We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : Plane Geometry

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 #455 : Triangles

Find the area.

Possible Answers:

$9\sqrt3$

$\frac{9}{2}$

$18\sqrt3$

$\frac{9\sqrt3}{2}$

Correct answer:

$\frac{9\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{6^2-3^2}$

Simplify.

$\text{base}=\sqrt{27}$

Reduce.

$\text{base}=3\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{3 \times 3\sqrt3}{2}$

Solve.

$\text{Area}\frac{9\sqrt3}{2}$

Report an Error

Example Question #56 : How To Find The Area Of A Right Triangle

Find the area.

Possible Answers:

$12\sqrt3$

$15\sqrt3$

$8\sqrt3$

$4\sqrt3$

Correct answer:

$8\sqrt3$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{8^2-4^2}$

Simplify.

$\text{base}=\sqrt{48}$

Reduce.

$\text{base}=4\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{4 \times 4\sqrt3}{2}$

Solve.

$\text{Area}=8\sqrt3$

Report an Error

Example Question #1441 : Basic Geometry

Find the area.

Possible Answers:

$25\sqrt3$

125

$75\sqrt2$

$\frac{25\sqrt3}{2}$

Correct answer:

$\frac{25\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{10^2-5^2}$

Simplify.

$\text{base}=\sqrt{75}$

Reduce.

$\text{base}=5\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{5 \times 5\sqrt3}{2}$

Solve.

$\text{Area}=\frac{25\sqrt3}{2}$

Report an Error

Example Question #1442 : Basic Geometry

Find the area.

Possible Answers:

$64\sqrt3$

$48\sqrt3$

$60\sqrt3$

$32\sqrt3$

Correct answer:

$32\sqrt3$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{16^2-8^2}$

Simplify.

$\text{base}=\sqrt{192}$

Reduce.

$\text{base}=8\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{8 \times 8\sqrt3}{2}$

Solve.

$\text{Area}=32\sqrt3$

Report an Error

Example Question #262 : Right Triangles

Find the area.

Possible Answers:

$81\sqrt3$

$99\sqrt3$

$\frac{81\sqrt3}{2}$

$\frac{117\sqrt3}{2}$

Correct answer:

$\frac{81\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{18^2-9^2}$

Simplify.

$\text{base}=\sqrt{243}$

Reduce.

$\text{base}=9\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{9 \times 9\sqrt3}{2}$

Solve.

$\text{Area}=\frac{81\sqrt3}{2}$

Report an Error

Example Question #1441 : Plane Geometry

Find the area.

Possible Answers:

$110\sqrt3$

$\frac{121\sqrt3}{2}$

$121\sqrt3$

$243\sqrt3$

Correct answer:

$\frac{121\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{22^2-11^2}$

Simplify.

$\text{base}=\sqrt{363}$

Reduce.

$\text{base}=11\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{11 \times 11\sqrt3}{2}$

Solve.

$\text{Area}=\frac{121\sqrt3}{2}$

Report an Error

Example Question #1445 : Basic Geometry

Find the area.

Possible Answers:

144

$72\sqrt3$

$144\sqrt3$

$36\sqrt3$

Correct answer:

$72\sqrt3$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{24^2-12^2}$

Simplify.

$\text{base}=\sqrt{432}$

Reduce.

$\text{base}=12\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{12 \times 12\sqrt3}{2}$

Solve.

$\text{Area}=72\sqrt3$

Report an Error

Example Question #1442 : Plane Geometry

Find the area.

Possible Answers:

$27\sqrt3$

$40\sqrt3$

$36\sqrt3$

$18\sqrt3$

Correct answer:

$18\sqrt3$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{12^2-6^2}$

Simplify.

$\text{base}=\sqrt{108}$

Reduce.

$\text{base}=6\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{6 \times 6\sqrt3}{2}$

Solve.

$\text{Area}=18\sqrt3$

Report an Error

Example Question #271 : Right Triangles

Find the area.

Possible Answers:

$14\sqrt3$

$\frac{49\sqrt3}{2}$

$49\sqrt3$

$56\sqrt3$

Correct answer:

$\frac{49\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{14^2-7^2}$

Simplify.

$\text{base}=\sqrt{147}$

Reduce.

$\text{base}=7\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{7 \times 7\sqrt3}{2}$

Solve.

$\text{Area}=\frac{49\sqrt3}{2}$

Report an Error

Example Question #1448 : Basic Geometry

Find the area.

Possible Answers:

$25\sqrt3$

$\frac{75\sqrt3}{2}$

$50\sqrt3$

$100\sqrt3$

Correct answer:

$50\sqrt3$

Explanation:

Recall how to find the area of a triangle:

$\text{Area}=\frac{\text{base}\times\text{height}}{2}$

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{20^2-10^2}$

Simplify.

$\text{base}=\sqrt{300}$

Reduce.

$\text{base}=10\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{10 \times 10\sqrt3}{2}$

Solve.

$\text{Area}=50\sqrt3$

Report an Error

View Tutors

Amy
Certified Tutor

Montana State University-Northern, Bachelor of Science, Nursing (RN). Montana State University-Northern, Master of Science, N...

View Tutors

Monique
Certified Tutor

California State University-Dominguez Hills, Bachelor of Science, Political Science and Government. California State Universi...

View Tutors

Patricia
Certified Tutor

Old Dominion University, Bachelors, History and French. George Washington University, Masters, Behavioral Science.

All Basic Geometry Resources

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

Popular Subjects

Algebra Tutors in Denver, Statistics Tutors in San Diego, Math Tutors in San Diego, French Tutors in Dallas Fort Worth, GMAT Tutors in Philadelphia, Reading Tutors in San Diego, Statistics Tutors in Atlanta, Physics Tutors in Houston, SSAT Tutors in Miami, English Tutors in Houston

Popular Courses & Classes

ACT Courses & Classes in Chicago, Spanish Courses & Classes in New York City, SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Philadelphia, ACT Courses & Classes in Denver, GRE Courses & Classes in San Diego, Spanish Courses & Classes in Phoenix, GRE Courses & Classes in Philadelphia, SSAT Courses & Classes in Denver, ISEE Courses & Classes in Houston

Popular Test Prep

LSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Phoenix, SSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Boston, SAT Test Prep in Denver, SSAT Test Prep in Chicago, GMAT Test Prep in Chicago, MCAT Test Prep in Los Angeles, SAT Test Prep in Miami, ISEE Test Prep in Denver

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 : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #455 : Triangles

Example Question #56 : How To Find The Area Of A Right Triangle

Example Question #1441 : Basic Geometry

Example Question #1442 : Basic Geometry

Example Question #262 : Right Triangles

Example Question #1441 : Plane Geometry

Example Question #1445 : Basic Geometry

Example Question #1442 : Plane Geometry

Example Question #271 : Right Triangles

Example Question #1448 : Basic Geometry

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: