}=96\,\text{cm}^2\).

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

- 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 \(b_2=20\,\text{cm}\)

- 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) \(p=u\), \(q=v\)

- The crossing braces form two triangles with vertex \(Z\). - Match segments that lie on the same diagonal brace. - The ratio of the parallel beams equals the ratio of corresponding brace segments.

1. Since \(AB\parallel CD\), triangles \(\triangle ZAB\) and \(\triangle ZCD\) are similar by AA. 2. Corresponding brace segments satisfy \(\frac{ZA}{ZC}=\frac{ZB}{ZD}\). Substitute: \(\frac{1.2}{2.4}=\frac{1.0}{ZD}\). 3. Since \(\frac{1.2}{2.4}=\frac{1}{2}\), \(ZD=2.0\,\text{m}\). 4. Corresponding parallel beams satisfy \(\frac{AB}{CD}=\frac{ZA}{ZC}\). Substitute: \(\frac{0.8}{CD}=\frac{1.2}{2.4}=\frac{1}{2}\). 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 \(e=5\)

- Identify the two right triangles that share the same shadow tip. - Find the flagpole’s total shadow length. - Match each height with its corresponding shadow length. - Use the fact that the sun’s rays are parallel.

1. The flagpole’s full shadow length is \(2+8=10\,\text{m}\). 2. The student and 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.

- Use the ratio of the parallel segments to relate \(x\) and \(y\). - Combine that proportion with \(y-x=2\). - The same ratio applies to the corresponding distances on the second intersecting line.

1. The ratio of the parallel segments is \(\frac{a}{b}=\frac{6}{9}=\frac{2}{3}\). 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, \(\frac{5}{v}=\frac{2}{3}\), 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) \(z=a+b\)

- Find the ratio of the two parallel segments. - Match distances on opposite sides of \(Z\). - Write one proportion for each unknown.

1. The ratio of the parallel segments 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 \(y=5\,\text{cm}\)

- Find the full distance \(ZB\). - Use corresponding distances from \(Z\) to find \(x\). - Use the same ratio for the parallel segment lengths to find \(y\).

1. The full distance is \(ZB=ZA+AB=4+6=10\,\text{cm}\). 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 similar triangles formed by the crossed lines and the parallel segments. - Match distances measured from \(S\) to find \(u\). - Use the same scale factor for the parallel segments to find \(v\).

1. Since \(AC\parallel BD\), \(\triangle ASC\sim\triangle BSD\). 2. Corresponding distances satisfy \(\frac{SA}{SB}=\frac{SC}{SD}\). Thus, \(\frac{3}{4.5}=\frac{2}{u}\), 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 \(f=4\,\text{cm}\)

- Use the parallel bases to identify similar triangles. - Determine how the diagonals divide each other. - Relate the distance from \(S\) to \(AD\) to the base \(AB\) using similar triangles.

1. Because \(AB\parallel CD\), \(\triangle ABS\sim\triangle CDS\). The ratio of the bases is \(\frac{CD}{AB}=\frac{4}{12}=\frac{1}{3}\). 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 \(DE=6\,\text{cm}\)

- 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) \(6.25\)

- First use the pair \(AB\parallel CD\). - Apply the same scale factor to the corresponding parallel segments. - Then use the pair \(CD\parallel EF\) to find the remaining values.

1. 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 \(EF=4.5\,\text{cm}\)

- The altitude to the hypotenuse creates three similar right triangles. - Use the relationship between a leg, its adjacent hypotenuse segment, and the entire hypotenuse. - After finding the full hypotenuse, subtract to find the other segment, then use the similarity relationships for the remaining lengths.

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

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

- The altitude to the hypotenuse creates three similar right triangles. - Relate the given leg to its adjacent hypotenuse segment and the full hypotenuse. - The two hypotenuse segments add to \(c\), and the altitude is the geometric mean of those segments.

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

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

- First find the hypotenuse with the Pythagorean theorem. - Use the similar-triangle relationships to find the two hypotenuse segments. - Check that \(p+q=c\).

1. Find the hypotenuse: \(c=\sqrt{5^2+12^2}=13\,\text{cm}\). 2. The similar triangles give \(a^2=cp\), so \(p=\frac{25}{13}\,\text{cm}\approx 1.92\,\text{cm}\). 3. Similarly, \(b^2=cq\), so \(q=\frac{144}{13}\,\text{cm}\approx 11.08\,\text{cm}\). 4. The altitude satisfies \(h^2=pq\). Using exact values, \(h=\sqrt{\frac{25}{13}\cdot\frac{144}{13}}=\frac{60}{13}\,\text{cm}\approx 4.62\,\text{cm}\).

\(p\approx 1.92\,\text{cm}\), \(q\approx 11.08\,\text{cm}\), and \(h\approx 4.62\,\text{cm}\)

- Recall the converse relationship in the Triangle Midsegment Theorem. - What must be true about \(P\) if a segment through midpoint \(K\) is parallel to \(AC\)?

1. Suppose \(KP \parallel AC\). 2. By the converse relationship in the Triangle Midsegment Theorem, a line through the midpoint of one side of a triangle that is parallel to a second side must meet 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 \(16\).

- Set up a proportion in which \(x\) appears twice. - Cross-multiply to obtain a quadratic equation. - Choose the solution that is valid for a length.

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

\(x=6\)

- Use proportional corresponding intervals on the two rays. - Use the total-length equation \(x+y=16\). - Express one variable in terms of the other before solving.

1. 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 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 \(y=6\)

- Which pair of corresponding sides has both lengths given? - Use that pair to find the scale factor between the triangles. - Decide whether to multiply or divide by the scale factor for each unknown side.

1. Use the known pair of corresponding sides to find the scale factor from \(A_1B_1C_1\) to \(ABC\): \(k=\frac{AB}{A_1B_1}=\frac{12}{4}=3\). 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 \(y=6\)

- How are the perimeters of similar figures related to their scale factor? - First find the perimeter of the smaller triangle. - Compare the two perimeters, then apply that factor to each side.

1. The perimeter of \(\triangle ABC\) 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 \(z=20\)

- First find the missing hypotenuse segment. - The altitude is the geometric mean of the two hypotenuse segments.

1. The other hypotenuse segment is \(q=13-4=9\,\text{cm}\). 2. The similar triangles formed by the altitude give \(h^2=pq\). 3. Thus, \(h=\sqrt{4\cdot 9}=6\,\text{cm}\).

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

- Relate the given leg to the full hypotenuse and its adjacent segment. - After finding both hypotenuse segments, use their geometric mean to find the altitude. - Check that the two segments add to \(10\,\text{cm}\).

1. The similar triangles give \(a^2=cq\), so \(q=\frac{8^2}{10}=6.4\,\text{cm}\). 2. The other segment is \(p=10-6.4=3.6\,\text{cm}\). 3. The altitude satisfies \(h^2=pq\), so \(h=\sqrt{3.6\cdot 6.4}=4.8\,\text{cm}\).

\(q=6.4\,\text{cm}\) 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 area ratio before taking its square root. - How is the length scale factor related to the area ratio of similar figures?

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 corresponding side length is \(x=9\,\text{cm}\cdot\frac{5}{3}=15\,\text{cm}\).

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

- Write a proportion using corresponding sides of the similar triangles. - Cross-multiply to clear the fractions.

1. Similar triangles give \(\frac{AD}{AB}=\frac{AE}{AC}\). 2. Substitute the expressions: \(\frac{x}{x+2}=\frac{6}{10}\). 3. Cross-multiply: \(10x=6(x+2)\). 4. Solve: \(10x=6x+12\), so \(4x=12\) and \(x=3\).

\(x=3\)

- 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, \(x=10\cdot4=40\,\text{cm}^2\).

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

- Use the side-length ratio to find the area scale factor between the two similar triangles. - After finding the area of the whole triangle, subtract the area of the smaller triangle.

1. The linear scale factor from \(\triangle ADE\) to \(\triangle ABC\) is \(3\). 2. The area scale factor is \(3^2=9\), so the area of \(\triangle ABC\) is \(5\cdot9=45\,\text{cm}^2\). 3. Subtract the area of the smaller triangle: \(x=45-5=40\,\text{cm}^2\).

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

- How can you obtain a linear scale factor from an area ratio? - Perimeters scale by the same factor as corresponding side lengths.

1. The area ratio is \(\frac{27}{12}=\frac{9}{4}\). 2. The linear scale factor is the square root of the area ratio: \(k=\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5\). 3. Perimeters scale by the same linear factor, so \(y=1.5x\). 4. Use the perimeter sum: \(x+1.5x=45\), so \(2.5x=45\) and \(x=18\,\text{cm}\). 5. Then \(y=45-18=27\,\text{cm}\).

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

- 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, \(x=6\).

\(x=6\)

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

1. Corresponding sides give \(\frac{PQ}{QS}=\frac{QS}{RS}\). 2. Substitute the known values: \(\frac{8}{x}=\frac{x}{18}\). 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 one angle and have another pair of congruent angles? - Use their similarity to write a proportion for the full length \(PQ\).

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

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

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

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

- Use the marked angles to match corresponding vertices. - Find the scale factor from the pair of known corresponding sides. - Write a separate proportion for each unknown.

1. The angle information gives 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 \(\frac{5}{10}=0.5\). 3. Since \(MN\) corresponds to \(CA\), \(\frac{y}{12}=0.5\), so \(y=6\). 4. Since \(KM\) corresponds to \(BC\), \(\frac{x}{x+3}=0.5\). Solving gives \(x=3\).

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

- Identify the similar triangles on opposite sides of \(O\). - Combine the side ratio with \(x+y=20\). - Use the same similarity ratio for \(QO\) and \(ON\).

1. Since \(MQ\parallel LN\), \(\triangle MOQ\sim\triangle LON\). 2. Corresponding sides satisfy \(\frac{x}{y}=\frac{MO}{OL}=\frac{12}{8}=1.5\), so \(x=1.5y\). 3. Use \(x+y=20\): \(1.5y+y=20\), so \(2.5y=20\) and \(y=8\). 4. Therefore, \(x=12\). 5. Also, \(\frac{QO}{ON}=\frac{MO}{OL}=1.5\). Thus, \(\frac{15}{ON}=1.5\), 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

- 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

- Find the scale factor using \(AM\) and \(AD\). - Apply it 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

- Use the ratio of the parallel sides to find the similarity scale factor. - Express each full side as the sum of its two labeled parts. - Solve one proportion for each unknown.

1. Similar triangles give \(\frac{DE}{AC}=\frac{BD}{BA}\). 2. Substitute the labeled lengths: \(\frac{15}{25}=\frac{x}{x+8}\). 3. Solve: \(0.6(x+8)=x\), so \(0.4x=4.8\) and \(x=12\) units. 4. The same scale factor gives \(\frac{BE}{BC}=0.6\). Thus, \(\frac{y}{y+10}=0.6\). 5. Solving gives \(0.4y=6\), so \(y=15\) units.

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

- Which angles in the two triangles are congruent? - Use the Pythagorean theorem to find the hypotenuse of the large triangle. - Match corresponding sides before writing a proportion.

1. For part a), \(\angle ADC\) and \(\angle ACB\) are right angles, and the triangles share \(\angle A\). Therefore, \(\triangle ACD\sim\triangle ABC\) by AA. 2. Use the Pythagorean theorem in \(\triangle ABC\): \(AB=\sqrt{6^2+8^2}=10\,\text{cm}\). 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 each has a right angle. b) \(CD=4.8\,\text{cm}\)

- Look for triangles containing the given and unknown altitudes. - Which right angles are formed by the altitudes? - Which angle at \(C\) belongs to both triangles? - Match corresponding sides before writing a proportion.

1. For part a), triangles \(ADC\) and \(BEC\) each have a right angle and share the angle at \(C\). Therefore, \(\triangle ADC\sim\triangle BEC\) by AA. 2. Corresponding sides give \(\frac{AD}{AC}=\frac{BE}{BC}\). 3. Therefore, \(AD=\frac{6\cdot10}{15}=4\,\text{cm}\).

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 \(4\)

- 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 \(p=\frac{2.4^2}{1.8}=3.2\,\text{cm}\). 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 \(y=9.6\,\text{cm}\)

- 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 \(1\).

- How can you represent the two side lengths when their ratio is known? - What happens to area each time a rectangle is cut in half? - What side-length relationship must similar rectangles have? - How many A4 sheets have the same total area as one A0 sheet?

1. Let \(x\) be the shorter side in centimeters. Then the longer side is \(\sqrt{2}x\). Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), \(x(\sqrt{2}x)=10{,}000\). Thus, \(x^2=\frac{10000}{\sqrt{2}}\), so \(x\approx84.1\,\text{cm}\). The longer side is \(\sqrt{2}x\approx118.9\,\text{cm}\). 2. Each halving cuts the area in half. After four halvings, the area is \(\frac{1}{2^4}=\frac{1}{16}\) of the original area. Therefore, the A4 area is \(\frac{10000\,\text{cm}^2}{16}=625\,\text{cm}^2\). 3. Halving the longer side changes dimensions in the ratio \(\sqrt{2}:1\) to dimensions in the ratio \(1:\frac{\sqrt{2}}{2}\). Rotating the smaller rectangle gives a longer-side-to-shorter-side ratio of \(\frac{1}{\sqrt{2}/2}=\sqrt{2}\). The side-length ratio and all right angles are preserved, so each new sheet is similar to the previous one and A4 is similar to A0.

a) The side lengths are approximately \(84.1\,\text{cm}\) and \(118.9\,\text{cm}\). b) The area of an A4 sheet is \(625\,\text{cm}^2\). c) Each halving preserves the \(\sqrt{2}\)-to-\(1\) side-length ratio after the new rectangle is rotated, so A4 and A0 are similar.

- Compare the angles in the two triangles. - Identify the hypotenuse and the leg adjacent to \(\angle B\) in each triangle. - Write a proportion using a leg and the hypotenuse in both triangles.

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) \(q=3.6\,\text{cm}\)

- 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.

- An increase of \(125\%\) gives what new total percent? - Use the surface-area factor to find the linear factor. - Cube the linear factor to obtain the volume factor. - Distinguish the new total percent from the percent increase.

1. A surface-area increase of \(125\%\) means the new surface area is \(225\%\), or \(2.25\) times the original. Thus, \(k^2=2.25\). 2. The linear scale factor is \(k=\sqrt{2.25}=1.5\). 3. The volume factor is \(k^3=1.5^3=3.375\). 4. The new volume is \(337.5\%\) of the original, so the increase is \(237.5\%\).

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

- The diagonals form two similar triangles on opposite sides of \(S\). - Which altitudes correspond to the bases \(AB\) and \(CD\)? - How do the two partial heights combine to form the 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: \(H=h_1+h_2=6\,\text{cm}+4\,\text{cm}=10\,\text{cm}\).

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

- Use the Pythagorean theorem to find \(AB\). - How is the area ratio related to the length scale factor? - Which four segments form the perimeter of trapezoid \(ABED\)?

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 \(22\,\text{cm}\).

- Identify the parallelogram and use its congruent opposite sides. - Use the similar triangles \(RST\) and \(RPQ\). - 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 \(ST\parallel PQ\), \(\triangle RST\sim\triangle RPQ\). Thus, \(\frac{ST}{PQ}=\frac{RT}{RQ}=\frac{10}{15}=\frac{2}{3}\). 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 \(QU=6\,\text{cm}\)

- Identify the posts, cables, and crossing point. - Use the smaller similar triangles formed by each cable and the vertical through the crossing point. - Express the crossing height using each post height. - The final height does not depend on the distance between the posts.

1. 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: \(\frac{h}{4}+\frac{h}{12}=\frac{d-x+x}{d}=1\). 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.

- Which angles are congruent because \(AB\parallel FD\)? - To find area, you need a base and its perpendicular height. - A coordinate system can be used to find the height of \(S\) above \(AB\). - Write equations for lines \(AD\) and \(FB\).

1. Segments \(AB\) and \(FD\) are parallel. Also, \(A,S,D\) are collinear and \(F,S,B\) are collinear. Therefore, \(\angle SAB\cong\angle SDF\) and \(\angle SBA\cong\angle SFD\), so the triangles are similar by AA. 2. Place \(A=(0, 0)\), \(B=(a, 0)\), \(F=(0, a)\), and \(D=(2a, a)\). Line \(AD\) has equation \(y=\frac{1}{2}x\), and line \(FB\) has equation \(y=-x+a\). 3. At their intersection, \(\frac{1}{2}x=-x+a\), so \(x_S=\frac{2}{3}a\) and \(y_S=\frac{1}{3}a\). 4. Triangle \(ABS\) has base \(AB=a\) and height \(\frac{1}{3}a\). Thus, its area is \(\frac{1}{2}\cdot a\cdot\frac{1}{3}a=\frac{1}{6}a^2\).

a) Yes. The triangles are similar by AA. b) The area of triangle \(ABS\) is \(\frac{1}{6}a^2\).

- Use the ratio of the perimeters to find the scale factor between the triangles. - Apply that factor to \(c\) to find \(c_1\). - Split the remaining part of the smaller perimeter in the given 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. 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 \(c_1=10\,\text{cm}\)

- Why is each shaded triangle similar to the large triangle? - How does an area ratio relate to the corresponding linear scale factor? - Think about how corresponding side segments from the three smaller triangles combine to make one full side of the 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 \(S=10^2=100\,\text{cm}^2\).

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

- Square the side-length ratio to obtain the area ratio. - Express one area in terms of the other, then use the given difference.

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

- 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

- Compare the two right triangles formed by the altitude. - Use the given angle congruence to establish AA similarity.

1. Triangles \(QPS\) and \(PSR\) are right triangles. Since \(S\) lies on \(QR\), \(\angle PRQ=\angle PRS\). 2. The given angle congruence therefore gives \(\angle QPS\cong\angle PRS\), so \(\triangle QPS\sim\triangle PRS\) by AA. 3. Corresponding sides give \(\frac{QS}{PS}=\frac{PS}{SR}\). 4. Substitute: \(\frac{4}{x}=\frac{x}{9}\). Then \(x^2=36\), so \(x=6\).

\(x=6\)

- Identify the two similar triangles formed by the transversals and parallel lines. - The two perpendicular distances add to \(10\,\text{cm}\). - Relate the distances using the ratio of the two base lengths.

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. Corresponding heights and bases have the same ratio: \(\frac{d_g}{d_h}=\frac{12}{8}=1.5\). 4. Thus, \(d_g=1.5d_h\). Substitute into the sum: \(1.5d_h+d_h=10\). 5. Therefore, \(2.5d_h=10\), so \(d_h=4\,\text{cm}\).

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

- Find the full length \(BC\). - Identify two triangles with two pairs of congruent angles. - Use the similarity statement to match corresponding sides correctly.

1. The full side length is \(BC=BD+DC=3+9=12\,\text{cm}\). 2. Triangles \(ABD\) and \(CBA\) share the angle at \(B\), and it is given that \(\angle BAD\cong\angle BCA\). 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, \(AB=6\,\text{cm}\).

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

- Use the angle bisector and the given angle congruence to identify an isosceles triangle. - Find two triangles with two pairs of congruent angles. - Match corresponding sides carefully, and use \(AC=CD+DA\).

1. Since \(BD\) bisects \(\angle B\), \(\angle ABD\cong\angle CBD\). The given condition then implies \(\angle ABD\cong\angle A\), so \(\triangle ABD\) is isosceles and \(BD=AD=5\,\text{cm}\). 2. Triangles \(BCD\) and \(ACB\) share the angle at \(C\), and \(\angle CBD\cong\angle A\). 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 \(OD=12\,\text{cm}\)

- Which two triangles share the angle at \(M\)? - Use the given angle congruence to establish similarity. - Match corresponding sides before writing each proportion.

1. Triangles \(MNL\) and \(MKN\) share the angle at \(M\), and \(\angle MNL\cong\angle MKN\). 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 two triangles with a shared angle and the given pair of congruent angles. - Match corresponding sides before writing the proportions.

1. Triangles \(RKE\) and \(RTK\) share the angle at \(R\), and \(\angle RKE\cong\angle RTK\). 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 right angles to identify another pair of congruent acute angles. - Write a proportion for the unknown leg \(x\). - Use the Pythagorean theorem to find the hypotenuse \(y\).

1. The right angles at \(R\) and \(L\), together with the right angle between \(KO\) and \(OM\), 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 find the full hypotenuse \(KM\). - Use the similar triangles to express \(EF\), \(FM\), and \(NF\). - Combine \(EF=NF\) with the Pythagorean Theorem.

1. The hypotenuse is \(KM=40+30=70\). 2. Since \(EF\parallel KN\), \(\triangle EFM\sim\triangle KNM\), with scale factor \(\frac{EM}{KM}=\frac{30}{70}=\frac{3}{7}\). 3. Therefore, \(EF=\frac{3}{7}KN\) and \(FM=\frac{3}{7}NM\). Thus, \(NF=NM-FM=\frac{4}{7}NM\). 4. Since \(EF=NF\), \(\frac{3}{7}KN=\frac{4}{7}NM\), 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