Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : Squares

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 #221 : Squares

Find the perimeter of a square given area is .

Possible Answers:

Correct answer:

Explanation:

To find perimeter, we must first find the side length and then multiply is by .

$A=s^2\Rightarrow s=\sqrt{A}=\sqrt{49}=7$

$P=4s=4\cdot7=28$

Report an Error

Example Question #872 : Plane Geometry

Find the area of the square.

Possible Answers:

1200

1600

800

900

Correct answer:

800

Explanation:

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

Thus, we can use the Pythagorean Theorem to find the length of the sides of the square.

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

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

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

Recall how to find the area of a square.

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

Now, substitute in the value of the diagonal to find the area of the square.

$\text{Area}=\frac{40^2}{2}$

Solve.

$\text{Area}=800$

Report an Error

Example Question #35 : How To Find The Perimeter Of A Square

A square has an area of 49in^2 . Find the perimeter.

Possible Answers:

25in

49in

14in

21in

28in

Correct answer:

28in

Explanation:

Since a square has four equal sides, finding the side lengths from the area is quite simple.

Area = Length(Width)

Since all the sides are equal in a square, the length and width are equal, meaning that Area = side ^2 . Plugging in the numbers gives us:

49 = side^2

$\sqrt{}49 = side$

7 = side

Now that we have found the length of each of the sides of the square, we simply add all the sides together, or multiply by 4 (because there are 4 sides in a square).

7(4) = 28in

Report an Error

Example Question #881 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of $2\sqrt2$ .

Possible Answers:

$8\sqrt2$

Correct answer:

Explanation:

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{2\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=2$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(2)$

Solve.

$\text{Perimeter}=8$

Report an Error

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

Find the perimeter of a square inscribed in a circle that has a diameter of $3\sqrt2$ .

Possible Answers:

$12\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{3\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=3$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(3)$

Solve.

$\text{Perimeter}=12$

Report an Error

Example Question #38 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $4\sqrt2$ .

Possible Answers:

$16\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{4\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=4$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(4)$

Solve.

$\text{Perimeter}=16$

Report an Error

Example Question #39 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $5\sqrt2$ .

Possible Answers:

$20\sqrt2$

$25\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{5\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=5$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(5)$

Solve.

$\text{Perimeter}=20$

Report an Error

Example Question #40 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $7\sqrt2$ .

Possible Answers:

$28\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{7\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=7$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(7)$

Solve.

$\text{Perimeter}=28$

Report an Error

Example Question #222 : Squares

Find the perimeter of a square inscribed in a circle that has a diameter of $9\sqrt2$ .

Possible Answers:

$18\sqrt2$

$36\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{9\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=9$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(9)$

Solve.

$\text{Perimeter}=36$

Report an Error

Example Question #882 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of $11\sqrt2$ .

Possible Answers:

$121\sqrt2$

$22\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{11\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=11$

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

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(11)$

Solve.

$\text{Perimeter}=44$

Report an Error

View Tutors

Robert
Certified Tutor

Daytona State College, Bachelor in Arts, Teaching English as a Second Language (ESL).

View Tutors

Adriana
Certified Tutor

CUNY Hunter College, Bachelor in Arts, Mathematics. University of Connecticut, Master of Science, Mathematics.

View Tutors

Daria
Certified Tutor

Rutgers University-Newark, Bachelor of Science, Information Resources Management.

All Basic Geometry Resources

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

Popular Subjects

Algebra Tutors in Boston, English Tutors in Los Angeles, French Tutors in Dallas Fort Worth, Physics Tutors in Boston, English Tutors in San Diego, Statistics Tutors in Philadelphia, Biology Tutors in Los Angeles, Reading Tutors in New York City, GMAT Tutors in Atlanta, Spanish Tutors in Houston

Popular Courses & Classes

GRE Courses & Classes in New York City, SSAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Miami, SSAT Courses & Classes in Philadelphia, Spanish Courses & Classes in New York City, GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Philadelphia, ACT Courses & Classes in Phoenix, Spanish Courses & Classes in San Diego, GRE Courses & Classes in Denver

Popular Test Prep

SSAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Houston, SSAT Test Prep in Chicago, GMAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Los Angeles, ACT Test Prep in Los Angeles, SSAT Test Prep in Houston, LSAT Test Prep in Los Angeles, LSAT Test Prep in Atlanta, SAT Test Prep in Los Angeles

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 : Squares

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #221 : Squares

Example Question #872 : Plane Geometry

Example Question #35 : How To Find The Perimeter Of A Square

Example Question #881 : Plane Geometry

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

Example Question #38 : How To Find The Perimeter Of A Square

Example Question #39 : How To Find The Perimeter Of A Square

Example Question #40 : How To Find The Perimeter Of A Square

Example Question #222 : Squares

Example Question #882 : Plane Geometry

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: