GMAT Math : Geometry

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Example Questions

Example Question #1 : Prisms

Jenny wants to make a cube out of sheet metal. What is the length of one side of the cube?

I) The cube will require 294 square inches of material.

II) The cube will hold 343 cubic inches.

Possible Answers:

Both statements are needed to answer the question.

Statement II is sufficient to answer the question, but statement I is not sufficient to answer the question.

Statement I is sufficient to answer the question, but statement II is not sufficient to answer the question.

Neither statement is sufficient to answer the question. More information is needed.

Either statement is sufficient to answer the question.

Correct answer:

Either statement is sufficient to answer the question.

Explanation:

The length of an edge of a cube can be used to find either volume or surface area, and vice versa.

I) Gives us the surface area thus, we are able to calculate the length of an edge using the formula,

$SA=6\cdot s^2$

$294=6s^2 \rightarrow 49=s^2 \rightarrow 7=s$ .

II) Gives us the volume thus, we are able to calculate the length of an edge using the formula,

V=s^3

$343=s^3 \rightarrow 7=s$

Either can be used to find the side length.

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Example Question #4 : Prisms

What is the surface area of a rectangular prism that is 4 inches long, 6 inches wide, and 5 inches high?

Possible Answers:

154

112

120

172

148

Correct answer:

148

Explanation:

SA=2lw+2lh+2hw=2*4*6+2*4*5+2*5*6=48+40+60=148

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Example Question #5 : Prisms

A box with dimensions 8 inches, 10 inches, and 5 inches needs to be gift wrapped. Gift wrapping is priced at $0.10 per square inch of surface of a box. How much will it cost to wrap the gift?

Possible Answers:

$\$17$

$\$20$

$\$40$

$\$34$

Correct answer:

$\$34$

Explanation:

Find the surface area of the box by summing the area of all six faces: two 8 by 10, two 8 by 5, two 10 by 5.

$2(80)+2(40)+2(50)=340\ in^{2}$

Since the price is $.10 for each square inch,

$Total\ cost=340(.10)=\$34$

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Example Question #1 : Calculating The Surface Area Of A Prism

The sum of the length, the width, and the height of a rectangular prism is one meter. The length of the prism is sixteen centimeters greater than its width, which is three times its height. What is the surface area of this prism?

Possible Answers:

$5,856 \textrm{ cm}^{2}$

$22,464 \textrm{ cm}^{2}$

$7,672 \textrm{ cm}^{2}$

$25,872 \textrm{ cm}^{2}$

$6,104 \textrm{ cm}^{2}$

Correct answer:

$5,856 \textrm{ cm}^{2}$

Explanation:

Let be the height of the prism. Then the width is , and the length is 3h+ 16 . Since the sum of the three dimensions is one meter, or 100 centimeters, we solve for in this equation:

h + 3h + (3h+16) = 100

7h+16 = 100

7h+16-16 = 100-16

7h = 84

$7h \div 7 = 84 \div 7$

h = 12

The height of the prism is 12 cm; the width is three times this, or 36 cm; the length is sixteen centimeters greater than the width, which is 52 cm.

Set l= 52,w=36,h=12 in the formula for the surface area of a rectangular prism:

A = 2lw + 2wh + 2lh

$= 2 \cdot 52 \cdot 36 + 2 \cdot 36 \cdot 12 + 2 \cdot 52 \cdot 12$

= 3,744 + 864 + 1,248

= 5,856 square centimeters

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Example Question #2 : Calculating The Surface Area Of A Prism

The volume of the rectangular solid above is 120. The area of ABFE is 20 and the area of ABCD is 30. What is the area of BFGC?

Possible Answers:

Correct answer:

Explanation:

We can set the lengths of three sides to be , , , respectively. The volume is 120 means that xyz=120 . Also we know the areas of two sides, so we can use to represent the area of 20 and to represent the area of 30. Now the question is to figure out . Then we can use

xy * xz * yz = x^2 * y^2 * z^2 =(xyz)^2 to solve .

20 * 30 * yz = 120^2

yz=24

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Example Question #5 : Calculating The Surface Area Of A Prism

A right prism has as its bases two right triangles, each of whose legs have lengths 12 and 16. The height of the prism is half the perimeter of a base. Give the surface area of the prism.

Possible Answers:

1,152

2,304

1,344

768

584

Correct answer:

1,344

Explanation:

The area of a right triangle is equal to half the product of its legs, so each base has area

$\frac{1}{2} \cdot 12 \cdot 16 = 96$

The measure of the hypotenuse of each base is determined using the Pythagorean Theorem:

$c = \sqrt{12^{2}+16^{2}}= \sqrt{144+256} = \sqrt{400}= 20$

Therefore, the perimeter of each base is

P=12+16+20 = 48 ,

and the height of the prism is half this, or $h = \frac{1}{2} \cdot 48 = 24$ .

The lateral area of the prism is the product of its height and the perimeter of a base; this is

$L= Ph= 24 \cdot 48 = 1,152$

The surface area is the sum of the lateral area and the two base areas, or

S= L+B+B =1,152+96+96 =1,344 .

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Example Question #11 : Prisms

A right prism has as its bases two isosceles right triangles, each of whose hypotenuse has length 10. The height of the prism is the length of one leg of a base. Give the surface area of the prism.

Possible Answers:

$150+ 50 \sqrt{2}$

250

$150\sqrt{2}$

$200\sqrt{2}$

$200+ 50 \sqrt{2}$

Correct answer:

$150+ 50 \sqrt{2}$

Explanation:

By the 45-45-90 Theorem, dividing the length of the hypotenuse of an isosceles right triangle by $\sqrt{2}$ yields the length of one leg; therefore, the length of one leg of each base is

$\frac{10}{\sqrt{2}} = \frac{10 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10 \sqrt{2}}{2} = 5\sqrt{2}$ .

The area of a right triangle is half the product of its legs, so the area of each base is

$B = \frac{1}{2} \cdot 5 \sqrt{2} \cdot 5 \sqrt{2} = \frac{1}{2} \cdot 25 \cdot 2 = 25$

The perimeter of each base is the sum of its sides, which here is

$P = 5 \sqrt{2}+ 5 \sqrt{2}+ 10 = 10 + 10 \sqrt{2}$

The height of the prism is the length of one leg of a base, which is $h = 5\sqrt{2}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$= \left (10+10 \sqrt{2} \right ) \cdot 5 \sqrt{2}$

$= 10 \cdot 5 \sqrt{2} +10 \sqrt{2} \cdot 5 \sqrt{2}$

$= 50 \sqrt{2} +10\cdot 5 \cdot 2$

$= 100+ 50 \sqrt{2}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 100+ 50 \sqrt{2}+25+25 = 150+ 50 \sqrt{2}$

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Example Question #7 : Calculating The Surface Area Of A Prism

A right prism has as its bases two equilateral triangles, each of whose sides has length 6. The height of the prism is three times the perimeter of a base. Give the surface area of the prism.

Possible Answers:

$972 + 9 \sqrt{3}$

$486 \sqrt{3}$

972

$972 + 18 \sqrt{3}$

$486\sqrt{2}$

Correct answer:

$972 + 18 \sqrt{3}$

Explanation:

The area of each equilateral triangle base can be determined by setting s = 6 in the formula

$B= \frac{s^{2}\sqrt{3}}{4} = \frac{6^{2}\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}$

The perimeter of each base is $P=6 \cdot 3 = 18$ , and the height of the prism is three times this, or $h = 18 \cdot 3 = 54$ . The lateral area of a prism is equal to the perimeter of a base multiplied by the height, so

$L =Ph= 18 \cdot 54 = 972$

Add this to the areas of two bases - the surface area is

$S= L+ B + B= 972 + 9\sqrt{3}+ 9\sqrt{3}= 972 + 18 \sqrt{3}$ .

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Example Question #4 : Calculating The Surface Area Of A Prism

A right prism has as its bases two isosceles right triangles, each of whose legs has length 16. The height of the prism is the length of the hypotenuse of a base. Give the surface area of the prism.

Possible Answers:

$768 + 512 \sqrt{2}$

$512+ 512 \sqrt{2}$

$384+ 512 \sqrt{2}$

$640+ 512 \sqrt{2}$

$256+ 512 \sqrt{2}$

Correct answer:

$768 + 512 \sqrt{2}$

Explanation:

By the 45-45-90 Theorem, multiplying the length of a leg of an isosceles right triangle by $\sqrt{2}$ yields the length of its hypotenuse; therefore, the length of the hypotenuse of each base is $16 \sqrt{2}$ .

The area of a right triangle is half the product of its legs, so the area of each base is

$B = \frac{1}{2} \cdot 16 \cdot 16 =128$

The perimeter of each base is the sum of its sides, which here is

$P =16+16+16 \sqrt{2} = 32 + 16 \sqrt{2}$

The height of the prism is the length of the hypotenuse of the base, which is $h = 16\sqrt{2}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$= \left ( 32 + 16 \sqrt{2}\right ) \cdot 16 \sqrt{2}$

$= 32 \cdot16 \sqrt{2} +16 \sqrt{2} \cdot 16 \sqrt{2}$

$= 32 \cdot16 \sqrt{2} +16 \cdot 16 \cdot \sqrt{2} \cdot \sqrt{2}$

$= 512 \sqrt{2} +16 \cdot 16 \cdot 2$

$= 512+ 512 \sqrt{2}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 512+ 512 \sqrt{2} + 128 +128 = 768 + 512 \sqrt{2}$

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Example Question #774 : Problem Solving Questions

A right prism has as its bases two right triangles, each of which has a hypotenuse of length 20 and a leg of length 10. The height of the prism is equal to the length of the longer leg of a base. Give the surface area of the prism.

Possible Answers:

500

$500+300 \sqrt{3}$

600

$300+400 \sqrt{3}$

$400+300 \sqrt{3}$

Correct answer:

$300+400 \sqrt{3}$

Explanation:

A right triangle with one leg half the length of the hypotenuse is a 30-60-90 triangle. Its other leg has measure $\sqrt{3}$ times the length of the first leg, which in the case of each base is $10 \sqrt{3}$ .

The area of each base is half the product of the legs, or

$B = \frac{1}{2} \cdot 10 \cdot 10 \sqrt{3} = 50 \sqrt{3}$ .

The perimeter of each base is

$P = 20+10+10 \sqrt {3} = 30 + 10 \sqrt {3}$

and the height of the prism is equal to the length of the longer leg, or

$h= 10 \sqrt{3}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$=\left ( 30 + 10 \sqrt {3} \right ) \cdot 10 \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \sqrt {3}\cdot 10 \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot \sqrt {3}\cdot \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot 3$

$= 300+ 30 0 \sqrt {3}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 300+300 \sqrt{3} + 50 \sqrt{3} + 50 \sqrt{3}= 300+400 \sqrt{3}$

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Prisms

Example Question #4 : Prisms

Example Question #5 : Prisms

Example Question #1 : Calculating The Surface Area Of A Prism

Example Question #2 : Calculating The Surface Area Of A Prism

Example Question #5 : Calculating The Surface Area Of A Prism

Example Question #11 : Prisms

Example Question #7 : Calculating The Surface Area Of A Prism

Example Question #4 : Calculating The Surface Area Of A Prism

Example Question #774 : Problem Solving Questions

All GMAT Math Resources

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