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

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

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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.

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

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Use the following formula, with :

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

What is the area of the trapezoid?

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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$$

What is the area of the above trapezoid?

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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}$$

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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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.

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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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.

Find the area of the trapezoid above.

Note: Image not drawn to scale.

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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 .

Find the area of the trapezoid:

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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)$

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

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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$

$A=30:cm^{2}$

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

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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$

$A=88:cm^{2}$

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

Tap to see back →

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.

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

Tap to see back →

Use the following formula, with :

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

What is the area of the trapezoid?

Tap to see back →

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$$

What is the area of the above trapezoid?

Tap to see back →

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}$$

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 see back →

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.

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 see back →

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.

Find the area of the trapezoid above.

Note: Image not drawn to scale.

Tap to see back →

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 .

Find the area of the trapezoid:

Tap to see back →

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)$

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

Tap to see back →

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$

$A=30:cm^{2}$

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

Tap to see back →

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$

$A=88:cm^{2}$