GMAT Math : Prisms

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Example Questions

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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Example Question #10 : Calculating The Surface Area Of A Prism

The length and width of a rectangular solid are 18 and 20; its volume is 5,400. Calculate its surface area.

Possible Answers:

1,620

2,160

1,800

1,898

1,860

Correct answer:

1,860

Explanation:

The volume of a rectangular solid is the product of its length, height, and width. Since L = 18 , W = 20 , and H = 5,400 ,

LWH = V

$18 \cdot 20 \cdot H =5,400$

360H = 5,400

H = 15

The surface area of a rectangular solid is

S = 2LH + 2LW + 2WH

$= 2 \cdot 18 \cdot 15 + 2 \cdot 18 \cdot 20 + 2 \cdot 20\cdot 15$

= 540+720 + 600

= 1,860

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Example Question #775 : Problem Solving Questions

The length of a rectangular prism is twice its width and five times its height. If is the width, give the surface area in terms of .

Possible Answers:

$\frac{13}{5}W^{2}$

$\frac{32}{5}W^{2}$

$\frac{13}{10}W^{2}$

$16W^{2}$

$\frac{16}{5}W^{2}$

Correct answer:

$\frac{32}{5}W^{2}$

Explanation:

If we let be the width, then, since the length is twice this,

L = 2W .

Since the length is five times the height, the height is one fifth the length, so $H = \frac{1}{5}L = \frac{1}{5} \cdot 2W = \frac{2}{5} W$ .

The surface area of a rectangular solid is

S = 2LH + 2LW + 2WH .

which can be rewritten as

$S = 2 \cdot 2W \cdot \frac{2}{5} W + 2 \cdot 2W \cdot W + 2W \cdot \frac{2}{5} W$

$S = \frac{8}{5} W^{2} + 4W^{2} + \frac{4}{5} W^{2} = \frac{32}{5}W^{2}$

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Example Question #61 : Rectangular Solids & Cylinders

Each base of a right prism is a regular hexagon with sidelength 6. Its height is two thirds the perimeter of a base. Give the surface area of the prism.

Possible Answers:

$864+ 108\sqrt{3}$

972

$972\sqrt{3}$

$1,296\sqrt{3}$

$864+ 54\sqrt{3}$

Correct answer:

$864+ 108\sqrt{3}$

Explanation:

The perimeter of a regular hexagon with sidelength 6 is

$P= 6 \cdot 6 = 36$

The height of the prism is two thirds of this, so

$h = \frac{2}{3} \cdot 36 = 24$ .

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

$L =Ph= 36 \cdot 24 = 864$

The area of each base can be calculated using the area formula for a regular hexagon:

$B = \frac{3s^{2}\sqrt{3}}{2}$

$B = \frac{3\cdot 6^{2}\sqrt{3}}{2}$

$B = \frac{108\sqrt{3}}{2}$

$B = 54\sqrt{3}$

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

$S=L+B+B= 864+ 54\sqrt{3} + 54\sqrt{3} = 864+ 108\sqrt{3}$

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Example Question #782 : Problem Solving Questions

Tetra_1

Refer to the above diagram. The perimeter of $\bigtriangleup DEG$ is 30. What is the surface area of the cube shown?

Possible Answers:

300

$600 \sqrt{2}$

Insufficient information is given to answer the question.

600

$300 \sqrt{2}$

Correct answer:

300

Explanation:

Since each side of $\bigtriangleup DEG$ is a diagonal of one of three congruent squares, each side has the same length; since the total perimeter is 30, each side measures one third of this, or 10.

Each square side, therefore, has diagonal 10, and, by the 45-45-90 Theorem, each side of each square - and each edge of the cube - measures $s= \frac{10}{\sqrt{2}}$ . We use the surface area formula:

$A= 6s^{2}$

$A= 6 \left ( \frac{10}{\sqrt{2}} \right )^{2} = 6 \cdot \frac{100}{2} = 300$

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Example Question #783 : Problem Solving Questions

A right prism has as its bases two triangles, each of which has a hypotenuse of length 25 and a leg of length 7. The height of the prism is one fourth the perimeter of a base. Give the surface area of the prism.

Possible Answers:

424

868

434

784

952

Correct answer:

952

Explanation:

The second leg of a right triangle with hypotenuse of length 25 and one leg of length 7 has length

$\sqrt{25^{2}-7^{2}}= \sqrt{625-49}= \sqrt{576}= 24$ .

The area of this right triangle is half the product of the lengths of the legs, which is

$B = \frac{1}{2} \cdot 7 \cdot 24 = 84$ .

The perimeter of each base is

P = 7+ 24 + 25 = 56 ,

and the height is one fourth this, or

$h = \frac{1}{4} \cdot P = \frac{1}{4} \cdot 56 =14$

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

$L = 14 \cdot 56 = 784$ .

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

S= L + B + B = 784+84+84 =952 .

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Example Question #15 : Prisms

The length of a cube is increased by 20%, and the width is decreased by 20%. Which of the following must happen to the height so that the resulting rectangular prism will have the same surface area as the original cube?

Possible Answers:

The height must be increased by 4%.

The height must remain the same.

The height must be decreased by 4%.

The height must be decreased by 2%.

The height must be increased by 2%.

Correct answer:

The height must be increased by 2%.

Explanation:

To look at this more easily, assume the cube has sides of length 100; this argument generalizes to any size. The surface area of the cube is

$S= 6s^{2}= 6 \cdot 100^{2}= 6 \cdot 10,000 = 60,000$ .

After the changes, the resulting rectangular prism will have length

100 (1+0.2) = 120

and width

100 (1-0.2) = 80

The surface area of a rectangular prism is

S = 2LH + 2LW + 2WH . We can call L = 80 , W = 120 , and S= 60,000 and solve for :

2(80)H + 2(80)(120)+ 2(120)H= 60,000

160H + 19,200+240H= 60,000

400H + 19,200 = 60,000

400H = 40,800

H = 102

This means that the height of the prism must be 102% of the height of the cube - equivalently, the height must be increased by 2%.

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GMAT Math : Prisms

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

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

Example Question #10 : Calculating The Surface Area Of A Prism

Example Question #775 : Problem Solving Questions

Example Question #61 : Rectangular Solids & Cylinders

Example Question #782 : Problem Solving Questions

Example Question #783 : Problem Solving Questions

Example Question #15 : Prisms

All GMAT Math Resources

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