Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : 45/45/90 Right Isosceles Triangles

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 #1132 : Plane Geometry

Find the perimeter.

Possible Answers:

111.67

115.29

112.67

109.76

Correct answer:

112.67

Explanation:

Notice that the given triangle is a right isosceles triangle. The two legs with the tick marks are the same length.

The lengths of the legs in the given triangle are then .

Next, find the length of the hypotenuse by using the Pythagorean Theorem.

a^2+a^2=b^2

b^2=2a^2

$b=a\sqrt2$

Plug in the value of the length of a leg to find the length of the hypotenuse.

$b=33\sqrt2$

Finally, recall how to find the perimeter of a triangle:

$\text{Perimeter}=a+a+b$

Plug in the values for this triangle to find its perimeter.

$\text{Perimeter}=33+33+33\sqrt2=112.67$

Make sure to round to two places after the decimal.

Report an Error

Example Question #152 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

55.29

49.08

52.97

54.20

Correct answer:

52.97

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{12\sqrt2}{2}$

$\text{leg}=6\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=12+12+12+6\sqrt2+6\sqrt2$

$\text{Perimeter}=52.97$

Report an Error

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

8.10

8.83

9.41

10.06

Correct answer:

8.83

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{2\sqrt2}{2}$

$\text{leg}=\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=2+2+2+\sqrt2+\sqrt2$

$\text{Perimeter}=8.83$

Report an Error

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

19.07

17.66

16.92

18.92

Correct answer:

17.66

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{4\sqrt2}{2}$

$\text{leg}=2\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=4+4+4+2\sqrt2+2\sqrt2$

$\text{Perimeter}=17.66$

Report an Error

Example Question #152 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

31.22

26.49

30.09

27.81

Correct answer:

26.49

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{6\sqrt2}{2}$

$\text{leg}=3\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=6+6+6+3\sqrt2+3\sqrt2$

$\text{Perimeter}=26.49$

Report an Error

Example Question #153 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

36.19

35.31

35.59

34.10

Correct answer:

35.31

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{8\sqrt2}{2}$

$\text{leg}=4\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=8+8+8+4\sqrt2+4\sqrt2$

$\text{Perimeter}=35.31$

Report an Error

Example Question #154 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

42.99

44.14

45.61

48.20

Correct answer:

44.14

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{10\sqrt2}{2}$

$\text{leg}=5\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=10+10+10+5\sqrt2+5\sqrt2$

$\text{Perimeter}=44.14$

Report an Error

Example Question #1141 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

72.59

74.09

69.39

61.80

Correct answer:

61.80

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{14\sqrt2}{2}$

$\text{leg}=7\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=14+14+14+7\sqrt2+7\sqrt2$

$\text{Perimeter}=61.80$

Report an Error

Example Question #1142 : Basic Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

48.20

69.59

70.63

80.05

Correct answer:

70.63

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{16\sqrt2}{2}$

$\text{leg}=8\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=16+16+16+8\sqrt2+8\sqrt2$

$\text{Perimeter}=70.63$

Report an Error

Example Question #1142 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

77.65

79.46

84.19

81.20

Correct answer:

79.46

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

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

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

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

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

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{18\sqrt2}{2}$

$\text{leg}=9\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=18+18+18+9\sqrt2+9\sqrt2$

$\text{Perimeter}=79.46$

Report an Error

View Tutors

Mark
Certified Tutor

Rochester Institute of Technology, Bachelor of Science, Game and Interactive Media Design. Rochester Institute of Technology,...

View Tutors

Jeremy
Certified Tutor

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

View Tutors

Jeffrey
Certified Tutor

All Basic Geometry Resources

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

Popular Subjects

SAT Tutors in Atlanta, GRE Tutors in Dallas Fort Worth, ISEE Tutors in New York City, French Tutors in San Francisco-Bay Area, French Tutors in Dallas Fort Worth, SSAT Tutors in Dallas Fort Worth, LSAT Tutors in Denver, GRE Tutors in Miami, GMAT Tutors in Boston, ISEE Tutors in Chicago

Popular Courses & Classes

LSAT Courses & Classes in Denver, MCAT Courses & Classes in Washington DC, ACT Courses & Classes in Washington DC, GRE Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Houston, ACT Courses & Classes in San Diego, SAT Courses & Classes in Chicago, SAT Courses & Classes in Boston, GRE Courses & Classes in Denver, Spanish Courses & Classes in Atlanta

Popular Test Prep

LSAT Test Prep in Dallas Fort Worth, SAT Test Prep in Dallas Fort Worth, GRE Test Prep in Chicago, ISEE Test Prep in Seattle, MCAT Test Prep in Boston, SSAT Test Prep in Phoenix, GMAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Seattle, MCAT Test Prep in San Diego, 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 : 45/45/90 Right Isosceles Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1132 : Plane Geometry

Example Question #152 : Triangles

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

Example Question #152 : Triangles

Example Question #153 : Triangles

Example Question #154 : Triangles

Example Question #1141 : Plane Geometry

Example Question #1142 : Basic Geometry

Example Question #1142 : Plane Geometry

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: