We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Advanced Geometry : How to find the surface area of a cone

Study concepts, example questions & explanations for Advanced Geometry

Create An Account

All Advanced Geometry Resources

6 Diagnostic Tests 57 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #541 : Geometry

What is the surface area of a cone with a radius of 4 and a height of 3?

Possible Answers:

$25\pi$

$40\pi$

$36\pi$

$16\pi$

$48\pi$

Correct answer:

$36\pi$

Explanation:

Here we simply need to remember the formula for the surface area of a cone and plug in our values for the radius and height.

$\Pi r^{2} + \Pi r\sqrt{r^{2} + h^{2}}= \Pi\ast 4^{2} + \Pi \ast 4\sqrt{4^{2} + 3^{2}} = 16\Pi + 4\Pi \sqrt{25} = 16\Pi + 20\Pi = 36\Pi$

Report an Error

Example Question #31 : Solid Geometry

The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?

Possible Answers:

90π

9π

54π

27π

81π

Correct answer:

81π

Explanation:

Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant height

LA = 2B

π(r)(l) = 2π(r²)

rl = 2r²

l = 2r

Cone

From the diagram, we can see that r² + h² = l². Since h = 9 and l = 2r, some substitution yields

r² + 9² = (2r)²

r² + 81 = 4r²

81 = 3r²

27 = r²

B = π(r²) = 27π

LA = 2B = 2(27π) = 54π

SA = B + LA = 81π

Report an Error

Example Question #11 : Surface Area

A right cone has a radius of 4R and a height of 3R. What is the ratio of the total surface area of the cone to the surface area of just the base?

Possible Answers:

$\frac{9}{5}$

$\frac{7}{6}$

$\frac{3}{2}$

$\frac{9}{4}$

Correct answer:

$\frac{9}{4}$

Explanation:

We need to find total surface area of the cone and the area of the base.

The area of the base of a cone is equal to the area of a circle. The formula for the area of a circle is given below:

$A=\pi r^2$ , where r is the length of the radius.

In the case of this cone, the radius is equal to 4R, so we must replace r with 4R.

$A=\pi(4R)^2=\pi(4)^2(R)^2=16\pi R^2$

To find the total area of the cone, we need the area of the base and the lateral surface area of the cone. The lateral surface area (LA) of a cone is given by the following formula:

$LA=\frac{1}{2}(2\pi r)(l)$ , where r is the radius and l is the slant height.

We know that r = 4R. What we need now is the slant height, which is the distance from the edge of the base of the cone to the tip.

In order to find the slant height, we need to construct a right triangle with the legs equal to the height and the radius of the cone. The slant height will be the hypotenuse of this triangle. We can use the Pythagorean Theorem to find an expression for l. According to the Pythagorean Theorem, the sum of the squares of the legs (which are 4R and 3R in this case) is equal to the square of the hypotenuse (which is the slant height). According to the Pythagorean Theorem, we can write the following equation:

(4R)^2+(3R)^2=l^2

16R^2+9R^2=l^2

25R^2=l^2

l=5R

Let's go back to the formula for the lateral surface area (LA).

$LA=\frac{1}{2}(2\pi r)(l)$

$LA=\frac{1}{2}(2\pi(4R))(5R)$

$LA=20\pi R^2$

To find the total surface area (TA), we must add the lateral area and the area of the base.

$TA=20\pi R^2+16\pi R^2=36\pi R^2$

The problem requires us to find the ratio of the total surface area to the area of the base. This means we must find the following ratio:

$\frac{36\pi R^2}{16\pi R^2}$

We can cancel $\pi R^2$ , which leaves us with 36/16.

Simplifying 36/16 gives 9/4.

The answer is 9/4.

Report an Error

Example Question #551 : Sat Subject Test In Math I

What is the surface area of a cone with a radius of 6 in and a height of 8 in?

Possible Answers:

96π in²

36π in²

60π in²

112π in²

66π in²

Correct answer:

96π in²

Explanation:

Find the slant height of the cone using the Pythagorean theorem: r² + h² = s² resulting in 6² + 8² = s² leading to s² = 100 or s = 10 in

SA = πrs + πr² = π(6)(10) + π(6)² = 60π + 36π = 96π in²

60π in² is the area of the cone without the base.

36π in² is the area of the base only.

Report an Error

Example Question #1 : Cones

Use the following formula to answer the question.

$SA=\pi r^2+\pi rl$

The slant height of a right circular cone is 6cm . The radius is 2cm , and the height is 4cm . Determine the surface area of the cone.

Possible Answers:

$16\pi cm^2$

$24\pi cm^2$

$18\pi cm^2$

$20\pi cm^2$

$12\pi cm^2$

Correct answer:

$16\pi cm^2$

Explanation:

Notice that the height of the cone is not needed to answer this question and is simply extraneous information. We are told that the radius is 2cm , and the slant height is 6cm .

First plug these numbers into the equation provided.

$SA=\pi r^2+\pi rl=\pi (2)^2+\pi (2)(6)=4\pi cm^2+12\pi cm^2$

Then simplify by combining like terms.

$4\pi cm^2+12\pi cm^2=16\pi cm^2$

Report an Error

Example Question #1 : Cones

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

Possible Answers:

$\frac{ \pi L^{2}}{20}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{6\pi L^{2}}{25}$

$\frac{11\pi L^{2}}{100}$

Correct answer:

$\frac{11\pi L^{2}}{100}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

Report an Error

Example Question #11 : How To Find The Surface Area Of A Cone

The radius of the base of a cone is ; its slant height is two-thirds of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$\frac{\left (6- \sqrt{7} \right )\pi r ^{2}}{3}$

$\frac{7\pi r ^{2}}{3}$

$\frac{4\pi r ^{2}}{3}$

$\frac{5\pi r ^{2}}{3}$

$\frac{\left (3- \sqrt{5} \right )\pi r ^{2}}{3}$

Correct answer:

$\frac{7\pi r ^{2}}{3}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice radius , or , and its slant height is two-thirds of this diameter, which is $\frac{2}{3}\cdot 2r = \frac{4}{3}r$ . Substitute this for in the formula:

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

$A = \pi r^{2} + \frac{4}{3} \pi r ^{2}$

$A = \frac{7}{3} \pi r ^{2} = \frac{7\pi r ^{2}}{3}$

Report an Error

Example Question #2 : Cones

The radius of the base of a cone is ; its height is twice of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

$\left ( 1+2 \sqrt{5} \right ) \pi r^{2}$

$3 \pi r^{2}$

$\left ( 1+ \sqrt{5} \right ) \pi r^{2}$

$5 \pi r^{2}$

Correct answer:

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The base has radius and diameter . The height is twice the diamter, which is . Its slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{r^{2}+ (4r)^{2}}$

$l = \sqrt{r^{2}+ 16r^{2}}$

$l = \sqrt{17r^{2}}$

$l = r \sqrt{17 }$

Substitute $r \sqrt{17 }$ for in the surface area formula:

$A = \pi r^{2} + \pi r \cdot r \sqrt{17}$

$A = \pi r^{2} + \pi r^{2} \sqrt{17}$

$A = \left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Report an Error

Example Question #12 : How To Find The Surface Area Of A Cone

The height of a cone is ; the diameter of its base is twice the height. Give its surface area in terms of .

Possible Answers:

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

$5 \pi h^{2}$

$4 \pi h^{2}$

$\left (2 +2 \sqrt{2} \right ) \pi h^{2}$

$10 \pi h^{2}$

Correct answer:

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice the height, which is ; the radius is half this, which is .

The slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{h^{2}+ h^{2}}$

$l = \sqrt{2h^{2}}$

$l =h \sqrt{2}$

Substitute $h \sqrt{2}$ for and for in the surface area formula:

$A = \pi h^{2} + \pi h \cdot h \sqrt{2}$

$A = \pi h^{2} + \pi h ^{2} \sqrt{2}$

$A = \left (1 + \sqrt{2} \right ) \pi h^{2}$

Report an Error

Example Question #13 : How To Find The Surface Area Of A Cone

The circumference of the base of a cone is 80; the slant height of the cone is equal to twice the diameter of the base. Give the surface area of the cone (nearest whole number).

Possible Answers:

2,485

4,077

2,039

1,975

2,546

Correct answer:

2,546

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The slant height is twice the diameter, or, equivalently, four times the radius, so

l = 4r

and

$A = \pi r^{2} + \pi r \cdot 4r$

$A = \pi r^{2} +4 \pi r^{2}$

$A = 5 \pi r^{2}$

The radius of the base is the circumference divided by $2\pi$ , which is

$80 \div 2 \pi \approx 80 \div (2 \times 3.14 )\approx 12.73$

Substitute:

$A = 5 \pi r^{2}$

$A \approx 5 \cdot 3.14 \cdot (12.73)^{2}$

$A \approx 2,546$

Report an Error

View Tutors

Lydia
Certified Tutor

The Richard Stockton College of New Jersey, Bachelor of Science, Biochemistry and Molecular Biology.

View Tutors

Michelle
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mathematics.

View Tutors

Kaylee
Certified Tutor

Brigham Young University, Bachelor, MFA - Animation.

All Advanced Geometry Resources

6 Diagnostic Tests 57 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Los Angeles, LSAT Tutors in Los Angeles, Calculus Tutors in New York City, SAT Tutors in San Diego, GRE Tutors in San Francisco-Bay Area, MCAT Tutors in Phoenix, Statistics Tutors in Chicago, Biology Tutors in Washington DC, English Tutors in Seattle, Math Tutors in Seattle

Popular Courses & Classes

Spanish Courses & Classes in Chicago, Spanish Courses & Classes in Miami, ISEE Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Phoenix, GMAT Courses & Classes in Boston, SSAT Courses & Classes in Miami, ACT Courses & Classes in San Diego, MCAT Courses & Classes in Denver, ISEE Courses & Classes in New York City, LSAT Courses & Classes in Houston

Popular Test Prep

SAT Test Prep in Washington DC, SAT Test Prep in Miami, GMAT Test Prep in Boston, MCAT Test Prep in Philadelphia, GRE Test Prep in Seattle, ISEE Test Prep in Denver, ACT Test Prep in San Diego, LSAT Test Prep in Philadelphia, ACT Test Prep in Phoenix, LSAT Test Prep in San Francisco-Bay Area

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

Advanced Geometry : How to find the surface area of a cone

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #541 : Geometry

Example Question #31 : Solid Geometry

Example Question #11 : Surface Area

Example Question #551 : Sat Subject Test In Math I

Example Question #1 : Cones

Example Question #1 : Cones

Example Question #11 : How To Find The Surface Area Of A Cone

Example Question #2 : Cones

Example Question #12 : How To Find The Surface Area Of A Cone

Example Question #13 : How To Find The Surface Area Of A Cone

All Advanced Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: