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SSAT Middle Level Math : Quadrilaterals

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Example Questions

Example Question #131 : Quadrilaterals

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

Possible Answers:

$30\:cm^{2}$ $\displaystyle 30\:cm^{2}$

$52\:cm^{2}$ $\displaystyle 52\:cm^{2}$

$88\:cm$ $\displaystyle 88\:cm$

$30\:cm$ $\displaystyle 30\:cm$

$88\:cm^{2}$ $\displaystyle 88\:cm^{2}$

Correct answer:

$88\:cm^{2}$ $\displaystyle 88\:cm^{2}$

Explanation:

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

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

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

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

$A=88\:cm^{2}$ $\displaystyle A=88\:cm^{2}$

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Example Question #31 : Geometry

Trapezoid

What is the area of the above trapezoid?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

Find the area of the trapezoid:

Question_7

Possible Answers:

$\displaystyle 56$

$\displaystyle 35$

$\displaystyle 28$

$\displaystyle 49$

Correct answer:

$\displaystyle 28$

Explanation:

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

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

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

$\displaystyle A=(7)(4)=28$

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Example Question #251 : Geometry

Trapezoid

What is the area of the trapezoid?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

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

Trapezoid

The above diagram depicts a rectangle RECT $\displaystyle RECT$ with isosceles triangle $\Delta ECM$ $\displaystyle \Delta ECM$. If $\displaystyle M$ is the midpoint of $\overline{CT}$ $\displaystyle \overline{CT}$, and the area of the orange region is $\displaystyle 72$, then what is the length of one leg of $\Delta ECM$ $\displaystyle \Delta ECM$ ?

Possible Answers:

$\sqrt {48}$ $\displaystyle \sqrt {48}$

$\sqrt {96}$ $\displaystyle \sqrt {96}$

$\sqrt {108}$ $\displaystyle \sqrt {108}$

$\sqrt {54}$ $\displaystyle \sqrt {54}$

$\displaystyle 6$

Correct answer:

$\sqrt {48}$ $\displaystyle \sqrt {48}$

Explanation:

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

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

$\displaystyle RE = CT = CM + MT = h + h = 2h$

Therefore, the orange region is a trapezoid with bases $\displaystyle h$ and $\displaystyle 2h$ and height $\displaystyle h$. Its area is 72, so we can set up and solve this equation using the area formula for a trapezoid:

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

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

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

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

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

$h^{2}= 48$ $\displaystyle h^{2}= 48$

$h = \sqrt {48}$ $\displaystyle h = \sqrt {48}$

This is the length of one leg of the triangle.

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

A trapezoid has a height of $\displaystyle 25$ inches and bases measuring $\displaystyle 24$ inches and $\displaystyle 36$ inches. What is its area?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #2 : 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:

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

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

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

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

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

Correct answer:

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

Explanation:

Set B=200 $\displaystyle B=200$, b=100 $\displaystyle b=100$, h = 20 $\displaystyle h = 20$.

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

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

The area is 3,000 square inches.

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SSAT Middle Level Math : Quadrilaterals

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

All SSAT Middle Level Math Resources

Example Questions

Example Question #131 : Quadrilaterals

Example Question #31 : Geometry

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

Example Question #251 : Geometry

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

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

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

All SSAT Middle Level Math Resources

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