GMAT Math : Other Quadrilaterals

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Example Questions

Example Question #1 : Other Quadrilaterals

What is the area of a trapezoid with a height of 7, a base of 5, and another base of 13?

Possible Answers:

$\dpi{100} \small 39$

$\dpi{100} \small 63$

$\dpi{100} \small 51$

$\dpi{100} \small 43$

$\dpi{100} \small 29$

Correct answer:

$\dpi{100} \small 63$

Explanation:

$area = \frac{(b_{1}+ b_{2}\cdot h)}{2} = \frac{(5 + 13)\cdot 7}{2} = \frac{18\cdot 7}{2} = \frac{126}{2} = 63$

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Example Question #2 : Other Quadrilaterals

A circle can be circumscribed about each of the following figures except:

Possible Answers:

A regular pentagon

A regular hexagon

A triangle with sides 30, 40, 50.

Each of the figures given in the other choices can have a circle circumscribed about it.

A $30^{\circ } -60 ^{\circ }-90^{\circ }$ triangle

Correct answer:

Each of the figures given in the other choices can have a circle circumscribed about it.

Explanation:

A circle can be circumscribed about any triangle regardless of its sidelengths or angle measures, so we can eliminate the two triangle choices.

A circle can be circumscribed about any regular polygon, so we can eliminate those two choices as well.

The correct choice is that each figure can have a circle circumscribed about it.

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Example Question #2 : Calculating The Area Of A Quadrilateral

What is the area of a quadrilateral on the coordinate plane with vertices (0.0),(8.0),(3,4),(11,4) ?

Possible Answers:

Correct answer:

Explanation:

As can be seen from this diagram, this is a parallelogram with base 8 and height 4:

Parallelogram

The area of this parallelogram is the product of its base and its height:

$A = bh = 8 \cdot 4 = 32$

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Example Question #1 : Calculating The Area Of A Quadrilateral

What is the area of a quadrilateral on the coordinate plane with vertices (0,0), (0,10),(8,2), (8,7) ?

Possible Answers:

120

Correct answer:

Explanation:

As can be seen in this diagram, this is a trapezoid with bases 10 and 5 and height 8.

Trapezoid

Setting B = 10, b = 5, h = 8 in the following formula, we can calculate the area of the trapezoid:

$A = \frac{1}{2} (B + b) h = \frac{1}{2} \cdot (10 + 5)\cdot 8 = 60$

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Example Question #5 : Calculating The Area Of A Quadrilateral

Quad

Note: Figure NOT drawn to scale

What is the area of Quadrilateral QUAD , above?

Possible Answers:

$38 \frac{1}{2}$

$42 \frac{1}{2}$

Correct answer:

Explanation:

Quadrilateral QUAD is a composite of two right triangles, $\Delta ADU$ and $\Delta DUQ$ , so we find the area of each and add the areas. First, we need to find and , since the area of a right triangle is half the product of the lengths of its legs.

By the Pythagorean Theorem:

$\left (AD \right )^{2} +\left ( AU\right ) ^{2} =\left ( DU\right ) ^{2}$

$\left (AD \right )^{2} +4 ^{2} =5 ^{2}$

$\left (AD \right )^{2} +16 =25$

$\left (AD \right )^{2} +16 -16 =25 -16$

$\left (AD \right )^{2} =9$

AD = 3

Also by the Pythagorean Theorem:

$\left ( QU\right ) ^{2} + \left ( DU\right ) ^{2}=\left ( DQ\right ) ^{2}$

$\left ( QU\right ) ^{2} + 5^{2}=13^{2}$

$\left ( QU\right ) ^{2} + 25 =169$

$\left ( QU\right ) ^{2} + 25 -25=169 -25$

$\left ( QU\right ) ^{2} =144$

QU = 12

The area of $\Delta ADU$ is $\frac{1}{2} \cdot AD \cdot AU = \frac{1}{2} \cdot 3 \cdot 4 = 6$ .

The area of $\Delta DUQ$ is $\frac{1}{2} \cdot DU \cdot QU = \frac{1}{2} \cdot 5 \cdot 12 = 30$ .

Add the areas to get 6 + 30 = 36 , the area of Quadrilateral QUAD .

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Example Question #3 : Calculating The Area Of A Quadrilateral

What is the area of the quadrilateral on the coordinate plane with vertices (0,0), (9,0), (4, 7), (7,7) ?

Possible Answers:

Correct answer:

Explanation:

The quadrilateral formed is a trapezoid with two horizontal bases. One base connects (0,0) and (9,0) and therefore has length 9 - 0 = 9 ; the other connects (4,7) and (7,7) and has length 7-4 = 3 . The height is the vertical distance between the two bases, which is the difference of the coorindates: 7 - 0 = 7 . Therefore, the area of the trapezoid is

$A = \frac{1}{2} (B + b ) h = \frac{1}{2} \cdot (9+3 ) \cdot 7 = 42$

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Example Question #6 : Calculating The Area Of A Quadrilateral

What is the area of the quadrilateral on the coordinate plane with vertices (-2, -1), (8, -1), (-1, 5), (5,5) .

Possible Answers:

Correct answer:

Explanation:

The quadrilateral is a trapezoid with horizontal bases; one connects (-2, -1) and (8, -1) and has length 8 - (-2) = 10 , and the other connects (-1, 5) and (5,5) and has length 5 - (-1) = 6 . The height is the vertical distance between the bases, which is the difference of the -coordinates; this is . Substitute B = 10, b = 6, h = 6 in the formula for the area of a trapezoid:

$A = \frac{1}{2} (B + b)h = \frac{1}{2}\cdot (10+6) \cdot 6 = 48$

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Example Question #5 : Other Quadrilaterals

What is the area of the quadrilateral on the coordinate plane with vertices (-4, 8) , (-4, 15), (-9, 6), (-9, 13) ?

Possible Answers:

Correct answer:

Explanation:

The quadrilateral is a parallelogram with two vertical bases, each with length 15-8 = 13-6 = 7 . Its height is the distance between the bases, which is the difference of the -coordinates: -4 - (-9) = 5 . The area of the parallelogram is the product of its base and its height:

$7 \times 5 = 35$

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Example Question #201 : Geometry

Parallelogram1

Give the area of the above parallelogram if BC = 16 .

Possible Answers:

$128\sqrt{2}$

$64\sqrt{3}$

128

$64\sqrt{2}$

Correct answer:

128

Explanation:

Multiply height by base to get the area.

By the 45-45-90 Theorem,

$BD=CD = BC \div \sqrt{2} = \frac{16}{\sqrt{2}} = \frac{16\cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}$ .

Since the product of the height and the base of a parallelogram is its area,

$A = BD \cdot CD$

$= 8 \sqrt{2} \cdot 8 \sqrt{2}$

$= 8\cdot 8 \cdot \sqrt{2} \cdot \sqrt{2}$

$= 8\cdot 8 \cdot 2$

= 128

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Example Question #1 : Calculating The Area Of A Quadrilateral

Parallelogram2

Give the area of the above parallelogram if BC = 10 .

Possible Answers:

$25\sqrt{2}$

$50\sqrt{2}$

$25\sqrt{3}$

Correct answer:

$25\sqrt{3}$

Explanation:

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$CD= \frac{BC}{2} = \frac{10}{2}} = 5$ .

$BD =CD \sqrt{3} = 5\sqrt{3}$

The area is therefore

$BD \cdot CD = 5\sqrt{3} \cdot 5= 25\sqrt{3}$

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GMAT Math : Other Quadrilaterals

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Other Quadrilaterals

Example Question #2 : Other Quadrilaterals

Example Question #2 : Calculating The Area Of A Quadrilateral

Example Question #1 : Calculating The Area Of A Quadrilateral

Example Question #5 : Calculating The Area Of A Quadrilateral

Example Question #3 : Calculating The Area Of A Quadrilateral

Example Question #6 : Calculating The Area Of A Quadrilateral

Example Question #5 : Other Quadrilaterals

Example Question #201 : Geometry

Example Question #1 : Calculating The Area Of A Quadrilateral

All GMAT Math Resources

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