Call Now to Set Up Tutoring:

(888) 888-0446

Basic Geometry : 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 #4 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $\sqrt6$ , what is the length of a side of the triangle?

Possible Answers:

$\sqrt3$

$2\sqrt3$

Correct answer:

$\sqrt3$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(\sqrt6)(\sqrt2)}{2}$

Simplify.

$a=\frac{\sqrt{12}}{2}$

Reduce.

$a=\sqrt3$

Report an Error

Example Question #7 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $2\sqrt2$ , what is the length of a side of this triangle?

Possible Answers:

$\sqrt3$

$\sqrt2$

Correct answer:

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(2\sqrt2)(\sqrt2)}{2}$

Simplify.

$a=\frac{\sqrt{4}}{2}$

Reduce.

a=2

Report an Error

Example Question #4 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $\sqrt{10}$ , what is the length of a side of this triangle?

Possible Answers:

$\sqrt5$

$\sqrt3$

$\sqrt6$

Correct answer:

$\sqrt5$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(\sqrt{10})(\sqrt2)}{2}$

Simplify.

$a=\frac{\sqrt{20}}{2}$

Reduce.

$a=\sqrt5$

Report an Error

Example Question #171 : Triangles

If the hypotenuse of a right isosceles triangle is $\sqrt{14}$ , what is the length of a side of the triangle?

Possible Answers:

$\sqrt7$

$2\sqrt2$

$2\sqrt{7}$

$\sqrt{15}$

Correct answer:

$\sqrt7$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(\sqrt{14})(\sqrt2)}{2}$

Simplify.

$a=\frac{\sqrt{28}}{2}$

Reduce.

$a=\sqrt7$

Report an Error

Example Question #172 : Triangles

If the hypotenuse of a right isosceles triangle is $12\sqrt2$ , what is the length of a side of the triangle?

Possible Answers:

Correct answer:

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(12\sqrt2)(\sqrt2)}{2}$

Simplify.

$a=\frac{24}{2}$

Reduce.

a=12

Report an Error

Example Question #173 : Triangles

If the hypotenuse of a right isosceles triangle is $2\sqrt5$ , what is the length of a side of the triangle?

Possible Answers:

$4\sqrt2$

$\sqrt{10}$

$2\sqrt3$

Correct answer:

$\sqrt{10}$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(2\sqrt5)(\sqrt2)}{2}$

Simplify.

$a=\frac{2\sqrt{10}}{2}$

Reduce.

$a=\sqrt{10}$

Report an Error

Example Question #174 : Triangles

If the hypotenuse of a right isosceles triangle is $12\sqrt{7}$ , what is the length of one side of the triangle?

Possible Answers:

$6\sqrt{14}$

$3\sqrt{22}$

$8\sqrt{10}$

Correct answer:

$6\sqrt{14}$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(12\sqrt7)(\sqrt2)}{2}$

Simplify.

$a=\frac{12\sqrt{14}}{2}$

Reduce.

$a=6\sqrt{14}$

Report an Error

Example Question #175 : Triangles

If the hypotenuse of a right isosceles triangle is $4\sqrt{3}$ , what is the length of one side of the triangle?

Possible Answers:

$2\sqrt6$

$8\sqrt3$

$6\sqrt2$

$4\sqrt6$

Correct answer:

$2\sqrt6$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(4\sqrt3)(\sqrt2)}{2}$

Simplify.

$a=\frac{4\sqrt{6}}{2}$

Reduce.

$a=2\sqrt6$

Report an Error

Example Question #176 : Triangles

If the hypotenuse of a right isosceles triangle is $9\sqrt6$ , what is the length of one side of the triangle?

Possible Answers:

$9\sqrt2$

$9\sqrt3$

$\frac{9\sqrt2}{2}$

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

Correct answer:

$9\sqrt3$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(9\sqrt6)(\sqrt2)}{2}$

Simplify.

$a=\frac{9\sqrt{12}}{2}$

$a=\frac{18\sqrt3}{2}$

Reduce.

$a=9\sqrt3$

Report an Error

Example Question #177 : Triangles

If the hypotenuse of a right isosceles triangle is 128 , what is the length of a side of the triangle?

Possible Answers:

$64\sqrt2$

128

$32\sqrt3$

Correct answer:

$64\sqrt2$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

$a^2=\frac{c^2}{2}$

Simplify.

$a=\sqrt{\frac{c^2}{2}}$

$a=\frac{c}{\sqrt2}$

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

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$a=\frac{c\sqrt2}{2}$

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{128(\sqrt2)}{2}$

Simplify.

$a=64\sqrt2$

Report an Error

View Tutors

Matthew
Certified Tutor

Bluffton University, Bachelor of Education, Mathematics. Bluffton University, Masters in Education, Education.

View Tutors

Manqoba
Certified Tutor

Marist College, Bachelor of Science, Business Administration and Management.

View Tutors

Paul
Certified Tutor

University of Phoenix-Albuquerque Campus, Bachelor of Science, Computer and Information Systems Security.

All Basic Geometry Resources

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

Popular Subjects

GMAT Tutors in Atlanta, ISEE Tutors in Dallas Fort Worth, Statistics Tutors in Miami, Physics Tutors in Miami, GMAT Tutors in San Francisco-Bay Area, Computer Science Tutors in Atlanta, GRE Tutors in Washington DC, Computer Science Tutors in Los Angeles, Statistics Tutors in Dallas Fort Worth, Chemistry Tutors in Los Angeles

Popular Courses & Classes

MCAT Courses & Classes in Boston, SSAT Courses & Classes in Denver, MCAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Chicago, MCAT Courses & Classes in Denver, LSAT Courses & Classes in Los Angeles, SAT Courses & Classes in Miami, Spanish Courses & Classes in Denver, LSAT Courses & Classes in Chicago, ISEE Courses & Classes in Boston

Popular Test Prep

MCAT Test Prep in Phoenix, GRE Test Prep in San Diego, GRE Test Prep in Houston, MCAT Test Prep in Los Angeles, SAT Test Prep in Seattle, MCAT Test Prep in Chicago, SAT Test Prep in Chicago, LSAT Test Prep in Boston, MCAT Test Prep in San Diego, SSAT Test Prep in Houston

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 : Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #4 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #7 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #4 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #171 : Triangles

Example Question #172 : Triangles

Example Question #173 : Triangles

Example Question #174 : Triangles

Example Question #175 : Triangles

Example Question #176 : Triangles

Example Question #177 : Triangles

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: