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SSAT Middle Level Math : Trapezoids

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Example Questions

Example Question #2 : How To Find The Area Of A Trapezoid

Find the area of the trapezoid above.

Note: Image not drawn to scale.

Possible Answers:

$80\: cm$

$200\: cm^{2}$

$120\: cm^{2}$

$160\: cm$

$120\: cm$

Correct answer:

$120\: cm^{2}$

Explanation:

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 $10\: cm$ and $20\: cm.$ Their average is $15\: cm$ . The height of the trapezoid is $8\: cm.$

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

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

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Example Question #3 : How To Find The Area Of A Trapezoid

Find the area of a trapezoid with a height of $5\:cm$ and base lengths of $4\:cm$ and $8\:cm$ , respectively.

Possible Answers:

$17\:cm$

$15\:cm^{2}$

$22\:cm^{2}$

$30\:cm^{2}$

$10\:cm^{2}$

Correct answer:

$30\:cm^{2}$

Explanation:

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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Example Question #3 : How To Find The Area Of A Trapezoid

Find the area of a trapezoid with a height of $8\:cm$ and base lengths of $10\:cm$ and $12\:cm$ , respectively.

Possible Answers:

$30\:cm$

$30\:cm^{2}$

$88\:cm$

$52\:cm^{2}$

$88\:cm^{2}$

Correct answer:

$88\:cm^{2}$

Explanation:

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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Example Question #1 : How To Find The Area Of A Trapezoid

Trapezoid

What is the area of the above trapezoid?

Possible Answers:

$96.48\textrm{ m}^{2}$

$218.36\textrm{ m}^{2}$

$109.18\textrm{ m}^{2}$

$142.04\textrm{ m}^{2}$

$76.32\textrm{ m}^{2}$

Correct answer:

$109.18\textrm{ m}^{2}$

Explanation:

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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Example Question #131 : Quadrilaterals

Find the area of the trapezoid:

Question_7

Possible Answers:

Correct answer:

Explanation:

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

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

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

A=(7)(4)=28

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Example Question #132 : Quadrilaterals

Trapezoid

What is the area of the trapezoid?

Possible Answers:

$198\textrm{ m}^{2}$

$63\textrm{ m}^{2}$

$99\textrm{ m}^{2}$

$135\textrm{ m}^{2}$

$105\textrm{ m}^{2}$

Correct answer:

$99\textrm{ m}^{2}$

Explanation:

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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Example Question #133 : Quadrilaterals

Trapezoid

The above diagram depicts a rectangle RECT 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 $\Delta ECM$ ?

Possible Answers:

$\sqrt {96}$

$\sqrt {108}$

$\sqrt {48}$

$\sqrt {54}$

Correct answer:

$\sqrt {48}$

Explanation:

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 CM = h , and since is the midpoint of $\overline{CT}$ , MT = h . Also, since opposite sides of a rectangle are congruent,

RE = CT = CM + MT = h + h = 2h

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:

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

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

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

$\frac{3}{2}h^{2} = 72$

$\frac{3}{2}h^{2} \cdot \frac{2}{3} = 72 \cdot \frac{2}{3}$

$h^{2}= 48$

$h = \sqrt {48}$

This is the length of one leg of the triangle.

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Example Question #134 : Quadrilaterals

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

Possible Answers:

$750\; \textrm{in}^{2}$

$864\; \textrm{in}^{2}$

$1,500\; \textrm{in}^{2}$

$900\; \textrm{in}^{2}$

$600\; \textrm{in}^{2}$

Correct answer:

$750\; \textrm{in}^{2}$

Explanation:

Use the following formula, with B = 36,b = 24,h=25 :

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

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Example Question #4 : How To Find The Area Of A Trapezoid

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

Possible Answers:

$2,000\; \textrm{in}^{2}$

$6,000\; \textrm{in}^{2}$

$4,000\; \textrm{in}^{2}$

$1,000\; \textrm{in}^{2}$

$3,000\; \textrm{in}^{2}$

Correct answer:

$3,000\; \textrm{in}^{2}$

Explanation:

Set B=200 , b=100 , h = 20 .

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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SSAT Middle Level Math : Trapezoids

Study concepts, example questions & explanations for SSAT Middle Level Math

All SSAT Middle Level Math Resources

Example Questions

Example Question #2 : How To Find The Area Of A Trapezoid

Example Question #3 : How To Find The Area Of A Trapezoid

Example Question #3 : How To Find The Area Of A Trapezoid

Example Question #1 : How To Find The Area Of A Trapezoid

Example Question #131 : Quadrilaterals

Example Question #132 : Quadrilaterals

Example Question #133 : Quadrilaterals

Example Question #134 : Quadrilaterals

Example Question #4 : How To Find The Area Of A Trapezoid

All SSAT Middle Level Math Resources

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