How to find the area of a trapezoid - SSAT Middle Level Quantitative

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Question

What is the area of a trapezoid with height 20 inches and bases of length 100 and 200?

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Answer

Set , , .

The area of a trapezoid can be found using this formula:

$A = $\frac{1}{2}$ (B+b)h = $\frac{1}{2}$ (200+100) \cdot 20 = 3,000$

The area is 3,000 square inches.

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Question

A trapezoid has a height of inches and bases measuring inches and inches. What is its area?

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Answer

Use the following formula, with :

$A = $\frac{1}{2}$ (B+b)h = $\frac{1}{2}$ (36+24) \cdot 25 = 750$

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Question

What is the area of the trapezoid?

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Answer

To find the area of a trapezoid, multiply the sum of the bases (the parallel sides) by the height (the perpendicular distance between the bases), and then divide by 2.

$A = $\frac{1}{2}$ \cdot (7 + 15) \cdot 9 = $\frac{1}{2}$ \cdot 22 \cdot 9 = 99 \textrm{ $m}^2$$

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Question

What is the area of the above trapezoid?

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Answer

To find the area of a trapezoid, multiply one half (or 0.5, since we are working with decimals) by the sum of the lengths of its bases (the parallel sides) by its height (the perpendicular distance between the bases). This quantity is

$A = 0.5 \cdot (7.2 + 13.4) \cdot 10.6 =0.5 \cdot 20.6 \cdot 10.6 = 109.18\textrm{ $m}^{2}$$

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Question

The above diagram depicts a rectangle with isosceles triangle $\Delta ECM$ . If is the midpoint of $\overline{CT}$ , and the area of the orange region is , then what is the length of one leg of $\Delta ECM$ ?

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Answer

The length of a leg of $\Delta ECM$ is equal to the height of the orange region, which is a trapezoid. Call this length/height .

Since the triangle is isosceles, then , and since is the midpoint of $\overline{CT}$ , . Also, since opposite sides of a rectangle are congruent,

Therefore, the orange region is a trapezoid with bases and and height . Its area is 72, so we can set up and solve this equation using the area formula for a trapezoid:

$$\frac{1}{2}$ (B + b)h = 72$

$$\frac{1}{2}$ (2h + h)h = 72$

$$\frac{1}{2}$ (3h )h = 72$

$h = \sqrt {48}$

This is the length of one leg of the triangle.

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Question

The above diagram depicts a rectangle with isosceles triangle $\Delta ECM$ . is the midpoint of $\overline{CT}$ . What is the ratio of the area of the orange trapezoid to that of the white triangle?

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Answer

We can simplify this problem by supposing that the length of one leg of a triangle is 2. Then the other leg is 2, and the area of the triangle is

$A = \frac {1}{2} \cdot 2 \cdot 2 = 2$

Since is the midpoint of $\overline{CT}$ , . Also, since opposite sides of a rectangle are congruent,

.

This makes the trapezoid one with height 2 and bases 2 and 4, so

$A = \frac {1}{2} \cdot (2 + 4) \cdot 2 = 6$

The ratio of the area of the trapezoid to that of the triangle is 6 to 2, which simplifies to 3 to 1.

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Question

Find the area of the trapezoid above.

Note: Image not drawn to scale.

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Answer

The area of a trapezoid is equal to the average of the length of the two bases multiplied by the height.

The formula to find the area of a trapezoid is: $\left $($\frac{base_{1}$$ + $base_{2}$}{2}\right )* height$

In this problem, the lengths of the bases are and Their average is . The height of the trapezoid is

$15: cm\ast 8: cm=120: cm$

Remember: the answer to the problem should have units in cm2 .

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Question

Find the area of the trapezoid:

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Answer

The area of a trapezoid can be determined using the equation $A=\frac{1}{2}$(b_1+b_2)h$ .

$A=\frac{1}{2}$(6+8)(4)$

$A=\frac{1}{2}$(14)(4)$

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Question

Find the area of a trapezoid with a height of and base lengths of and , respectively.

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Answer

The area of a trapezoid is equal to the average of its two bases ( $b_{1}$ and $b_{2}$ ) multiplied by its height . Therefore:

$A=\frac{1}{2}(b_{1}+b_{2})h$

$A=\frac{1}{2}(4:cm+8:cm)5:cm$

$A=\(6:cm)5:cm$

$A=30:cm^{2}$

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Question

Find the area of a trapezoid with a height of and base lengths of and , respectively.

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Answer

The area of a trapezoid is equal to the average of its two bases ( $b_{1}$ and $b_{2}$ ) multiplied by its height . Therefore:

$A=\frac{1}{2}(b_{1}+b_{2})h$

$A=\frac{1}{2}(10:cm+12:cm)8:cm$

$A=\(11:cm)8:cm$

$A=88:cm^{2}$

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Question

What is the area of a trapezoid with height 20 inches and bases of length 100 and 200?

Tap to reveal answer

Answer

Set , , .

The area of a trapezoid can be found using this formula:

$A = $\frac{1}{2}$ (B+b)h = $\frac{1}{2}$ (200+100) \cdot 20 = 3,000$

The area is 3,000 square inches.

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Question

A trapezoid has a height of inches and bases measuring inches and inches. What is its area?

Tap to reveal answer

Answer

Use the following formula, with :

$A = $\frac{1}{2}$ (B+b)h = $\frac{1}{2}$ (36+24) \cdot 25 = 750$

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Question

What is the area of the trapezoid?

Tap to reveal answer

Answer

To find the area of a trapezoid, multiply the sum of the bases (the parallel sides) by the height (the perpendicular distance between the bases), and then divide by 2.

$A = $\frac{1}{2}$ \cdot (7 + 15) \cdot 9 = $\frac{1}{2}$ \cdot 22 \cdot 9 = 99 \textrm{ $m}^2$$

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Question

What is the area of the above trapezoid?

Tap to reveal answer

Answer

To find the area of a trapezoid, multiply one half (or 0.5, since we are working with decimals) by the sum of the lengths of its bases (the parallel sides) by its height (the perpendicular distance between the bases). This quantity is

$A = 0.5 \cdot (7.2 + 13.4) \cdot 10.6 =0.5 \cdot 20.6 \cdot 10.6 = 109.18\textrm{ $m}^{2}$$

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Question

The above diagram depicts a rectangle with isosceles triangle $\Delta ECM$ . If is the midpoint of $\overline{CT}$ , and the area of the orange region is , then what is the length of one leg of $\Delta ECM$ ?

Tap to reveal answer

Answer

The length of a leg of $\Delta ECM$ is equal to the height of the orange region, which is a trapezoid. Call this length/height .

Since the triangle is isosceles, then , and since is the midpoint of $\overline{CT}$ , . Also, since opposite sides of a rectangle are congruent,

Therefore, the orange region is a trapezoid with bases and and height . Its area is 72, so we can set up and solve this equation using the area formula for a trapezoid:

$$\frac{1}{2}$ (B + b)h = 72$

$$\frac{1}{2}$ (2h + h)h = 72$

$$\frac{1}{2}$ (3h )h = 72$

$h = \sqrt {48}$

This is the length of one leg of the triangle.

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Question

The above diagram depicts a rectangle with isosceles triangle $\Delta ECM$ . is the midpoint of $\overline{CT}$ . What is the ratio of the area of the orange trapezoid to that of the white triangle?

Tap to reveal answer

Answer

We can simplify this problem by supposing that the length of one leg of a triangle is 2. Then the other leg is 2, and the area of the triangle is

$A = \frac {1}{2} \cdot 2 \cdot 2 = 2$

Since is the midpoint of $\overline{CT}$ , . Also, since opposite sides of a rectangle are congruent,

.

This makes the trapezoid one with height 2 and bases 2 and 4, so

$A = \frac {1}{2} \cdot (2 + 4) \cdot 2 = 6$

The ratio of the area of the trapezoid to that of the triangle is 6 to 2, which simplifies to 3 to 1.

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Question

Find the area of the trapezoid above.

Note: Image not drawn to scale.

Tap to reveal answer

Answer

The area of a trapezoid is equal to the average of the length of the two bases multiplied by the height.

The formula to find the area of a trapezoid is: $\left $($\frac{base_{1}$$ + $base_{2}$}{2}\right )* height$

In this problem, the lengths of the bases are and Their average is . The height of the trapezoid is

$15: cm\ast 8: cm=120: cm$

Remember: the answer to the problem should have units in cm2 .

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Question

Find the area of the trapezoid:

Tap to reveal answer

Answer

The area of a trapezoid can be determined using the equation $A=\frac{1}{2}$(b_1+b_2)h$ .

$A=\frac{1}{2}$(6+8)(4)$

$A=\frac{1}{2}$(14)(4)$

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Question

Find the area of a trapezoid with a height of and base lengths of and , respectively.

Tap to reveal answer

Answer

The area of a trapezoid is equal to the average of its two bases ( $b_{1}$ and $b_{2}$ ) multiplied by its height . Therefore:

$A=\frac{1}{2}(b_{1}+b_{2})h$

$A=\frac{1}{2}(4:cm+8:cm)5:cm$

$A=\(6:cm)5:cm$

$A=30:cm^{2}$

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Question

Find the area of a trapezoid with a height of and base lengths of and , respectively.

Tap to reveal answer

Answer

The area of a trapezoid is equal to the average of its two bases ( $b_{1}$ and $b_{2}$ ) multiplied by its height . Therefore:

$A=\frac{1}{2}(b_{1}+b_{2})h$

$A=\frac{1}{2}(10:cm+12:cm)8:cm$

$A=\(11:cm)8:cm$

$A=88:cm^{2}$

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