We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : Circles

Study concepts, example questions & explanations for Basic Geometry

Create An Account

All Basic Geometry Resources

9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #281 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$39\sqrt2\pi$

$13\sqrt2\pi$

$26\sqrt2\pi$

Correct answer:

$13\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=13\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=13\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=13\sqrt2\pi$

Report an Error

Example Question #91 : How To Find Circumference

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$12\sqrt2\pi$

$6\sqrt2\pi$

$24\sqrt2\pi$

Correct answer:

$12\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=12\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=12\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=12\sqrt2\pi$

Report an Error

Example Question #282 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$22\sqrt2$

$121\sqrt2\pi$

$11\sqrt2\pi$

$44\sqrt2\pi$

Correct answer:

$11\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=11\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=11\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=11\sqrt2\pi$

Report an Error

Example Question #281 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$20\sqrt2\pi$

$10\sqrt2\pi$

$10\sqrt2$

$30\sqrt2\pi$

Correct answer:

$10\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=10\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=10\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=10\sqrt2\pi$

Report an Error

Example Question #284 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$16\sqrt2\pi$

$64\pi$

$8\sqrt2\pi$

$128\pi$

Correct answer:

$8\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=8\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=8\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=8\sqrt2\pi$

Report an Error

Example Question #285 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$9\sqrt2\pi$

$18\sqrt2\pi$

$81\pi$

$18\pi$

Correct answer:

$9\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=9\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=9\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=9\sqrt2\pi$

Report an Error

Example Question #286 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$98\pi$

$49\sqrt2\pi$

$14\sqrt2\pi$

$7\sqrt2\pi$

Correct answer:

$7\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=7\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=7\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=7\sqrt2\pi$

Report an Error

Example Question #287 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$6\sqrt2\pi$

$3\sqrt2\pi$

$36\sqrt2\pi$

$12\sqrt2\pi$

Correct answer:

$6\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=6\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=6\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=6\sqrt2\pi$

Report an Error

Example Question #281 : Basic Geometry

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$10\sqrt2\pi$

$25\pi$

$5\sqrt3\pi$

$5\sqrt2\pi$

Correct answer:

$5\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=5\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=5\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=5\sqrt2\pi$

Report an Error

Example Question #289 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are .

Possible Answers:

$4\pi$

$4\sqrt2\pi$

$6\sqrt2\pi$

$8\pi$

Correct answer:

$4\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=4\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=4\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=4\sqrt2\pi$

Report an Error

View Tutors

Claryse
Certified Tutor

St. Mary's University, Bachelor of Education, Education.

View Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View Tutors

Jecintha
Certified Tutor

Metropolitan College of New York, Bachelor in Arts, Human Services.

All Basic Geometry Resources

9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in Dallas Fort Worth, Math Tutors in Phoenix, GRE Tutors in San Diego, Algebra Tutors in Dallas Fort Worth, SSAT Tutors in Dallas Fort Worth, SSAT Tutors in Philadelphia, ACT Tutors in San Francisco-Bay Area, SAT Tutors in Philadelphia, ISEE Tutors in Denver, GMAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

ACT Courses & Classes in Atlanta, SSAT Courses & Classes in New York City, SSAT Courses & Classes in Boston, GRE Courses & Classes in Atlanta, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Chicago, GMAT Courses & Classes in Houston, MCAT Courses & Classes in Los Angeles, SAT Courses & Classes in Los Angeles, ACT Courses & Classes in Philadelphia

Popular Test Prep

ACT Test Prep in Philadelphia, SSAT Test Prep in Washington DC, LSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Denver, ACT Test Prep in Atlanta, GRE Test Prep in Houston, SSAT Test Prep in Chicago, ISEE Test Prep in Washington DC, GMAT Test Prep in Houston, MCAT Test Prep in Boston

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

Basic Geometry : Circles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #281 : Circles

Example Question #91 : How To Find Circumference

Example Question #282 : Circles

Example Question #281 : Circles

Example Question #284 : Circles

Example Question #285 : Circles

Example Question #286 : Circles

Example Question #287 : Circles

Example Question #281 : Basic Geometry

Example Question #289 : Circles

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: