GMAT Math : Problem-Solving Questions

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Example Questions

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 #11 : Calculating The Area Of A Quadrilateral

Triangle

Note: Figure NOT drawn to scale.

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

Possible Answers:

154

The correct answer is not among the other choices.

160

150

144

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 : How To Find The Area Of A Rhombus

Rhombus_1

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

Possible Answers:

120

$40\sqrt{61}$

$40 \sqrt{11}$

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 #201 : Geometry

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

Possible Answers:

144

$72 \sqrt{3}$

$144 \sqrt{3}$

$72 \sqrt{2}$

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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Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

While walking with some friends on campus, you come across an open grassy rectangular area. You shout "Pythagoras!" and run across the rectangular area from one corner to another. If the rectangle measures meters by meters, what distance did you cover?

Possible Answers:

65 m

70 m

45 m

85 m

50 m

Correct answer:

65 m

Explanation:

This problem can be solved in a variety of ways, however you do it, it may be worth eliminating any options shorter than either side of the quad. The diagonal distance is the hypotenuse of a triangle, so it must be longer than 25 or 60 meters.

Then, we can either find our hypotenuse via Pythagorean Theorem, or our knowledge of Pythagorean Triples.

Using Pythagorean Theorem:

c^2=a^2+b^2

$c=\sqrt{25^2+60^2}=\sqrt{4225}=65$

So 65 meters.

Alternatively, recognize the triangle as a 5x/12x/13x triangle

5x=25

12x=60

13x=?

x=5 so 13x=65

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Example Question #11 : Other Quadrilaterals

Given the area of a square is 9x^2 , what is the diagonal?

Possible Answers:

$\frac{9x\sqrt2}{2}$

$\frac{3x\sqrt2}{2}$

$6x\sqrt2$

$3x^2\sqrt2$

$3x\sqrt2$

Correct answer:

$3x\sqrt2$

Explanation:

Write the formula for finding the area of a square given the diagonal.

$A=\frac{d^2}{2}$

Rearrange the equation so that the diagonal term is isolated.

$d=\sqrt{2A}$

Substitute the known area and simplify.

$d=\sqrt{2(9x^2)} = \sqrt{18x^2} = 3x\sqrt2$

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Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

What is the length of the diagonal for a square with a side length of ?

Possible Answers:

$4\sqrt{3}$

$4\sqrt{2}$

$8\sqrt{2}$

Correct answer:

$4\sqrt{2}$

Explanation:

The diagonal of a square is simply the hypotenuse of a right triangle whose other two sides are the length and width of the square. Because all sides of a square are equal in length, this means the length and width are both , which gives us a right triangle with a base of and a height of , for which the hypotenuse is the diagonal of the square. Applying the Pythagorean Theorem to find the length of the diagonal, we have:

a^2+b^2=c^2

4^2+4^2=c^2

$c^2=32\rightarrow c=\sqrt{32}=\sqrt{2*16}=4\sqrt{2}$

So the length of the diagonal for a square with a side length of is $4\sqrt{2}$ . In general, we could check the length of the diagonal for any square with side length , and we would see that the diagonal length is always $a\sqrt{2}$ .

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #91 : Quadrilaterals

Example Question #11 : Calculating The Area Of A Quadrilateral

Example Question #11 : How To Find The Area Of A Rhombus

Example Question #201 : Geometry

Example Question #14 : Other Quadrilaterals

Example Question #15 : Other Quadrilaterals

Example Question #16 : Other Quadrilaterals

Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

Example Question #11 : Other Quadrilaterals

Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

All GMAT Math Resources

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