GMAT Math : Calculating the area of a quadrilateral

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Example Question #91 : Quadrilaterals

Parallelogram2

Give the area of the above parallelogram if BC = 30 .

Possible Answers:

$225\sqrt{3}$

$75\sqrt{2}$

$225\sqrt{2}$

$75\sqrt{3}$

225

Correct answer:

$225\sqrt{3}$

Explanation:

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$BD= \frac{BC}{2} = \frac{30}{2}} = 15$ .

$CD =BD \sqrt{3} = 15\sqrt{3}$

The area is therefore

$BD \cdot CD =15 \cdot 15\sqrt{3} = 225\sqrt{3}$

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Example Question #92 : Quadrilaterals

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the area of Quadrilateral ABCD .

Possible Answers:

154

144

160

150

The correct answer is not among the other choices.

Correct answer:

144

Explanation:

Apply the Pythagorean Theorem twice here.

$BD = \sqrt{6^{2}+ 8^{2}} = \sqrt{36+64 } = \sqrt{100}= 10$

$BC = \sqrt{26^{2}-10^{2}} = \sqrt{676-100} = \sqrt{576} = 24$

The quadrilateral is a composite of two right triangles, each of whose area is half the product of its legs:

Area of $\Delta ABD$ : $\frac{1}{2} \times 6 \times 8 = 24$

Area of $\Delta BCD$ : $\frac{1}{2} \times 10 \times 24= 120$

Add:

A = 24 + 120 = 144

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Example Question #11 : Calculating The Area Of A Quadrilateral

Rhombus_1

The above figure shows a rhombus RHOM . Give its area.

Possible Answers:

120

$40 \sqrt{11}$

$40\sqrt{61}$

Correct answer:

$40 \sqrt{11}$

Explanation:

Construct the other diagonal of the rhombus, which, along with the first one, form a pair of mutual perpendicular bisectors.

Rhombus_1

By the Pythagorean Theorem,

$n = \sqrt{12^{2}- 10^{2}} =\sqrt{144- 100} = \sqrt{44}= \sqrt{4}\cdot \sqrt{11} = 2 \sqrt{11}$

The rhombus can be seen as the composite of four congruent right triangles, each with legs 10 and $2 \sqrt{11}$ , so the area of the rhombus is

$A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}$ .

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Example Question #12 : Calculating The Area Of A Quadrilateral

Rhombus ABCD has perimeter 48; AC = 12 . What is the area of Rhombus ABCD ?

Possible Answers:

$72 \sqrt{3}$

$144 \sqrt{3}$

$72 \sqrt{2}$

144

Correct answer:

$72 \sqrt{3}$

Explanation:

Each side of a rhombus is congruent, so if it has perimeter 48, it has sidelength 12. Also, the diagonals of a rhombus are each other's perpendicular bisectors, so if they are both constructed, and their point of intersection is called , then AM = 6 . The following figure is formed by the rhombus and its diagonals.

Untitled

$\bigtriangleup AMB$ is a right triangle with its short leg half the length of its hypotenuse, so it is a 30-60-90 triangle, and its long leg measures $6\sqrt{3}$ by the 30-60-90 Theorem. Therefore, $BD = 12\sqrt{3}$ . The area of a rhombus is half the product of the lengths of its diagonals:

$\frac{1}{2} \times 12 \times 12 \sqrt{3} = 72 \sqrt{3}$

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Example Question #14 : Other Quadrilaterals

Trapezoid 2

Note: figure NOT drawn to scale.

Give the area of the above trapezoid.

Possible Answers:

$5 \cdot 2^{2x }$

$5 \cdot 2^{2x-1 }$

$3 \cdot 2^{2x }$

$2^{2x+1 }$

$3 \cdot 2^{2x-1 }$

Correct answer:

$3 \cdot 2^{2x-1 }$

Explanation:

The area of a trapezoid with height and bases of length and is

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

Setting $h= b = 2^{x}, B = 2^{x+1}$ :

$A = \frac{1}{2} ( 2^{x+1} + 2^{x}) \cdot 2^{x}$

$= \frac{1}{2} ( 2 \cdot 2^{x } + 2^{x}) \cdot 2^{x}$

$= 2^{-1}\cdot 2^{x} \cdot ( 2 \cdot 2^{x } + 2^{x})$

$= 2^{x-1} \cdot ( 2 \cdot 2^{x } + 2^{x})$

$= 2^{x-1} \cdot 3 \cdot 2^{x }$

$= 3 \cdot 2^{x+x-1 }$

$= 3 \cdot 2^{2x-1 }$

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Example Question #15 : Other Quadrilaterals

Trapezoid 1

Note: figure NOT drawn to scale.

Give the area of the above trapezoid.

Possible Answers:

$2^{2x+2}$

$2^{2x+1}$

$2^{x^{2}+1}$

$2^{2x }$

$2^{x^{2}+x}$

Correct answer:

$2^{2x+1}$

Explanation:

The area of a trapezoid with height and bases of length and is

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

Setting $h= b = 2^{x}, B = 3 \cdot 2^{x}$ :

$A = \frac{1}{2} (3 \cdot 2^{x} + 2^{x}) \cdot 2^{x}$

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

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

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

$= 2^{1+x+x}$

$= 2^{2x+1}$

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Example Question #16 : Other Quadrilaterals

Rhombus

Note: figure NOT drawn to scale.

The above figure depicts a rhombus, RHOM .

$HM = 2^{X}$

RO = 2 (HM)

Give the area of Rhombus RHOM .

Possible Answers:

$2^{x}$

$2^{2x-1}$

$2^{2x}$

$2^{3x}$

$2^{2x+1}$

Correct answer:

$2^{2x}$

Explanation:

The area of a rhombus is half the product of the lengths of its diagonals, so

$A = \frac{1}{2} \cdot RO \cdot HM$

$= \frac{1}{2} \cdot 2 (HM)\cdot HM$

$= HM \cdot HM$

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

$= 2^{x+x}$

$= 2^{2x}$

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GMAT Math : Calculating the area of a quadrilateral

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #91 : Quadrilaterals

Example Question #92 : Quadrilaterals

Example Question #11 : Calculating The Area Of A Quadrilateral

Example Question #12 : Calculating The Area Of A Quadrilateral

Example Question #14 : Other Quadrilaterals

Example Question #15 : Other Quadrilaterals

Example Question #16 : Other Quadrilaterals

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