We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : Basic 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 #311 : Quadrilaterals

If the perimeter of a square is 220 , what is the area of the square?

Possible Answers:

2225

4065

3025

5025

Correct answer:

3025

Explanation:

First, recall how to find the perimeter of a square.

$\text{Perimeter}=4\times\text{side length}$

From this equation, we can solve for the side length of a square.

$\text{Side length}=\frac{\text{Perimeter}}{4}$

For the given square,

$\text{Side length}=\frac{220}{4}=55$

Now, recall how to find the area of a square.

$\text{Area}=\text{side length}^2$

For the square in question, we can plug in the side length we found from the perimeter to find the area.

$\text{Area}=55^2=3025$

Report an Error

Example Question #721 : Basic Geometry

If the diagonal of a square is $6\sqrt2$ , what is the area of the square?

Possible Answers:

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{6\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=6$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}=6^2$

Solve.

$\text{Area}=36$

Report an Error

Example Question #722 : Basic Geometry

If the diagonal of a square is $5\sqrt2$ , what is the area of the square?

Possible Answers:

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{5\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=5$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 5 ^ 2$

Solve.

$\text{Area}= 25$

Report an Error

Example Question #37 : How To Find The Area Of A Square

If the diagonal of a square is $10\sqrt2$ , what is the area of the square?

Possible Answers:

200

100

150

Correct answer:

100

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{10\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=10$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 10 ^2$

Solve.

$\text{Area}= 100$

Report an Error

Example Question #314 : Quadrilaterals

If the diagonal of a square is $8\sqrt2$ , what is the area of the square?

Possible Answers:

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{8\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=8$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 8 ^2$

Solve.

$\text{Area}=64$

Report an Error

Example Question #41 : How To Find The Area Of A Square

If the diagonal of a square is $12\sqrt2$ , what is the area of the square?

Possible Answers:

121

144

Correct answer:

144

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{12\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=12$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 12 ^2$

Solve.

$\text{Area}= 144$

Report an Error

Example Question #316 : Quadrilaterals

If the diagonal of a square is $13\sqrt2$ , what is the area of the square?

Possible Answers:

169

Correct answer:

169

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{13\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=13$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 13 ^2$

Solve.

$\text{Area}= 169$

Report an Error

Example Question #43 : How To Find The Area Of A Square

If the diagonal of a square is $\frac{\sqrt2}{2}$ , what is the area of the square?

Possible Answers:

$\frac{\sqrt2}{4}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{2\sqrt2}{3}$

Correct answer:

$\frac{1}{4}$

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{\frac{\sqrt2}{2}}{\sqrt2}$

Dividing a fraction by a number is the same as multiplying by its reciprocal.

$\text{Side length}=\frac{\sqrt2}{2}\times\frac{1}{\sqrt2}$

Simplify.

$\text{Side length}=\frac{1}{2}$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= \left ( \frac{1}{2} \right )^2$

Solve.

$\text{Area}= \frac{1}{4}$

Report an Error

Example Question #321 : Quadrilaterals

If the diagonal of a square is $9\sqrt2$ , what is the area of the square?

Possible Answers:

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{9\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=9$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 9 ^2$

Solve.

$\text{Area}= 81$

Report an Error

Example Question #321 : Quadrilaterals

If the diagonal of a square is $11\sqrt2$ , what is the area of the square?

Possible Answers:

121

132

Correct answer:

121

Explanation:

The diagonal of a square is also the hypotenuse of a right triangle that has the side lengths of the square as its legs.

We can then use the Pythgorean Theorem to write the following equations:

$\text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

Now, use this formula and substitute using the given values to find the side length of the square.

$\text{Side length}=\frac{11\sqrt2}{\sqrt2}$

Simplify.

$\text{Side length}=11$

Now, recall how to find the area of a square.

$\text{Area}=(\text{side length})^2$

For this square in question,

$\text{Area}= 11 ^2$

Solve.

$\text{Area}=121$

Report an Error

View Tutors

David
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Molecular Biology. University of California-San Diego, Master of Scie...

View Tutors

James
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Accounting.

View Tutors

Wendie
Certified Tutor

East Tennessee State University, Bachelor of Science, Kinesiotherapy.

All Basic Geometry Resources

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

Popular Subjects

French Tutors in New York City, LSAT Tutors in Los Angeles, English Tutors in Atlanta, SAT Tutors in Philadelphia, MCAT Tutors in Philadelphia, ACT Tutors in San Diego, MCAT Tutors in Miami, Chemistry Tutors in Washington DC, Computer Science Tutors in Dallas Fort Worth, ACT Tutors in Los Angeles

Popular Courses & Classes

SSAT Courses & Classes in New York City, LSAT Courses & Classes in Washington DC, ACT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in New York City, SSAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Miami, LSAT Courses & Classes in New York City, MCAT Courses & Classes in Chicago, SAT Courses & Classes in Chicago

Popular Test Prep

ACT Test Prep in Dallas Fort Worth, LSAT Test Prep in New York City, GMAT Test Prep in Houston, SSAT Test Prep in Philadelphia, ACT Test Prep in Boston, ACT Test Prep in Seattle, ISEE Test Prep in San Francisco-Bay Area, ISEE Test Prep in Boston, SAT Test Prep in San Diego, SAT Test Prep in Atlanta

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 : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #311 : Quadrilaterals

Example Question #721 : Basic Geometry

Example Question #722 : Basic Geometry

Example Question #37 : How To Find The Area Of A Square

Example Question #314 : Quadrilaterals

Example Question #41 : How To Find The Area Of A Square

Example Question #316 : Quadrilaterals

Example Question #43 : How To Find The Area Of A Square

Example Question #321 : Quadrilaterals

Example Question #321 : Quadrilaterals

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: