We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

SAT Math : Solid Geometry

Study concepts, example questions & explanations for SAT Math

Create An Account

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 … 8 9 10 11 12 13 14 15 16 17 18 19 Next →

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

At x = 3, the line y = 4x + 12 intersects the surface of a sphere that passes through the xy-plane. The sphere is centered at the point at which the line passes through the x-axis. What is the volume, in cubic units, of the sphere?

Possible Answers:

None of the other answers

816π√(11)

4896π√(17)

4896π

2040π√(7)

Correct answer:

4896π√(17)

Explanation:

We need to ascertain two values: The center point and the point of intersection with the surface. Let's do that first:

The center is defined by the x-intercept. To find that, set the line equation equal to 0 (y = 0 at the x-intercept):

0 = 4x + 12; 4x = –12; x = –3; Therefore, the center is at (–3,0)

Next, we need to find the point at which the line intersects with the sphere's surface. To do this, solve for the point with x-coordinate at 3:

y = 4 * 3 + 12; y = 12 + 12; y = 24; therefore, the point of intersection is at (3,24)

Reviewing our data so far, this means that the radius of the sphere runs from the center, (–3,0), to the edge, (3,24). If we find the distance between these two points, we can ascertain the length of the radius. From that, we will be able to calculate the volume of the sphere.

The distance between these two points is defined by the distance formula:

d = √( (x₁ – x₀)² + (y₁ – y₀)²)

For our data, that is:

√( (3 + 3)² + (24 – 0)²) = √( 6² + 24²) = √(36 + 576) = √612 = √(2 * 2 * 3 * 3 * 17) = 6√(17)

Now, the volume of a sphere is defined by: V = (4/3)πr³

For our data, that would be: (4/3)π * (6√(17))³ = (4/3) * 6³ * 17√(17) * π = 4 * 2 * 6² * 17√(17) * π = 4896π√(17)

Report an Error

Example Question #1 : Spheres

The surface area of a sphere is 676π in². How many cubic inches is the volume of the same sphere?

Possible Answers:

8788π

(8788π)/3

(2197π)/3

2028π

(2028π)/3

Correct answer:

(8788π)/3

Explanation:

To begin, we must solve for the radius of our sphere. To do this, recall the equation for the surface area of a sphere: A = 4πr²

For our data, that is: 676π = 4πr²; 169 = r²; r = 13

From this, it is easy to solve for the volume of the sphere. Recall the equation:

V = (4/3)πr³

For our data, this is: V = (4/3)π * 13³ = (4π * 2197)/3 = (8788π)/3

Report an Error

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

What is the difference between the volume and surface area of a sphere with a radius of 6?

Possible Answers:

144π

216π

288π

720π

133π

Correct answer:

144π

Explanation:

Surface Area = 4πr² = 4 * π * 6² = 144π

Volume = 4πr³/3 = 4 * π * 6³ / 3 = 288π

Volume – Surface Area = 288π – 144π = 144π

Report an Error

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

A sphere is perfectly contained within a cube that has a volume of 216 units. What is the volume of the sphere?

Possible Answers:

$12\pi$

$288\pi$

$36\pi$

$48\pi$

$9\pi$

Correct answer:

$36\pi$

Explanation:

To begin, we must determine the dimensions of the cube. This is done by solving the simple equation: V=s^3

We know the volume is 216, allowing us to solve for the length of a side of the cube.

216=s^3

Taking the cube root of both sides, we get s = 6.

The diameter of the sphere will be equal to side of the cube, since the question states that the sphere is perfectly contained. The diameter of the sphere will be 6, and the radius will be 3.

We can plug this into the equation for volume of a sphere:

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

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

We can cancel out the 3 in the denominator.

$V=4\pi (3)^2$

Simplify.

$V=4\pi (9)$

$V=36\pi$

Report an Error

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

The surface area of a sphere is $36\pi\ mm^2$ . Find the volume of the sphere in cubic millimeters.

Possible Answers:

$9\pi\ mm^3$

$36\pi\ mm^3$

$4\pi\ mm^3$

$6\pi\ mm^3$

$3\pi\ mm^3$

Correct answer:

$36\pi\ mm^3$

Explanation:

$SA=4\pi r^2$

$36\pi=4\pi r^2$

36=4r^2

9=r^2

3=r

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

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

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

$V=4(9)\pi$

$V=36\pi$

Report an Error

Example Question #181 : Psat Mathematics

A solid hemisphere has a radius of length r. Let S be the number of square units, in terms of r, of the hemisphere's surface area. Let V be the number of cubic units, in terms of r, of the hemisphere's volume. What is the ratio of S to V?

Possible Answers:

$\frac{4r}{3}$

$\frac{9r}{2}$

$\frac{9}{2r}$

$\frac{3}{r}$

$3r$

Correct answer:

$\frac{9}{2r}$

Explanation:

First, let's find the surface area of the hemisphere. Because the hemisphere is basically a full sphere cut in half, we need to find half of the surface area of a full sphere. However, because the hemisphere also has a circular base, we must then add the area of the base.

$S = \frac{1}{2}\cdot$ (surface area of sphere) + (surface area of base)

The surface area of a sphere with radius r is equal to $4\pi r^2$ . The surface area of the base is just equal to the surface area of a circle, which is $\pi r^2$ .

$S=\frac{1}{2}\cdot 4\pi r^2+\pi r^2=2\pi r^2+\pi r^2=3\pi r^2$

The volume of the hemisphere is going to be half of the volume of an entire sphere. The volume for a full sphere is $\frac{4}{3}\pi r^3$ .

$V = \frac{1}{2}\cdot$ (volume of sphere)

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

Ultimately, the question asks us to find the ratio of S to V. To do this, we can write S to V as a fraction.

$\frac{S}{V}=\frac{3\pi r^2}{\frac{2}{3}\pi r^3}$

In order to simplify this, let's multiply the numerator and denominator both by 3.

$\frac{S}{V}=\frac{3\pi r^2}{\frac{2}{3}\pi r^3}$ = $\frac{9\pi r^2}{2\pi r^3}=\frac{9}{2r}$

The answer is $\frac{9}{2r}$ .

Report an Error

Example Question #11 : How To Find The Volume Of A Sphere

Find the volume of a sphere with a radius of $\frac{3}{5}$ .

Possible Answers:

$9\pi$

$\frac{36}{125}\pi$

$\frac{36}{5}\pi$

$\frac{12}{25}\pi$

$\frac{36}{25}\pi$

Correct answer:

$\frac{36}{125}\pi$

Explanation:

Write the formula to find the volume of the sphere.

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

Substitute the radius and solve for the volume.

$V=\frac{4}{3}\pi (\frac{3}{5})^3=\frac{4}{3}\pi (\frac{27}{125})=\frac{36}{125}\pi$

Report an Error

Example Question #32 : Spheres

Six spheres have volumes that form an arithmetic sequence. The two smallest spheres have radii 4 and 6. Give the volume of the largest sphere.

Possible Answers:

$\frac{3,296}{3} \pi$

$784 \pi$

$196 \pi$

$\frac{10, 976 }{3}\pi$

$464 \pi$

Correct answer:

$\frac{3,296}{3} \pi$

Explanation:

The volume of a sphere with radius can be determined using the formula

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

The smallest sphere, with radius r = 4 , has volume

$V_{1} = \frac{4}{3} \pi \cdot 4 ^{3} = \frac{4}{3} \pi \cdot 64 = \frac{256}{3} \pi$ .

The second-smallest sphere, with radius r = 6 , has volume

$V_{2} = \frac{4}{3} \pi \cdot 6 ^{3} = \frac{4}{3} \pi \cdot 216= \frac{864}{3} \pi$ .

The volumes are in an arithmetic sequence; their common difference is the difference of these two volumes, or

$d = V_{2} - V_{1}= \frac{864}{3} \pi - \frac{256}{3} \pi = \frac{608}{3} \pi$

Since the six volumes are in an arithmetic sequence, the volume of the largest of the six spheres - that is, the sixth-smallest sphere - is

$V_{6}= V_{1}+ 5d$

$= \frac{256}{3} \pi + 5 \cdot \frac{608}{3} \pi$

$= \frac{256}{3} \pi + \frac{3,040}{3} \pi$

$= \frac{3,296}{3} \pi$

Report an Error

Example Question #33 : Spheres

The radii of six spheres form an arithmetic sequence. The smallest and largest spheres have radii 10 and 30, respectively. Give the volume of the second-largest sphere.

Possible Answers:

$2, 704 \pi$

$\frac{25,600}{9} \pi$

None of the other responses gives a correct answer.

$\frac{70, 304}{3} \pi$

$\frac{2,048,000}{81} \pi$

Correct answer:

$\frac{70, 304}{3} \pi$

Explanation:

The radii of the spheres form an arithmetic sequence, with

$r_{1} =1 0$ and $r_{6} =30$

The common difference can be computed as follows:

$r_{6} = (6-1) d + r_{1}$

30=5d + 10

5d= 20

d = 4

The second-largest - or fifth-smallest - sphere has radius:

$r_{5} = (5-1) d + r_{1} = 4 \cdot 4 + 10 = 16 + 10 = 26$

The volume of a sphere with radius can be determined using the formula

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

Set r = 26 :

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

$= \frac{4}{3} \pi \cdot 26^{3}$

$= \frac{4}{3} \pi \cdot 17,576$

$= \frac{70, 304}{3} \pi$

Report an Error

Example Question #34 : Spheres

A sphere with radius fits perfectly inside of a cube so that the sides of the cube are barely touching the sphere. What is the volume of the cube that is not occupied by the sphere?

Possible Answers:

$2744-\frac{4}{3}\pi$

$1000-\frac{4}{3}\pi$

$\frac{1372}{3}\pi$

$2744-\frac{1372}{3}\pi$

$\frac{1372}{3}\pi - 2744$

Correct answer:

$2744-\frac{1372}{3}\pi$

Explanation:

Because the sides of the interior of the cube are tangent to the sphere, we know that the length of each side is equal to the diameter of the sphere. Since the radius of this sphere is , then its diameter is .

To find the volume that is not occupied by the sphere, we will subtract the sphere's volume from the volume of the cube.

The volume of the cube is:

$V=lwh=14\cdot14\cdot14=2744$

The volume of the sphere is:

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\cdot\pi\cdot7^3 = \frac{1372}{3}\pi$

Therefore, with these values, the volume of the cube not occupied by the sphere is:

$2744-\frac{1372}{3}\pi$

Report an Error

← Previous 1 2 … 8 9 10 11 12 13 14 15 16 17 18 19 Next →

View SAT Mathematics Tutors

Erik
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Science, Physics. University of California-San Diego, Master of Scien...

View SAT Mathematics Tutors

Beulah
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Engineering, Electrical Engineering.

View SAT Mathematics Tutors

Christopher
Certified Tutor

Rose-Hulman Institute of Technology, Bachelor of Science, Physics. University of Washington, Master of Science, Physics.

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

SAT Math Tutors in Top Cities:

Atlanta SAT Math Tutors, Austin SAT Math Tutors, Boston SAT Math Tutors, Chicago SAT Math Tutors, Dallas Fort Worth SAT Math Tutors, Denver SAT Math Tutors, Houston SAT Math Tutors, Kansas City SAT Math Tutors, Los Angeles SAT Math Tutors, Miami SAT Math Tutors, New York City SAT Math Tutors, Philadelphia SAT Math Tutors, Phoenix SAT Math Tutors, San Diego SAT Math Tutors, San Francisco-Bay Area SAT Math Tutors, Seattle SAT Math Tutors, St. Louis SAT Math Tutors, Tucson SAT Math Tutors, Washington DC SAT Math Tutors

Popular Courses & Classes

SSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Seattle, SSAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Phoenix, ISEE Courses & Classes in Los Angeles, MCAT Courses & Classes in Chicago, GMAT Courses & Classes in Denver, LSAT Courses & Classes in Houston, GRE Courses & Classes in Boston, SSAT Courses & Classes in Houston

Popular Test Prep

SAT Test Prep in Denver, MCAT Test Prep in Chicago, GRE Test Prep in Miami, LSAT Test Prep in New York City, GRE Test Prep in Atlanta, SSAT Test Prep in Atlanta, SAT Test Prep in Seattle, GRE Test Prep in San Francisco-Bay Area, ACT Test Prep in Denver, SAT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

SAT Math : Solid Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

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

Example Question #1 : Spheres

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

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

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

Example Question #181 : Psat Mathematics

Example Question #11 : How To Find The Volume Of A Sphere

Example Question #32 : Spheres

Example Question #33 : Spheres

Example Question #34 : Spheres

All SAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: