Calculating the surface area of a prism - GMAT Quantitative

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Question

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?

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Answer

We can set the lengths of three sides to be , , , respectively. The volume is 120 means that . 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

to solve .

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Question

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

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Answer

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Question

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?

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Answer

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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Question

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?

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Answer

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

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

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 in the formula for the surface area of a rectangular prism:

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

square centimeters

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Question

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.

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Answer

The area of each equilateral triangle base can be determined by setting 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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Question

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.

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Answer

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

,

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

.

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Question

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.

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Answer

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

$= \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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Question

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.

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Answer

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

$= \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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Question

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.

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Answer

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

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

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

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

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

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

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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Question

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.

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Answer

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

.

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

,

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:

.

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Question

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.

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Answer

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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Question

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 .

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Answer

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

.

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

.

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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Question

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?

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Answer

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

and width

The surface area of a rectangular prism is

. We can call , , and and solve for :

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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Question

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

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Answer

The volume of a rectangular solid is the product of its length, height, and width. Since , , and ,

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

The surface area of a rectangular solid is

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

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Question

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

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Answer

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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Question

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?

Tap to reveal answer

Answer

We can set the lengths of three sides to be , , , respectively. The volume is 120 means that . 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

to solve .

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Question

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

Tap to reveal answer

Answer

← Didn't Know|Knew It →

Question

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?

Tap to reveal answer

Answer

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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Question

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?

Tap to reveal answer

Answer

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

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

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 in the formula for the surface area of a rectangular prism:

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

square centimeters

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Question

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.

Tap to reveal answer

Answer

The area of each equilateral triangle base can be determined by setting 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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