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ISEE Upper Level Math : Geometry

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Example Questions

Example Question #1 : How To Find The Volume Of A Pyramid

A 300 foot tall pyramid has a square base measuring feet on each side. What is the volume of the pyramid?

Possible Answers:

$160,000 ft^{3}$

$12,000 ft^{3}$

$480,000 ft^{3}$

$4,000 ft^{3}$

$160,000 ft^{2}$

Correct answer:

$160,000 ft^{3}$

Explanation:

In order to find the area of a triangle, we use the formula $\frac{1}{3}Bh$ . In this case, since the base is a square, we can replace with $s^{2}$ , so our formula for volume is $\frac{1}{3}s^{2}h$ .

Since the length of each side of the base is feet, we can substitute it in for .

$\frac{1}{3}\cdot (40)^{2}\cdot h$

We also know that the height is 300 feet, so we can substitute that in for .

$\frac{1}{3}\cdot (40)^{2}\cdot 300$

This gives us an answer of $160,000 ft^{3}$ .

It is important to remember that volume is expressed in units cubed.

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Example Question #3 : Pyramids

The height of a right pyramid is feet. Its base is a square with sidelength feet. Give its volume in cubic inches.

Possible Answers:

$15,552 \textrm{ in}^{3}$

$41,472 \textrm{ in}^{3}$

$10,368 \textrm{ in}^{3}$

$31,104 \textrm{ in}^{3}$

$62,208\textrm{ in}^{3}$

Correct answer:

$10,368 \textrm{ in}^{3}$

Explanation:

Convert each of the measurements from feet to inches by multiplying by .

Height: $2 \times 12 = 24$ inches

Sidelength of base: $3 \times 12 = 36$ inches

The base of the pyramid has area

$B = s^{2} = 36^{2} = 1,296$ square inches.

Substitute B = 1,296, h = 24 into the volume formula:

$V = \frac {1}{3} Bh$

$V = \frac {1}{3} \cdot 1,296 \cdot 24 =10,368$ cubic inches

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Example Question #1 : How To Find The Volume Of A Pyramid

The height of a right pyramid is inches. Its base is a square with sidelength inches. Give its volume in cubic feet.

Possible Answers:

$9 \frac {1}{3} \textrm{ ft}^3$

$4 \frac {2}{3} \textrm{ ft}^3$

$2 \frac {1}{3} \textrm{ ft}^3$

$14 \textrm{ ft}^3$

$4 \textrm{ ft}^3$

Correct answer:

$4 \frac {2}{3} \textrm{ ft}^3$

Explanation:

Convert each of the measurements from inches to feet by dividing by .

Height: $42 \div 12 = 3 \frac{1}{2}$ feet

Sidelength: $24 \div 12 = 2$ feet

The base of the pyramid has area

$B = s^{2} = 2^{2} = 4$ square feet.

Substitute $B = 4, h = 3 \frac{1}{2}$ into the volume formula:

$V = \frac {1}{3} Bh$

$V = \frac {1}{3}\cdot 4 \cdot 3 \frac{1}{2}$

$V = \frac {1}{3} \cdot \frac {4}{1} \cdot \frac{7}{2}$

$V = \frac {1}{3} \cdot \frac {2}{1} \cdot \frac{7}{1} = \frac {14}{3} = 4 \frac {2}{3}$ cubic feet

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Example Question #3 : Volume Of A Pyramid

The height of a right pyramid and the sidelength of its square base are equal. The perimeter of the base is 3 feet. Give its volume in cubic inches.

Possible Answers:

$243 \textrm{ in}^{3}$

$27 \textrm{ in}^{3}$

$576 \textrm{ in}^{3}$

$81 \textrm{ in}^{3}$

$729 \textrm{ in}^{3}$

Correct answer:

$243 \textrm{ in}^{3}$

Explanation:

The perimeter of the square base, feet, is equivalent to $3 \times 12 = 36$ inches; divide by to get the sidelength of the base - and the height: $36 \div 4 = 9$ inches.

The area of the base is therefore $B = s^{2} = 9^{2} = 81$ square inches.

In the formula for the volume of a pyramid, substitute B = 81,h=9 :

$V = \frac{1}{3} Bh = \frac{1}{3}\cdot 81 \cdot 9 = 243$ cubic inches.

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Example Question #2 : Pyramids

What is the volume of a pyramid with the following measurements?

$length=7; width=6;height=9;slant\ length=11$

Possible Answers:

154

126

462

378

Correct answer:

126

Explanation:

The volume of a pyramid can be determined using the following equation:

$V=\frac{1}{3}lwh=\frac{1}{3}(7)(6)(9)=126$

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Example Question #11 : Solid Geometry

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:

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

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

n = 20

n = 30

n = 10

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 #321 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Find the volume of a pyramid with the following measurements:

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

Possible Answers:

$24\text{cm}^3$

$126\text{cm}^3$

$288\text{cm}^3$

$154\text{cm}^3$

$96\text{cm}^3$

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:

$112\text{in}^3$

$426\text{in}^3$

$168\text{in}^3$

$336\text{in}^3$

$221\text{in}^3$

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 #322 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Possible Answers:

$426\text{in}^3$

$168\text{in}^3$

$221\text{in}^3$

$336\text{in}^3$

$112\text{in}^3$

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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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Geometry

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Volume Of A Pyramid

Example Question #3 : Pyramids

Example Question #1 : How To Find The Volume Of A Pyramid

Example Question #3 : Volume Of A Pyramid

Example Question #2 : Pyramids

Example Question #11 : Solid Geometry

Example Question #3 : Pyramids

Example Question #321 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #2 : Pyramids

Example Question #322 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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