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ISEE Upper Level Math : Solid Geometry

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Example Questions

Example Question #2 : Pyramids

A right regular pyramid with volume 1,000 has its vertices at the points

(0,0,0), (n,0, 0), (n,n,0), (0, n, n), (n,n,n)

where n > 0 .

Evaluate .

Possible Answers:

n = 30

$n = 10 \sqrt[3]{3 }$

n = 20

n = 10

$n = 10 \sqrt[3]{2}$

Correct answer:

$n = 10 \sqrt[3]{3 }$

Explanation:

The pyramid has a square base that is units by units, and its height is units, as can be seen from this diagram,

Pyramid

The square base has area $B = n^{2}$ ; the pyramid has volume $V = \frac{1}{3} \cdot n^{2} \cdot n = \frac{1}{3} n^{3}$

Since the volume is 1,000, we can set this equal to 1,000 and solve for :

$\frac{1}{3} n^{3} = 1,000$

$n^{3} = 3,000$

$n = \sqrt[3]{3,000} = \sqrt[3]{1,000} \cdot \sqrt[3]{3 } = 10 \sqrt[3]{3 }$

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Example Question #3 : Pyramids

Find the volume of a pyramid with the following measurements:

length = 4in
width = 3in
height = 5in

Possible Answers:

$80\text{in}^3$

$40\text{in}^3$

$20\text{in}^3$

$60\text{in}^3$

$12\text{in}^3$

Correct answer:

$20\text{in}^3$

Explanation:

To find the volume of a pyramid, we will use the following formula:

$V = \frac{l \cdot w \cdot h}{3}$

where l is the length, w is the width, and h is the height of the pyramid.

Now, we know the base of the pyramid has a length of 4in. We also know the base of the pyramid has a width of 3in. We also know the pyramid has a height of 5in.

Knowing this, we can substitute into the formula. We get

$V = \frac{4\text{in} \cdot 3\text{in} \cdot 5\text{in}}{3}$

$V = \frac{60\text{in}^3}{3}$

$V = 20\text{in}^3$

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Example Question #2 : How To Find The Volume Of A Pyramid

Find the volume of a pyramid with the following measurements:

length = 4cm
width = 9cm
height = 8cm

Possible Answers:

$24\text{cm}^3$

$154\text{cm}^3$

$288\text{cm}^3$

$96\text{cm}^3$

$126\text{cm}^3$

Correct answer:

$96\text{cm}^3$

Explanation:

To find the volume of a pyramid, we will use the following formula:

$V = \frac{l \cdot w \cdot h}{3}$

where l is the length, w is the width, and h is the height of the pyramid.

Now, we know the following measurements:

length = 4cm
width = 9cm
height = 8cm

Knowing this, we can substitute into the formula. We get

$V = \frac{4\text{cm} \cdot 9\text{cm} \cdot 8\text{cm}}{3}$

$V = \frac{288\text{cm}^3}{3}$

$V = 96\text{cm}^3$

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Example Question #2 : Pyramids

Find the volume of a pyramid with the following measurements:

length: 7in
width: 6in
height: 8in

Possible Answers:

$112\text{in}^3$

$336\text{in}^3$

$426\text{in}^3$

$221\text{in}^3$

$168\text{in}^3$

Correct answer:

$112\text{in}^3$

Explanation:

To find the volume of a pyramid, we will use the following formula:

$V = \frac{lwh}{3}$

where l is the length, w is the width, and h is the height of the pyramid.

Now, we know the following measurements:

length: 7in
width: 6in
height: 8in

So, we get

$V = \frac{ 7\text{in} \cdot 6\text{in} \cdot 8\text{in}}{3}$

$V = \frac{42\text{in}^2 \cdot 8\text{in}}{3}$

$V = \frac{336\text{in}^3}{3}$

$V = 112\text{in}^3$

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Example Question #11 : Solid Geometry

Find the volume of a pyramid with the following measurements: length= in, width= in, height= in

Possible Answers:

$112\text{in}^3$

$168\text{in}^3$

$336\text{in}^3$

$221\text{in}^3$

$426\text{in}^3$

Correct answer:

$112\text{in}^3$

Explanation:

To find the volume of a pyramid, we will use the following formula:

$V = \frac{lwh}{3}$

where l is the length, w is the width, and h is the height of the pyramid.

Now, we know the following measurements:

length: 7in
width: 6in
height: 8in

So, we get

$V = \frac{ 7\text{in} \cdot 6\text{in} \cdot 8\text{in}}{3}$

$V = \frac{42\text{in} \cdot 8\text{in}}{3}$

$V = \frac{336\text{in}}{3}$

$V = 112\text{in}^3$

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Example Question #921 : Grade 7

If a cube is inches tall, what is its volume?

Possible Answers:

512

500

Not enough information provided.

384

Correct answer:

512

Explanation:

To find the volume of a cube, we multiply length by width by height, which can be represented with the forumla $v=l\cdot w\cdot h$ . Since a cube has equal sides, we can use for all three values.

$8\cdot 8\cdot 8=512\: in ^{3}$

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Example Question #331 : Isee Upper Level (Grades 9 12) Mathematics Achievement

What is the volume of a cube with a side length equal to inches?

Possible Answers:

$53\: in^{3}$

$21\: in^{3}$

$49\: in^{3}$

$343\: in^{3}$

Correct answer:

$343\: in^{3}$

Explanation:

The volume of a a cube (or rectangular prism) can be solved using the following equation:

$V=l\times\ w\times\ h$

$V=7in\times 7in\times 7in=343in^3$

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Example Question #1 : Finding Volume Of A Cube

Give the volume of a cube with surface area 150 square inches.

Possible Answers:

$375\textup{ in}^{3}$

$250\sqrt{3}\textup{ in}^{3}$

$250\textup{ in}^{3}$

$125\textup{ in}^{3}$

$125\sqrt{3}\textup{ in}^{3}$

Correct answer:

$125\textup{ in}^{3}$

Explanation:

Let be the length of one edge of the cube. Since its surface area is 150 square inches, one face has one-sixth of this area, or $\frac{150}{6} = 25$ square inches. Therefore, $s^{2} = 25$ , and s = 5 .

The volume is the cube of this, or $V= 5^{3}=125$ cubic inches.

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Example Question #1 : Cubes

What is the volume of a cube in which the edge is equal to $\sqrt[3]{x}$ , and the value of is:

$x=5(4+3^{2})-3$

Possible Answers:

Correct answer:

Explanation:

First, the value of x must be solved for:

$x=5(4+3^{2})-3$

x=5(4+9)-3

x=5(13)-3

x=62

Given the edge of the cube is $\sqrt[3]{x}$ , plugging in the value of x results in $\sqrt[3]{62}$ . Thus, the area would be equal to this value cubed, which would result in 62.

Thus, 62 is the correct answer.

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Example Question #12 : Solid Geometry

The length of a diagonal of one face of a cube is . Give the volume of the cube.

Possible Answers:

$\frac{N^{3} \sqrt{3} }{9}$

$\frac{N^{3} \sqrt{3} }{3}$

$\frac{N^{3} \sqrt{2} }{2}$

$3N^{3}$

$\frac{N^{3} \sqrt{2} }{4}$

Correct answer:

$\frac{N^{3} \sqrt{2} }{4}$

Explanation:

Since a diagonal of a square face of the cube is, each side of each square has length $\frac{N}{\sqrt{2}}=\frac{N\sqrt{2}}{2}$ .

Cube this to get the volume of the cube:

$\left (\frac{N\sqrt{2}}{2} \right )^{3}$

$= \frac{N^{3}\left ( \sqrt{2} \right )^{3} }{2^{3} }$

$= \frac{N^{3} \cdot 2 \sqrt{2} }{8}$

$= \frac{N^{3} \sqrt{2} }{4}$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Solid Geometry

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #2 : Pyramids

Example Question #3 : Pyramids

Example Question #2 : How To Find The Volume Of A Pyramid

Example Question #2 : Pyramids

Example Question #11 : Solid Geometry

Example Question #921 : Grade 7

Example Question #331 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #1 : Finding Volume Of A Cube

Example Question #1 : Cubes

Example Question #12 : Solid Geometry

All ISEE Upper Level Math Resources

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