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 #81 : Squares

If the diagonal of a square is $30\sqrt2$ , what is the area of the square?

Possible Answers:

300

900

600

120

Correct answer:

900

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{30\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=30$

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

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

For the square in question,

$\text{Area }=30 ^2$

Solve.

$\text{Area }=900$

Report an Error

Example Question #81 : Squares

If the diagonal of a square is $31\sqrt2$ , what is the area of the square?

Possible Answers:

661

961

696

1204

Correct answer:

961

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{31\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=31$

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

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

For the square in question,

$\text{Area }=31 ^2$

Solve.

$\text{Area }=961$

Report an Error

Example Question #82 : Squares

If the diagonal of a square is $32\sqrt2$ , what is the area of the square?

Possible Answers:

900

964

1224

1024

Correct answer:

1024

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{32\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=32$

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

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

For the square in question,

$\text{Area }=32 ^2$

Solve.

$\text{Area }=1024$

Report an Error

Example Question #87 : Squares

If the diagonal of a square is $34\sqrt2$ , what is the area of the square?

Possible Answers:

1113

1156

1354

1264

Correct answer:

1156

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{34\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=34$

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

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

For the square in question,

$\text{Area }=34 ^2$

Solve.

$\text{Area }=1156$

Report an Error

Example Question #88 : Squares

If the diagonal of a square is $\frac{\sqrt2}{4}$ , what is the area of the square?

Possible Answers:

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{32}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{16}$

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{\frac{\sqrt2}{4}}{\sqrt2}=\frac{1}{4}$

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

$\text{side length}=\frac{\sqrt2}{4}} \times \frac{1}{\sqrt{2}}$

Simplify.

$\text{side length}=\frac{\sqrt{2}}{4\sqrt{2}}$

Multiply the fraction by one in the form: $\frac{\sqrt{2}}\sqrt{2}{}$ .

$\text{side length}=\frac{\sqrt{2}}{4\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$

Simplify.

$\text{side length}=\frac{2}{8}$

Reduce.

$\text{side length}=\frac{1}{4}$

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

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

For the square in question,

$\text{Area }=\left ( \frac{1}{4} \right )^2$

Solve.

$\text{Area }=\frac{1}{16}$

Report an Error

Example Question #741 : Basic Geometry

Find the area of a square whose side length is .

Possible Answers:

Correct answer:

Explanation:

To find the area of a square, use the following formula where is side length.

A=s^2=6^2=36

Report an Error

Example Question #90 : Squares

Find the area of a square if its diagonal is $5\sqrt3$

Possible Answers:

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

$75\sqrt6$

$75\sqrt2$

$\frac{75}{2}$

Correct answer:

$\frac{75}{2}$

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

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

Now, use the Pythagorean theorem to find the area of the square.

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

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

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(5\sqrt3)^2}{2}=\frac{75}{2}$

Report an Error

Example Question #332 : Quadrilaterals

Find the area of a square if its diagonal is $6\sqrt7$ .

Possible Answers:

126

$252\sqrt2$

$63\sqrt3$

$126\sqrt3$

Correct answer:

126

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

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

Now, use the Pythagorean theorem to find the area of the square.

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

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

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(6\sqrt7)^2}{2}=\frac{252}{2}=126$

Report an Error

Example Question #341 : Quadrilaterals

Find the area of a square if its diagonal is $7\sqrt5$ .

Possible Answers:

$46\sqrt5$

$71\sqrt3$

$\frac{245\sqrt2}{4}$

$\frac{245}{2}$

Correct answer:

$\frac{245}{2}$

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

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

Now, use the Pythagorean theorem to find the area of the square.

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

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

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(7\sqrt5)^2}{2}=\frac{245}{2}$

Report an Error

Example Question #342 : Quadrilaterals

Find the area of a square if its diagonal is $9\sqrt{10}$ .

Possible Answers:

$405\sqrt2$

$\frac{405}{2}$

$405\sqrt3$

405

Correct answer:

405

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

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

Now, use the Pythagorean theorem to find the area of the square.

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

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

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(9\sqrt{10})^2}{2}=\frac{810}{2}=405$

Report an Error

View Tutors

Verden
Certified Tutor

University of Mount Olive, Bachelor of Science, Biology, General. University of Mount Olive, Masters in Business Administrati...

View Tutors

Emma
Certified Tutor

Trinity Western University, Bachelor, History.

View Tutors

Joseph
Certified Tutor

Brown University, Bachelor of Science, Education.

All Basic Geometry Resources

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

Popular Subjects

Biology Tutors in Miami, LSAT Tutors in Chicago, SSAT Tutors in Chicago, English Tutors in Chicago, Reading Tutors in Seattle, Algebra Tutors in Boston, ISEE Tutors in Denver, LSAT Tutors in Houston, GMAT Tutors in Denver, ISEE Tutors in Seattle

Popular Courses & Classes

ACT Courses & Classes in Atlanta, GRE Courses & Classes in Atlanta, GMAT Courses & Classes in Washington DC, Spanish Courses & Classes in Boston, GRE Courses & Classes in Houston, ISEE Courses & Classes in New York City, SSAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Miami, LSAT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Atlanta, LSAT Test Prep in Houston, ACT Test Prep in Denver, MCAT Test Prep in Miami, GRE Test Prep in Houston, ISEE Test Prep in Denver, LSAT Test Prep in San Diego, GRE Test Prep in Dallas Fort Worth, LSAT Test Prep in New York City, ACT Test Prep in Dallas Fort Worth

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 #81 : Squares

Example Question #81 : Squares

Example Question #82 : Squares

Example Question #87 : Squares

Example Question #88 : Squares

Example Question #741 : Basic Geometry

Example Question #90 : Squares

Example Question #332 : Quadrilaterals

Example Question #341 : Quadrilaterals

Example Question #342 : Quadrilaterals

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: