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ISEE Upper Level Math : Solid Geometry

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Example Questions

Example Question #7 : Tetrahedrons

A regular tetrahedron comprises four faces, each of which is an equilateral triangle. Each edge has length 16. What is its surface area?

Possible Answers:

$256 \sqrt{2}$

$682\frac{2}{3}$

$341\frac{1}{3}$

256

$256 \sqrt{3}$

Correct answer:

$256 \sqrt{3}$

Explanation:

The area of each face of a regular tetrahedron, that face being an equilateral triangle, is

$A = \frac{s^{2}\sqrt{3}}{4}$

Substitute edge length 16 for :

$A = \frac{16^{2}\sqrt{3}}{4} = \frac{256\sqrt{3}}{4} = 64 \sqrt{3}$

The tetrahedron has four faces, so the total surface area is

$SA = 4 \cdot 64 \sqrt{3} = 256 \sqrt{3}$

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,0), (12,0,0), (0,12,0), (0,0, 12) .

What is the surface area of this tetrahedron?

Possible Answers:

$288 \sqrt{3}$

$288 \sqrt{2}$

$216+ 72 \sqrt{2}$

$216 \sqrt{2} + 72 \sqrt{3}$

$216+ 72 \sqrt{3}$

Correct answer:

$216+ 72 \sqrt{3}$

Explanation:

The tetrahedron looks like this:

Tetrahedron

is the origin and A,B,C are the other three points, which are each twelve units away from the origin on one of the three (mutually perpendicular) axes.

Three of the surfaces are right triangles with two legs of length 12, so the area of each is

$A = \frac{1}{2} \cdot 12 \cdot 12 = 72$ .

The fourth surface, $\Delta ABC$ , has three edges each of which is the hypotenuse of an isosceles right triangle with legs 12, so each has length $12\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\frac{(12 \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{144 \cdot 2 \cdot \sqrt{3}}{4} = 72 \sqrt{3}$

The surface area is therefore

$3 \times 72 + 72 \sqrt{3} = 216+ 72 \sqrt{3}$ .

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,0), (n,0,0), (0,n,0), (0,0, n) .

In terms of , give the surface area of this tetrahedron.

Possible Answers:

$2n^{2}$

$\frac{3n^{2}+ n^{2} \sqrt{3}}{2}$

$4n^{2}$

$2n^{2}\sqrt{3}$

$\frac{3n^{2}+ n^{2} \sqrt{2}}{2}$

Correct answer:

$\frac{3n^{2}+ n^{2} \sqrt{3}}{2}$

Explanation:

The tetrahedron looks like this:

Tetrahedron

is the origin and A,B,C are the other three points, which are units away from the origin, each along one of the three (perpendicular) axes.

Three of the surfaces are right triangles with two legs of length 12, so the area of each is

$A = \frac{1}{2} \cdot n\cdot n = \frac{n^{2}}{2}$ .

The fourth surface, $\Delta ABC$ , has three edges each of which is the hypotenuse of an isosceles right triangle with legs , so each has length $n\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\frac{(n \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{n^{2}\cdot 2 \cdot \sqrt{3}}{4} = \frac{n^{2} \sqrt{3}}{2}$

The surface area is therefore

$3 \cdot \frac{n^{2}}{2} + \frac{n^{2} \sqrt{3}}{2} = \frac{3n^{2}+ n^{2} \sqrt{3}}{2}$ .

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Example Question #1 : Spheres

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its diameter?

Possible Answers:

30ft

15ft

25 ft

22.5ft

Correct answer:

30ft

Explanation:

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its diameter?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$900 \pi ft^2=4 \pi r^2$

First divide both sides by $4\pi$ .

225ft^2=r^2

Then square root both sides to get our radius:

r=15ft

Now, because the question is asking for our diameter and not our radius, we need to double our radius to get our answer:

d=2r=2*15ft=30 ft

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Example Question #2 : Spheres

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Possible Answers:

Cannot be determined from the information provided

25 m

12.5m

15m

Correct answer:

15m

Explanation:

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4\pi r^2$

Now, plug in our surface area and solve with algebra:

$900 \pi m^2=4\pi r^2$

Get rid of the pi

900 m^2=4 r^2

Divide by 4

225 m^2= r^2

Square root both sides to get our answer:

r=15m

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Example Question #3 : Spheres

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its radius?

Possible Answers:

15 ft

25 ft

225 ft

45 ft

Correct answer:

15 ft

Explanation:

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its radius?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$900 \pi ft^2=4 \pi r^2$

First divide both sides by $4\pi$ .

225ft^2=r^2

Then square root both sides to get our answer:

r=15ft

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Example Question #1 : Volume Of A Sphere

In terms of $\pi$ , give the volume, in cubic inches, of a spherical water tank with a diameter of 20 feet.

Possible Answers:

$230,400 \pi\textrm{ in}^{3}$

$2,304,000 \pi \textrm{ in}^{3}$

$18,432,000\pi\textrm{ in}^{3}$

$307,200\pi\textrm{ in}^{3}$

$57,600 \pi\textrm{ in}^{3}$

Correct answer:

$2,304,000 \pi \textrm{ in}^{3}$

Explanation:

20 feet = $20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set r = 120 , substitute in the volume formula, and solve for :

$V = \frac{4}{3} \pi r^{3}$

$V = \frac{4}{3} \pi \cdot 120^{3}$

$V = \frac{4}{3} \pi \cdot 1,728,000$

$V = 2,304,000 \pi$ cubic inches

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Example Question #71 : Solid Geometry

A sphere has diameter 3 meters. Give its volume in cubic centimeters (leave in terms of $\pi$ ).

Possible Answers:

$36,000,000\pi \textrm{ cm}^{3}$

$1,125,000\pi \textrm{ cm}^{3}$

$4,500,000\pi \textrm{ cm}^{3}$

$3,375,000\pi \textrm{ cm}^{3}$

$27,000,000\pi \textrm{ cm}^{3}$

Correct answer:

$4,500,000\pi \textrm{ cm}^{3}$

Explanation:

The diameter of 3 meters is equal to $3 \times 100 = 300$ centimeters; the radius is half this, or 150 centimeters. Substitute r = 150 in the volume formula:

$V= \frac{4}{3} \pi r^{3}$

$V= \frac{4}{3} \cdot \pi \cdot 150^{3}$

$V= \frac{4}{3} \cdot 150 \cdot 150 \cdot 150\cdot \pi$

$V= 4,500,000\pi$ cubic centimeters

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Example Question #1 : Spheres

A spherical buoy has a radius of 5 meters. What is the volume of the buoy?

Possible Answers:

$165 \pi m^3$

$16\frac{2}{3}\pi m^3$

$166\frac{2}{3}\pi m^3$

$166\frac{2}{3} m^3$

Correct answer:

$166\frac{2}{3}\pi m^3$

Explanation:

A spherical buoy has a radius of 5 meters. What is the volume of the buoy?

To find the volume of a sphere, use the following formula:

$V_{sphere}=\frac{4}{3}\pi r^3$

All we have to do is plug in 5 meters and simplify:

$V_{sphere}=\frac{4}{3}\pi (5m)^3=\frac{4}{3}\pi*125m^3=166\frac{2}{3}\pi m^3$

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Example Question #2 : Spheres

You have a ball with a radius of 12 cm, what is its volume?

Possible Answers:

$144 \pi cm^3$

$192 \pi cm^3$

$404 \pi cm^3$

$2304 \pi cm^3$

Correct answer:

$2304 \pi cm^3$

Explanation:

You have a ball with a radius of 12 cm, what is its volume?

The volume of a sphere can be found via the following formula:

$V_{sphere}=\frac{4}{3}\pi r^3$

We know our radius, so all we need to do is plug in and simplify:

$V_{sphere}=\frac{4}{3}\pi (12cm)^3=2304 \pi cm^3$

So we have our answer:

$2304 \pi cm^3$

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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 #7 : Tetrahedrons

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Example Question #1 : Spheres

Example Question #2 : Spheres

Example Question #3 : Spheres

Example Question #1 : Volume Of A Sphere

Example Question #71 : Solid Geometry

Example Question #1 : Spheres

Example Question #2 : Spheres

All ISEE Upper Level Math Resources

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