Prisms - GMAT Quantitative

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What is the volume of a cube whose diagonal measures 10 inches?

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By an extension of the Pythagorean Theorem, if is the length of an edge of the cube and is its diagonal length,

$s$\sqrt{3 }$= d$

$s= $\frac{d}{\sqrt{3 }$}$

The volume is therefore

$V = $s^{3}$ = \left ($\frac{d}{\sqrt{3}$} \right $)^{3}$ = \left ($\frac{10}{\sqrt{3}$} \right $)^{3}$\approx 192.5$

A cube of iron has a mass of 4 kg. What is the mass of a rectangular prism of iron that is the same height but has a width and length that are twice as long as the cube?

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Let the length of the sides of the cube equal 1. The volume of the cube is then 1times 1times 1=1. Therefore, the volume of the prism is 2times 2times 1=4. Therefore, the mass of the prism must be 4 times greater than the mass of the cube.

(4)(4)= 16

The sum of the length, the width, and the height of a rectangular prism is one yard. The length of the prism is eleven inches greater than its width, and the width is twice its height. What is the volume of the prism?

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Let be the height of the prism. Then the width is , and the length is . Since the sum of the three dimensions is one yard, or 36 inches, we solve for in this equation:

$5h \div 5 = 25 \div 5$

The height is 5 inches; the width is twice this, or 10 inches; the length is eleven inches greater than the width, or 21 inches. The volume is the product of the three dimensions:

$V = lwh = 5 \cdot10\cdot21 = 1,050$ cubic inches.

The length, width, and height of a rectangular prism, in inches, are three different prime numbers. All three dimensions are between one yard and four feet. What is the volume of the prism?

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One yard is equal to 36 inches; four feet are equal to 48 inches. There are four prime numbers between 36 and 48 - 37, 41, 43, and 47. Since we are only given that the dimensions are three different prime numbers between 36 and 48, we have no way of knowing which three they are.

A company makes cubic metal blocks whose edges measure 5 inches. The blocks are supposed to be packed in crates 40 inches long, 30 inches wide, and 32 inches high. How many blocks can be packed in one of these crates?

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Each layer of blocks will be $40 \div 5 = 8$ blocks long and $30 \div 5 = 6$ blocks wide, making a layer of $8 \times 6 = 48$ blocks. The height is 32 feet, which divided by 5 is

$32 \div 5 = 6\textrm{ R }2$

so there will be 6 layers of blocks.

Therefore, $48 \times 6 = 288$ blocks can be packed.

A given rectangular prism has a length of , a width of , and a height of . What is the volume of the prism?

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The volume of a given rectangular prism is the product of its length , width , and height , or . Plugging in the values provided:

A given cylinder has circular bases of radius and height . What is the volume of the cylinder?

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The volume of a given cylinder is the product of its base area and height , or . Since the circular base area $A=\pi $r^{2}$$ , we can substitute and plug in the values provided:

$V=(\pi $r^{2}$)h$

$V=(\pi $(6)^{2}$)(12)$

$V=432\pi$

A given rectangular prism has a length of , a width of , and a height of . What is its volume?

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The volume of a given rectangular prism is the product of its length , width , and height , or . Plugging in the values provided:

Ron is making a wooden chest. The chest needs to measure 45 inches by 72 inches by 36 inches. What volume will the chest be able to hold in cubic feet?

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Because we are asked for an answer in cubic feet, convert all measurments to feet right away

45 inches: 3.5 feet

72 inches: 6 feet

36 inches: 3 feet

Volume of a rectangular prism is given by:

V=lwh

So

V=363.5=67.5 $ft^3$

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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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 .

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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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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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,

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

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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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

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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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}$$ .

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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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

.

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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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}$$

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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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}$$

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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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}$$

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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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:

.

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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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}$$