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

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Question

Note: Figure NOT drawn to scale

The above figure shows Square .

Which is the greater quantity?

(a) The area of Trapezoid

(b) The area of Trapezoid

Answer

The easiest way to answer the question is to locate on $\overline{SQ}$ such that :

Trapezoids and have the same height, which is . Their bases, by construction, have the same lengths - and . Therefore, Trapezoids and have the same area.

Since , it follows that , and . It follows that Trapezoid is greater in area than Trapezoids and , and Trapezoid is less in area.

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Question

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

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?

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?

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 $\Delta ECM$ ?

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:

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

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

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

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

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

$h^{2}= 48$

$h = \sqrt {48}$

This is the length of one leg of the triangle.

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Question

Find the area of the trapezoid:

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 bases equal to 7 and 9 and height is 2.

Answer

To solve, simply use the formula for the area of a trapezoid.

Thus,

$A=\frac{1}{2}(b1+b2)h=\frac{1}{2}(7+9)(2)=16$

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Question

Find the area of a trapezoid with bases of 10 centimeters and 8 centimeters, and a height of 4 centimeters.

Answer

The formula for area of a trapezoid is:

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

where

therefore the area equation becomes,

$A = \frac{1}{2} (10 + 8) \times 4$

$A = \frac{1}{2} (72)$

$A = 36cm^{2}$

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Question

You recently bought a new bookshelf with a base in the shape of an isosceles trapezoid. If the small base is 2 feet, the large base is 3 feet, and the depth is 8 inches, what is the area of the base of your new bookshelf?

Answer

You recently bought a new bookshelf with a base in the shape of an isosceles trapezoid. If the small base is 2 feet, the large base is 3 feet, and the depth is 8 inches, what is the area of the base of your new bookshelf?

To find the area of a trapezoid, we need to use the following formula:

$A_{trapezoid}=\frac{a+b}{2}h$

Where a and b are the lengths of the bases, and h is the perpendicular distance from one base to another.

We are given a and b, and then h will be the same as our depth. The tricky part is realizing that our depth is in inches, while our base lengths are in feet. We need to convert 8 inches to feet:

$8in\frac{1ft}{12in}=\frac{8}{12}=\frac{2}{3}ft$

Next, plug it all into our equation up above.

$A_{trapezoid}=\frac{2+3}{2}\frac{2}{3}=\frac{5}{2}\frac{2}{3}$

$\frac{5}{2}*\frac{2}{3}=\frac{5}{3}ft^2$

So our answer is:

$\frac{5}{3}ft^2$

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Question

The above diagram shows Rectangle , with midpoint of $\overline{CT}$ .

The area of Quadrilateral is . Evaluate .

Answer

The easiest way to see this problem is to note that Quadrilateral has as its area that of Rectangle minus that of $\bigtriangleup ECM$ .

The area of Rectangle is its length multiplied by its width:

$2x \cdot \frac{1}{2} x = 2 \cdot \frac{1}{2} \cdot x \cdot x = x^{2}$

is the midpoint of $\overline{CT}$ , so $\bigtriangleup ECM$ has as its base and height $\frac{1}{2} \cdot 2x =x$ and $\frac{1}{2} x$ , respectively;

its area is half their product, or

$\frac{1}{2} \cdot x \cdot \frac{1}{2} x = \frac{1}{4} x^{2}$

The area of Quadrilateral is

$x^{2} - \frac{1}{4} x^{2} = \frac{3}{4} x^{2} = 1,200$ , so

$\frac{3}{4} x^{2} \cdot \frac{4} {3} = 1,200\cdot \frac{4} {3}$

$x^{2} = 1,600$

$x =\sqrt{ 1,600} = 40$

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Question

Note: Figure NOT drawn to scale

The above figure shows Square .

Which is the greater quantity?

(a) The area of Trapezoid

(b) The area of Trapezoid

Answer

The easiest way to answer the question is to locate on $\overline{SQ}$ such that :

Trapezoids and have the same height, which is . Their bases, by construction, have the same lengths - and . Therefore, Trapezoids and have the same area.

Since , it follows that , and . It follows that Trapezoid is greater in area than Trapezoids and , and Trapezoid is less in area.

Compare your answer with the correct one above

Question

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

Answer

Use the following formula, with :

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

Compare your answer with the correct one above

Question

What is the area of the trapezoid?

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.

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

Compare your answer with the correct one above

Question

What is the area of the above trapezoid?

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

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

Compare your answer with the correct one above

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 $\Delta ECM$ ?

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:

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

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

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

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

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

$h^{2}= 48$

$h = \sqrt {48}$

This is the length of one leg of the triangle.

Compare your answer with the correct one above

Question

Find the area of the trapezoid:

Answer

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

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

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

Compare your answer with the correct one above

Question

Find the area of a trapezoid with bases equal to 7 and 9 and height is 2.

Answer

To solve, simply use the formula for the area of a trapezoid.

Thus,

$A=\frac{1}{2}(b1+b2)h=\frac{1}{2}(7+9)(2)=16$

Compare your answer with the correct one above

Question

Find the area of a trapezoid with bases of 10 centimeters and 8 centimeters, and a height of 4 centimeters.

Answer

The formula for area of a trapezoid is:

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

where

therefore the area equation becomes,

$A = \frac{1}{2} (10 + 8) \times 4$

$A = \frac{1}{2} (72)$

$A = 36cm^{2}$

Compare your answer with the correct one above

Question

You recently bought a new bookshelf with a base in the shape of an isosceles trapezoid. If the small base is 2 feet, the large base is 3 feet, and the depth is 8 inches, what is the area of the base of your new bookshelf?

Answer

You recently bought a new bookshelf with a base in the shape of an isosceles trapezoid. If the small base is 2 feet, the large base is 3 feet, and the depth is 8 inches, what is the area of the base of your new bookshelf?

To find the area of a trapezoid, we need to use the following formula:

$A_{trapezoid}=\frac{a+b}{2}h$

Where a and b are the lengths of the bases, and h is the perpendicular distance from one base to another.

We are given a and b, and then h will be the same as our depth. The tricky part is realizing that our depth is in inches, while our base lengths are in feet. We need to convert 8 inches to feet:

$8in\frac{1ft}{12in}=\frac{8}{12}=\frac{2}{3}ft$

Next, plug it all into our equation up above.

$A_{trapezoid}=\frac{2+3}{2}\frac{2}{3}=\frac{5}{2}\frac{2}{3}$

$\frac{5}{2}*\frac{2}{3}=\frac{5}{3}ft^2$

So our answer is:

$\frac{5}{3}ft^2$

Compare your answer with the correct one above

Question

The above diagram shows Rectangle , with midpoint of $\overline{CT}$ .

The area of Quadrilateral is . Evaluate .

Answer

The easiest way to see this problem is to note that Quadrilateral has as its area that of Rectangle minus that of $\bigtriangleup ECM$ .

The area of Rectangle is its length multiplied by its width:

$2x \cdot \frac{1}{2} x = 2 \cdot \frac{1}{2} \cdot x \cdot x = x^{2}$

is the midpoint of $\overline{CT}$ , so $\bigtriangleup ECM$ has as its base and height $\frac{1}{2} \cdot 2x =x$ and $\frac{1}{2} x$ , respectively;

its area is half their product, or

$\frac{1}{2} \cdot x \cdot \frac{1}{2} x = \frac{1}{4} x^{2}$

The area of Quadrilateral is

$x^{2} - \frac{1}{4} x^{2} = \frac{3}{4} x^{2} = 1,200$ , so

$\frac{3}{4} x^{2} \cdot \frac{4} {3} = 1,200\cdot \frac{4} {3}$

$x^{2} = 1,600$

$x =\sqrt{ 1,600} = 40$

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