- Which two measurements in the diagram are needed to find a triangle's area? - How is a triangle's area related to a rectangle with the same base and height? - Check that you are using a perpendicular height.

1. The diagram shows a base of \(15\,\text{cm}\) and a perpendicular height of \(10\,\text{cm}\). 2. Use the triangle area formula: \(A=\frac{1}{2}bh\). 3. Substitute the measurements: \(A=\frac{1}{2}\times15\times10=75\,\text{cm}^2\).

The area is \(75\,\text{cm}^2\).

- What is the area of the whole rectangle? - Does a diagonal of a rectangle make two equal-area parts? - Use one of those parts to find the triangle's area.

1. The rectangle has area \(6\times4=24\,\text{cm}^2\). 2. A diagonal cuts a rectangle into two equal-area triangles, so triangle \(ABC\) has area \(24\div2=12\,\text{cm}^2\).

\(12\,\text{cm}^2\)

- How should a height meet the chosen base? - Look for the right-angle mark, not just a side of the triangle.

1. The height must meet the line containing \(AB\) at a right angle. 2. The dashed segment \(CD\) is perpendicular to the base, so the height is \(4\,\text{cm}\). The slanted sides are not heights for this base.

The dashed segment \(CD\), with length \(4\,\text{cm}\).

- Must a height touch the base segment itself, or can it touch its extended line? - Which right angle shows the perpendicular distance from the opposite vertex?

1. The base \(AB\) lies on the same straight line as \(BF\). 2. Segment \(CF\) meets that line at a right angle. It is the height to \(AB\), even though it lies outside the triangle.

Yes. \(CF\) is perpendicular to the line containing \(AB\), so it is a height to that base.

- Does area measure a length or the amount of surface covered? - Compare the number Mia wrote with half of the rectangle's area.

1. The rectangle's area is \(4\times3=12\,\text{cm}^2\), and half of it is \(6\,\text{cm}^2\). 2. Mia has the correct number but used a length unit instead of a square unit.

Only the unit is wrong. The area is \(6\,\text{cm}^2\).

- Identify which measurements determine the area of a triangle. - Compare what changes and what stays fixed as \(C\) moves. - Think about how a line parallel to the base affects the perpendicular distance to the base.

1. The area is \(A=\frac{1}{2}bh=\frac{1}{2}\times10\times4=20\,\text{cm}^2\). 2. Every point on a line parallel to \(AB\) is the same perpendicular distance from \(AB\). 3. Therefore, both the base \(AB\) and the height remain constant, so the area remains \(20\,\text{cm}^2\).

a) \(20\,\text{cm}^2\) b) The height stays constant because the moving point remains on a line parallel to the base. Since the base also stays fixed, the area does not change.

- Which measurements appear in the standard triangle area formula? - Imagine changing the other two side lengths while keeping the base and perpendicular height fixed. What happens to the area?

1. Use the \(4\,\text{m}\) side as the base: \(A=\frac{1}{2}\times4\times2.2=4.4\,\text{m}^2\). 2. A triangle's area is determined by a chosen base and the perpendicular height to that base. The other two sides affect the triangle's shape but do not appear in this area calculation.

1. \(4.4\,\text{m}^2\) 2. The area formula uses the base and its perpendicular height, so the other two side lengths are not needed.

- How could a rectangle with the same base and height help you find the triangle's area? - Think about which measurements determine a triangle's area. - Try sketching two differently shaped triangles with the same base and perpendicular height.

1. The area is \(A=\frac{1}{2}\times10\times4=20\,\text{cm}^2\). 2. Moving \(C\) sideways without changing its perpendicular distance from \(AB\) leaves the base and height unchanged. 3. The area remains \(20\,\text{cm}^2\), even though the triangle's shape changes.

a) \(20\,\text{cm}^2\) b) No. The base and its perpendicular height stay the same, so the area stays \(20\,\text{cm}^2\).

- Read the coordinates of the three triangle vertices. - Which sides are perpendicular, and how long are they? - How could you compare the triangle with a rectangle made from those two lengths?

1. Read the points \(N=(2,0)\) and \(T=(0,1)\) from the graph. The triangle's third vertex is \(O=(0,0)\). 2. The two sides along the axes have lengths \(2\) and \(1\), and they meet at a right angle. 3. Use the triangle area formula: \(A=\frac{1}{2}\cdot2\cdot1=1\) square unit.

\(1\) square unit.

- Use the triangle area formula with a base and its corresponding height. - Count the spaces between pegs to determine lengths. - For triangle a), the two sides that form the right angle can be used as the base and height.

1. Triangle a) is a right triangle with perpendicular side lengths \(4\) units and \(5\) units. Its area is \(A_a=\frac{1}{2}\times4\times5=10\) square units. 2. Triangle b) has a base of \(5\) units and a height of \(4\) units. Its area is \(A_b=\frac{1}{2}\times5\times4=10\) square units. 3. The areas are equal, so neither triangle covers more area.

Both triangles have an area of \(10\) square units, so neither is larger.

- Use the grid to identify a base and its corresponding perpendicular height. - Substitute those measurements into the triangle area relationship. - Convert your final area to a decimal only after you have found it exactly.

1. The horizontal base is \(5\,\text{cm}\) long. 2. The corresponding height is \(3\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\times5\times3=7.5\,\text{cm}^2\).

The area is \(7.5\,\text{cm}^2\).

- Imagine adding a matching right triangle to complete a rectangle. - How are the two triangles related inside that rectangle? - Use the rectangle's dimensions after explaining the one-half relationship.

1. Complete the triangle to a \(7\,\text{cm}\times3\,\text{cm}\) rectangle. 2. A diagonal divides that rectangle into two congruent right triangles, so the shown triangle is exactly half of the rectangle. 3. The rectangle area is \(7\times3=21\,\text{cm}^2\). 4. The triangle area is \(21\div2=10.5\,\text{cm}^2\).

The triangle is half of a \(7\,\text{cm}\times3\,\text{cm}\) rectangle, so its area is \(10.5\,\text{cm}^2\).

- How many congruent right triangles make the completed rectangle? - Read the rectangle's dimensions from the diagram and find its area first. - Divide that area by the number of congruent triangles.

1. The completed rectangle has area \(8\times5=40\,\text{cm}^2\). 2. The right triangle is half of that rectangle, so its area is \(40\div2=20\,\text{cm}^2\).

\(20\,\text{cm}^2\)

- Read the base length from the diagram. - Substitute the known area and base into the triangle area relationship. - Rearrange the equation so the unknown height is alone.

1. Use \(A=\frac{1}{2}bh\): \(30=\frac{1}{2}\times12\times h=6h\). 2. Divide by \(6\): \(h=30\div6=5\,\text{cm}\).

The height is \(5\,\text{cm}\).

- Which side is shown as the base? - Recall the triangle area formula. - Use the height that is perpendicular to the chosen base.

1. The base is \(8\,\text{cm}\), and its corresponding height is \(4\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\times8\times4=16\,\text{cm}^2\).

The area of the triangle is \(16\,\text{cm}^2\).

- What does it mean for \(M\) to be the midpoint of \(\overline{BC}\)? - Compare the bases \(BM\) and \(MC\) and the heights of the two smaller triangles. - How can the two smaller triangle areas be combined to get the whole triangle?

1. Because \(M\) is the midpoint of \(\overline{BC}\), triangles \(ABM\) and \(ACM\) have equal base lengths and the same height from \(A\) to line \(BC\). 2. Therefore, the two smaller triangles have equal areas. 3. The total area is \(18+18=36\,\text{cm}^2\).

Triangle \(ABC\) has area \(36\,\text{cm}^2\).

- Compare the triangle's bottom edge with the rectangle's bottom edge. - Where does the triangle's top corner lie? - Think about making a second triangle of equal area.

1. The triangle and rectangle have the same base and perpendicular height. 2. The triangle therefore has half the area of the rectangle. 3. The rectangle's area is \(2\times14=28\,\text{cm}^2\).

\(28\,\text{cm}^2\)

- What does the height in the triangle area formula mean? - Does every side of a triangle work as the height to a chosen base? - Check Jay's calculation against the measurements in the drawing.

1. The \(5\,\text{ft}\) side slopes toward the bottom corner and is not perpendicular to the \(8\,\text{ft}\) base. 2. The dashed \(3\,\text{ft}\) segment meets that base at a right angle, so the correct area is \(\frac{1}{2}\times8\times3=12\,\text{ft}^2\).

No. He used a slanted side instead of the perpendicular height. The area is \(12\,\text{ft}^2\).

- Compare the two bases before calculating either area. - What part of the triangle-area formula is identical for both triangles? - After finding the area ratio, use it with one triangle area.

1. Triangles \(ABC\) and \(ADC\) use bases on the same line \(AB\), so they have the same perpendicular height. 2. With the same height, the ratio of their areas equals the ratio of their bases: \(\frac{A_{ADC}}{A_{ABC}}=\frac{AD}{AB}=\frac{3}{12}=\frac{1}{4}\). 3. The area of triangle \(ABC\) is \(\frac{1}{2}\times12\times5=30\,\text{cm}^2\). 4. Therefore, \(A_{ADC}=\frac{1}{4}\times30=7.5\,\text{cm}^2\).

a) \(\frac{A_{ADC}}{A_{ABC}}=\frac{1}{4}\) b) \(A_{ADC}=7.5\,\text{cm}^2\)

- First find the area shared by both triangles. - How long is the second triangle's base? - Use the triangle area formula to find the missing height. - Compare the two base-height pairs.

1. The first triangle has area \(A=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. The second triangle's base is \(10\div2=5\,\text{cm}\). 3. Set its area equal to \(30\): \(30=\frac{1}{2}\times5\times h_2=2.5h_2\). Thus, \(h_2=30\div2.5=12\,\text{cm}\). 4. When the base is halved, the height must double to keep the product of base and height constant.

a) \(30\,\text{cm}^2\) b) \(12\,\text{cm}\) c) The height must double.

- How are the perimeter and the three side lengths related? - What is the area formula for a triangle? - Which side should be used as the base to find the requested height?

1. The third side has length \(32-10-10=12\,\text{cm}\). 2. Let \(h\) be the height to the \(12\,\text{cm}\) side. From \(48=\frac{1}{2}\times12\times h\), \(48=6h\). 3. Therefore, \(h=48\div6=8\,\text{cm}\).

The third side is \(12\,\text{cm}\) long, and its corresponding height is \(8\,\text{cm}\).

- First express the measurements in compatible units. - Equal areas can be represented by an equation. - Find the first sail's area before solving for the missing height. - Convert the final height to centimeters.

1. Convert the first sail's height to meters: \(80\,\text{cm}=0.8\,\text{m}\). 2. The first sail has area \(A=\frac{1}{2}\times1.2\times0.8=0.48\,\text{m}^2\). 3. For the second sail, \(0.48=\frac{1}{2}\times1.5\times h=0.75h\). 4. Therefore, \(h=0.48\div0.75=0.64\,\text{m}=64\,\text{cm}\).

The second sail's height is \(64\,\text{cm}\).

- A triangle's area can be found using any side and its corresponding height. - The area is the same no matter which base-height pair you use. - Once two side lengths are known, use the perimeter to find the third.

1. Use the \(10\,\text{cm}\) side and its corresponding height to find the area: \(A=\frac{1}{2}\times10\times4.8=24\,\text{cm}^2\). 2. Let \(x\) be the side whose corresponding height is \(8\,\text{cm}\). Then \(24=\frac{1}{2}\times x\times8=4x\), so \(x=6\,\text{cm}\). 3. Let \(y\) be the remaining side. Use the perimeter: \(y=24-6-10=8\,\text{cm}\).

The second side is \(6\,\text{cm}\), and the remaining side is \(8\,\text{cm}\).

- What happens to a product if one factor doubles but the product must stay the same? - For part b), relate the original and new base-height pairs. - For part c), keep the factor of one-half in the triangle formula and compare the two expressions.

1. The original area is \(A=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. Keeping that area with base \(20\,\text{cm}\) gives \(30=\frac{1}{2}\times20\times h\), so \(h=3\,\text{cm}\). The height is halved. 3. With an original base \(b\) and height \(h\), the area after doubling both is \(\frac{1}{2}(2b)(2h)=4(\frac{1}{2}bh)\). The claim is always true.

a) \(30\,\text{cm}^2\) b) Halve the height to \(3\,\text{cm}\). c) Yes. \(\frac{1}{2}(2b)(2h)=4(\frac{1}{2}bh)\), so its area is four times as large.

- Find the area of the square first. - Equal-area figures have the same numerical area. - Substitute the known base and area into the triangle area formula. - Solve the resulting equation for the height.

1. The square's area is \(12\times12=144\,\text{m}^2\). 2. The triangle must also have area \(144\,\text{m}^2\), so \(144=\frac{1}{2}\times16\times h\). 3. Simplify to \(144=8h\), then divide: \(h=144\div8=18\,\text{m}\).

The triangle must have a height of \(18\,\text{m}\).

- Use the triangle area formula. - Find Leo's base and height before calculating his area. - What happens when one factor doubles and the other is halved? - In part b), substitute the target area and solve for the height.

1. Mia's triangle has area \(A_M=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. Leo's second triangle has base \(20\,\text{cm}\) and height \(3\,\text{cm}\), so \(A_L=\frac{1}{2}\times20\times3=30\,\text{cm}^2\). The areas are equal. 3. The third triangle needs area \(4\times30=120\,\text{cm}^2\). 4. Solve \(120=\frac{1}{2}\times20\times h=10h\). Thus, \(h=12\,\text{cm}\).

a) Both areas are \(30\,\text{cm}^2\). b) The height must be \(12\,\text{cm}\).

- Find the area of the triangular sail first. - The replacement sail must have the same area. - Divide the rectangle's area by its known length.

1. The triangular sail has area \(\frac{1}{2}\times5.40\times3.50=9.45\,\text{m}^2\). 2. The rectangle must have the same area, so its width is \(9.45\div4.20=2.25\,\text{m}\).

The rectangular sail must be \(2.25\,\text{m}\) wide.

- Compare the area formulas for a triangle and a rectangle. - One side length is the same in both figures. - Picture two congruent copies of the triangle forming a rectangle.

1. A triangle has half the area of a rectangle with the same base and height. 2. Since both figures have a \(12\,\text{m}\) side, the rectangle's other side must be half of \(18\,\text{m}\) to have the same area. 3. Therefore, the missing side length is \(18\div2=9\,\text{m}\).

The rectangle's other side is \(9\,\text{m}\). The factor \(\frac{1}{2}\) in the triangle area formula means the corresponding rectangle side must be half as long when the shared side is unchanged.

- Find the area of the original triangular stage. - Apply the required change to the area before working with the rectangle. - Divide the new area by the rectangle's known length.

1. The original stage has area \(\frac{1}{2}\times15\times6.40=48\,\text{m}^2\). 2. Doubling the area gives \(2\times48=96\,\text{m}^2\). 3. The rectangle's width is \(96\div12=8\,\text{m}\).

The new rectangular stage must be \(8\,\text{m}\) wide.

- What does the right angle tell you about the two legs and their corresponding heights? - Use the area you found to write a second area equation with the hypotenuse as the base. - Rearrange the triangle area formula to solve for a height.

1. The legs are perpendicular. Therefore, the height corresponding to the \(5\,\text{cm}\) leg is \(12\,\text{cm}\), and the height corresponding to the \(12\,\text{cm}\) leg is \(5\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\times5\times12=30\,\text{cm}^2\). 3. Let \(h\) be the height drawn to the hypotenuse. Using the hypotenuse as the base, \(30=\frac{1}{2}\times13\times h\). Thus, \(h=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

a) The two heights are \(12\,\text{cm}\) and \(5\,\text{cm}\), respectively. b) \(30\,\text{cm}^2\) c) \(\approx4.62\,\text{cm}\)

- Use the triangle-area relationship to recover each missing measurement. - For the general claim, represent the original base and height by letters. - Compare the original area expression with the expression after both changes.

1. For triangle A, \(24=\frac{1}{2}\times8\times h=4h\), so \(h=6\,\text{cm}\). 2. For triangle B, \(24=\frac{1}{2}\times b\times4=2b\), so \(b=12\,\text{cm}\). 3. A triangle with base \(b\) and corresponding height \(h\) has area \(A=\frac{1}{2}bh\). 4. After doubling the base and halving the height, the area is \(\frac{1}{2}(2b)\left(\frac{h}{2}\right)=\frac{1}{2}bh=A\). 5. Therefore, Putri's claim is always true.

a) \(6\,\text{cm}\) b) \(12\,\text{cm}\) c) Putri's claim is always true because \(\frac{1}{2}(2b)\left(\frac{h}{2}\right)=\frac{1}{2}bh\).

- What happens to a product when one factor is divided by \(2\) and the other is multiplied by \(2\)? - Find the new target area and new base separately. - Write an equation for the unknown height.

1. The original area is \(A=\frac{1}{2}\times12\times4=24\,\text{cm}^2\). 2. The changed measurements are a \(6\,\text{cm}\) base and an \(8\,\text{cm}\) height. The area is \(\frac{1}{2}\times6\times8=24\,\text{cm}^2\), so the statement is true. 3. The new base in part c) is \(12\div3=4\,\text{cm}\), and the target area is \(2\times24=48\,\text{cm}^2\). 4. Solve \(48=\frac{1}{2}\times4\times h=2h\). Thus, \(h=24\,\text{cm}\).

a) \(24\,\text{cm}^2\) b) The statement is true. c) \(24\,\text{cm}\)

- Rearrange the area formula to find a missing height. - Find two positive numbers whose product is twice the area. - Imagine moving the third vertex left or right without changing its distance from the base.

1. Use \(A=\frac{1}{2}bh\): \(15=\frac{1}{2}\times6\times h=3h\), so \(h=5\,\text{cm}\). 2. Any positive base-height pair with product \(30\,\text{cm}^2\) works. For example, \(b=10\,\text{cm}\) and \(h=3\,\text{cm}\). 3. The triangles do not have to have the same shape. The third vertex can move along a line parallel to the base, changing side lengths and angles while keeping the height and area fixed.

a) \(5\,\text{cm}\) b) One possible pair is \(10\,\text{cm}\) and \(3\,\text{cm}\). c) No. The third vertex can move parallel to the base without changing the height or area.

- How can the perpendicular legs be used as a base and height? - Any side can be a base if you use its corresponding height. - What happens to a product when one factor doubles?

1. The legs are perpendicular, so \(A=\frac{1}{2}\times6\times8=24\,\text{cm}^2\). 2. Let \(h\) be the height corresponding to the hypotenuse. Then \(24=\frac{1}{2}\times10\times h=5h\), so \(h=4.8\,\text{cm}\). 3. If the \(6\,\text{cm}\) leg doubles while the other leg stays fixed, one factor in \(\frac{1}{2}bh\) doubles. Therefore, the area doubles to \(48\,\text{cm}^2\).

a) \(24\,\text{cm}^2\) b) \(4.8\,\text{cm}\) c) The area doubles to \(48\,\text{cm}^2\).

- Find each triangle's area separately. - Identify which rectangle side is halved in each case. - Compare the two base-height products.

1. In Rectangle A, the triangle has base \(5\,\text{cm}\) and height \(6\,\text{cm}\). Its area is \(\frac{1}{2}\times5\times6=15\,\text{cm}^2\). 2. In Rectangle B, the triangle has base \(3\,\text{cm}\) and height \(10\,\text{cm}\). Its area is \(\frac{1}{2}\times3\times10=15\,\text{cm}^2\). 3. The triangles have equal areas because the base-height products are equal: \(5\times6=3\times10=30\).

Both triangles have area \(15\,\text{cm}^2\). Their shapes differ, but their base-height products are equal.

- First use the triangle area formula to find the missing height. - How does halving or tripling one factor affect a product? - You may calculate directly or use proportional reasoning.

1. Solve \(24=\frac{1}{2}\times8\times h=4h\), giving \(h=6\,\text{cm}\). 2. With base \(4\,\text{cm}\), the area is \(\frac{1}{2}\times4\times6=12\,\text{cm}^2\). 3. With height \(18\,\text{cm}\), the area is \(\frac{1}{2}\times8\times18=72\,\text{cm}^2\).

a) \(6\,\text{cm}\) b) \(12\,\text{cm}^2\) c) \(72\,\text{cm}^2\)

- Which base already has a known corresponding height? - How are base, height, and area related? - Can one fixed triangle have two different areas?

1. The area is \(A=\frac{1}{2}\times14\times5=35\,\text{cm}^2\). 2. Let \(h\) be the height corresponding to the \(10\,\text{cm}\) side. Then \(35=\frac{1}{2}\times10\times h=5h\). Thus, \(h=7\,\text{cm}\). 3. The triangle itself does not change when a different side is chosen as the base. Each base must be paired with its own perpendicular height, and every valid pair gives the same enclosed area.

a) \(35\,\text{cm}^2\) b) \(7\,\text{cm}\) c) Choosing a different base does not change the triangle; the corresponding height adjusts so the area remains the same.

- Recall the formula for the area of a triangle. - Substitute the known area and height, then simplify the coefficient of \(b\). - Divide both sides by the coefficient of \(b\).

1. Use the triangle area formula: \(A=\frac{1}{2}bh\). 2. Substitute the given values: \(18.6=\frac{1}{2}\times b\times4\). 3. Simplify: \(18.6=2b\). 4. Divide by \(2\): \(b=9.3\). 5. Check: \(\frac{1}{2}\times9.3\times4=18.6\).

The base is \(9.3\,\text{in.}\) long.

- Find a rectangle that exactly encloses the triangle. - Calculate the areas outside the triangle but inside the rectangle. - Subtract those areas from the rectangle area.

1. Enclose the triangle in a \(6 \times 4\) rectangle with area \(24\) square units. 2. The left outside triangle has area \(\frac{1}{2} \times 2 \times 4=4\) square units. The right outside triangle has area \(\frac{1}{2} \times 4 \times 4=8\) square units. 3. The triangle area is \(24-4-8=12\) square units.

\(12\) square units

- Use the two parallel-side labels to determine the triangle's horizontal base. - The dashed segment shows that the triangle has the same perpendicular height as the trapezoid. - Apply the triangle area relationship to those two measurements.

1. The triangle's base is \(100-70=30\,\text{cm}\). 2. Its height is the trapezoid's height, \(40\,\text{cm}\). 3. The area is \(\frac{1}{2}\times30\times40=600\,\text{cm}^2\).

The triangle has area \(600\,\text{cm}^2\).

- Which segment in the diagram is perpendicular to the line containing the base? - The height of an obtuse triangle can lie outside the triangle. - Which two diagram measurements determine the area?

1. The diagram shows base \(QR=5\,\text{cm}\) and a perpendicular height of \(6\,\text{cm}\) drawn to the line containing the base. 2. The area formula still applies: \(A=\frac{1}{2}\times5\times6=15\,\text{cm}^2\).

The area of the triangle is \(15\,\text{cm}^2\).

- Count the grid cells along the horizontal side. - Count the perpendicular distance from the opposite vertex to the base line. - Use the triangle area formula.

1. The horizontal base is \(3\,\text{cm}\) long. 2. The perpendicular height is \(3\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\times3\times3=4.5\,\text{cm}^2\).

The area of the triangle is \(4.5\,\text{cm}^2\).

- Find \(DC\) from the total length \(BC\). - Do the two triangles have the same height to line \(BC\)?

1. Find the remaining segment: \(DC=10-3=7\,\text{cm}\). 2. The two triangles have the same perpendicular height to line \(BC\), so their areas are proportional to their bases. 3. Therefore, the ratio is \(\frac{BD}{DC}=\frac{3}{7}\).

The ratio of the area of \(\triangle ABD\) to the area of \(\triangle ADC\) is \(\frac{3}{7}\).

- Compare the two triangular pieces inside the parallelogram. - What fraction of the parallelogram belongs to one triangle? - Express the parallelogram's area using its labeled base and perpendicular height before taking that fraction.

1. The diagonal separates the parallelogram into the two congruent triangle copies, so the two triangles have equal area and together make the entire parallelogram. 2. A parallelogram with base \(b\) and perpendicular height \(h\) has area \(bh\). 3. Because each congruent triangle is one half of the parallelogram, one triangle has area \(\frac{1}{2}bh\). 4. Any triangle can be copied and paired in this way, so the argument gives the general triangle-area relationship.

The two congruent triangles exactly make a parallelogram of area \(bh\). Each triangle is half of that area, so \(A=\frac{1}{2}bh\).

- First calculate the height needed to produce the required area. - Compare a triangle's height with the sides that connect the opposite vertex to the base. - Can an adjacent side be shorter than the perpendicular distance to the base line?

1. Let \(h\) be the height to the \(8\,\text{cm}\) base. From \(A=\frac{1}{2}bh\), \(20=\frac{1}{2}\times8\times h\). 2. This gives \(20=4h\), so \(h=5\,\text{cm}\). 3. The perpendicular distance from the opposite vertex to the base line cannot be longer than a segment from that vertex to an endpoint of the base. 4. Because the required height is \(5\,\text{cm}\) but the specified adjacent side is only \(4\,\text{cm}\), the triangle is impossible.

The required height is \(5\,\text{cm}\). Since an adjacent side is only \(4\,\text{cm}\), it cannot reach a point that is \(5\,\text{cm}\) from the base line. Therefore, the triangle cannot be constructed.

- Equal triangle areas impose a condition on the products of their bases and corresponding heights. - The whole-number condition suggests organizing possible factor pairs rather than guessing randomly. - Use the stated difference between base and height to select among the possible pairs.

1. Triangle A has area \(\frac{1}{2}\times14\times6=42\,\text{cm}^2\). 2. For triangle B to have the same area, its base-height product must be \(84\), because \(\frac{1}{2}bh=42\). 3. Check whole-number factor pairs of \(84\): \(1\times84\), \(2\times42\), \(3\times28\), \(4\times21\), \(6\times14\), and \(7\times12\). 4. Only \(7\) and \(12\) differ by \(5\). Since the base is longer, the height is \(7\,\text{cm}\) and the base is \(12\,\text{cm}\).

Base: \(12\,\text{cm}\) Height: \(7\,\text{cm}\)