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SAT II Math I : 3-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

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Example Question #91 : Geometry

What is the volume of a regular tetrahedron with an edge length of 6?

Possible Answers:

25.5

32.1

26.6

27.8

24.2

Correct answer:

25.5

Explanation:

The volume of a tetrahedron can be solved for by using the equation:

$V= \frac{a^3}{6\sqrt{2}}$

where is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for .

$V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}$

${\color{Blue} V=25.5}$

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Example Question #81 : Advanced Geometry

What is the volume of the tetrahedron shown below?

Screen shot 2015 10 21 at 7.16.10 pm

Possible Answers:

$\frac{256}{6\sqrt2}$

25.63

$\frac{256}{3\sqrt2}$

512

$\frac{36}{\sqrt2}$

Correct answer:

$\frac{256}{3\sqrt2}$

Explanation:

The volume of a tetrahedron is $V=\frac{s^3}{6\sqrt2}$ .

This tetrahedron has a side with a length of 8.

$V=\frac{8^3}{6\sqrt2}$ , which becomes $\frac{512}{6\sqrt2}$ .

You can reduce that answer further, so that it becomes

$\frac{256}{3\sqrt2}$ .

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Example Question #21 : Volume

Find the volume of sphere whose radius is 15.

Possible Answers:

$1200\pi$

$36000\pi$

$562.5\pi$

$300\pi$

$4500\pi$

Correct answer:

$4500\pi$

Explanation:

The volume of a sphere is given by the equation:

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

The problem says that the radius is 15, so plug in 15 for r and simplify.

$V=\frac{4}{3} \pi r^3=\frac{4}{3} \pi 15^3=4500\pi$

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Example Question #101 : Geometry

Find the volume of a sphere with the diameter of .

Possible Answers:

$\frac{9}{4}\pi$

$12\pi$

$\frac{9}{2}\pi$

$27\pi$

$9\pi$

Correct answer:

$\frac{9}{2}\pi$

Explanation:

Write the equation for the volume of a sphere.

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

The radius is half the diameter, or $\frac{3}{2}$ .

Substitute the radius into the equation.

$V=\frac{4}{3}\pi( \frac{3}{2})^3$

$V=\frac{4}{3}\pi( \frac{27}{8})$

Reduce the fractions.

The answer is: $\frac{9}{2}\pi$

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Example Question #21 : 3 Dimensional Geometry

Determine the volume of a rectangular prism if the length is $\frac{1}{2}$ , width is $\frac{2}{3}$ , and the height is $\frac{5}{6}$ .

Possible Answers:

$\frac{2}{3}$

$\frac{5}{18}$

$\frac{9}{5}$

$\frac{5}{9}$

$\frac{1}{3}$

Correct answer:

$\frac{5}{18}$

Explanation:

Write the formula for the volume of a rectangular prism.

$A=\textup{ (Length) (Width) (Height)}$

Substitute the dimensions into the formula.

$A = \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6} = \frac{10}{36}$

Reduce this fraction.

$\frac{5\times 2}{18\times 2}$

The answer is: $\frac{5}{18}$

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Example Question #21 : 3 Dimensional Geometry

The length of a box is half its height and two-thirds its width. The volume of the box is four cubic meters. Give the length of the box to the nearest centimeter.

Possible Answers:

$229\textrm{ cm}$

None of these

$144 \textrm{ cm}$

$175 \textrm{ cm}$

$110 \textrm{ cm}$

Correct answer:

$110 \textrm{ cm}$

Explanation:

Call , , and the length, width, and height of the crate.

The length of the crate is half its height, so

$L = \frac{1}{2} H$

$2 \cdot L = 2 \cdot \frac{1}{2} H$

2 L = H

The length of the crate is two-thirds its width, so

$L = \frac{2}{3} W$

$\frac{3} {2} \cdot L = \frac{3} {2} \cdot \frac{2}{3} W$

$\frac{3} {2} L = W$

The dimensions of the crate in terms of are , $\frac{3} {2} L$ , and . The volume is their product:

V= LWH

Substitute:

LWH = V

$L \cdot \frac{3} {2} L \cdot 2L = 4$

$3 L ^{3} = 4$

Solve for :

$3 L ^{3} = 4$

$\frac{3 L ^{3} }{3} = \frac{4 }{3}$

$L ^{3} = \frac{4 }{3}$

$L =\sqrt[3]{ \frac{4 }{3}}$

$L \approx 1.101$ meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

$1.101 \cdot 100 = 110.1$ ,

which rounds to 110 centimeters.

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Example Question #1 : Surface Area

A circular swimming pool has diameter 32 meters and depth meters throughout. Which of the following expressions gives the total area of the inside of the pool, in square meters?

Possible Answers:

None of the other responses is correct.

$256\pi + 16 \pi t$

$512\pi + 32 \pi t$

$256\pi + 32 \pi t$

$512\pi + 16 \pi t$

Correct answer:

$256\pi + 32 \pi t$

Explanation:

The bottom of the pool is a circle with diameter 32, and, subsequently, radius half this, or 16; its area is

$B = \pi r^{2} = \pi \times 16^{2} = 256 \pi$

The side of the pool is the lateral surface of a cylinder with radius 16 and height ; the area of this is

$LA = 2 \pi r h = 2 \pi \cdot 16 \cdot t = 32 \pi t$

The area of the inside of the pool is the sum of these two, or

$A = 256 \pi + 32 \pi t$

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Example Question #2 : Surface Area

A circular swimming pool at an apartment complex has diameter 50 feet and depth six feet throughout.

The apartment manager needs to get the interior of the swimming pool painted. The paint she wants to use covers 350 square feet per gallon. How many one-gallon cans of paint will she need to purchase?

Possible Answers:

The correct answer is not given among the other responses.

Correct answer:

Explanation:

The pool can be seen as a cylinder with depth (or height) six feet and a base with diameter 50 feet - and radius half this, or 25 feet.

The bottom of the pool - the base of the cylinder - is a circle with radius 25 feet, so its area is

$B = \pi r ^{2} = 3.14 \cdot 25 \cdot 25 = 1,962.5$ square feet.

Its side - the lateral face of the cylinder - has area

$LA = 2 \pi r h = 2 \cdot 3.14 \cdot 25 \cdot 6 = 942$ square feet.

Their sum - the total area to be painted - is 1,962.5 + 942 = 2,904.5 square feet. Since one gallon of paint covers 350 square feet, divide:

$2,904.5 \div 350 \approx 8.3$

Eight cans of paint and part of a ninth will be required, so the correct response is nine.

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Example Question #1 : Surface Area

Example cylinder

Figure not drawn to scale.

Find the surface area of the cylinder above.

Possible Answers:

94.25 in²

73.44 in²

87.25 in²

122.14 in²

56.55 in²

Correct answer:

94.25 in²

Explanation:

In order to find the surface area of a cylinder, you need to find the surface areas of the circles that are the top and bottom of the cylinder (2 x pi x radius²) and add it to the surface are of the rectangle that is the side of the cylinder (diameter x height).

$\small \small \small \small Surface Area=(2\pi r{^2})+(\pi2r\times h)$

$\small \small (2\pi\times 3{^2})+(6\pi\times 2)$

$\small \small (6.28\times 9)+(18.84\times 2)$

$\small 56.54 + 37.70=94.25$

The surface area of the cylinder is 94.25 in²

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Example Question #4 : Surface Area

Example images

Figure not drawn to scale

What is the surface area of the sphere above?

Possible Answers:

615.75 yd²

712.12 yd²

512.63 yd²

615.75 yd³

815.44 yd²

Correct answer:

615.75 yd²

Explanation:

In order to find the surface area of a sphere, you must use the equation below:

$\small Surface Area=4\pi r{^{2}}$

$\small 4\pi(7{^{2}})=4\pi (49)=196\pi=615.75$

The surface area of the sphere is 615.75 yd²

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SAT II Math I : 3-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #91 : Geometry

Example Question #81 : Advanced Geometry

Example Question #21 : Volume

Example Question #101 : Geometry

Example Question #21 : 3 Dimensional Geometry

Example Question #21 : 3 Dimensional Geometry

Example Question #1 : Surface Area

Example Question #2 : Surface Area

Example Question #1 : Surface Area

The surface area of the cylinder is 94.25 in²

Example Question #4 : Surface Area

What is the surface area of the sphere above?

The surface area of the sphere is 615.75 yd²

All SAT II Math I Resources

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HIGH SCHOOL

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K-8

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math tutoring

science tutoring

foreign languages

elementary tutoring

other

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SAT II Math I : 3-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #91 : Geometry

Example Question #81 : Advanced Geometry

Example Question #21 : Volume

Example Question #101 : Geometry

Example Question #21 : 3 Dimensional Geometry

Example Question #21 : 3 Dimensional Geometry

Example Question #1 : Surface Area

Example Question #2 : Surface Area

Example Question #1 : Surface Area

The surface area of the cylinder is 94.25 in2

Example Question #4 : Surface Area

What is the surface area of the sphere above?

The surface area of the sphere is 615.75 yd2

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

The surface area of the cylinder is 94.25 in²

The surface area of the sphere is 615.75 yd²