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 #71 : Triangles

If the hypotenuse of an isoceles right triangle is 144 , what is the length of the height?

Possible Answers:

$24\sqrt3$

$88\sqrt2$

$72\sqrt2$

Correct answer:

$72\sqrt2$

Explanation:

An isoceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean theorem:

$\text{base}^2+\text{height}^2=\text{hypotenuse}^2$

Because we are working with a 45-45-90 triangle, the base and the height have the same length. We can rewrite the above equation as the following:

$\text{height}^2+\text{height}^2=\text{hypotenuse}^2$

$2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}$

Simplify.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}$

Multiply the fraction by one in the form of:

$1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}$

Substitute.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$\text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{144\sqrt2}{2}$

Simplify.

$\text{height}=72\sqrt2$

Report an Error

Example Question #72 : Triangles

If the hypotenuse of an isoceles right triangle is , what is the length of the height of the triangle?

Possible Answers:

$95\sqrt2$

$\frac{95\sqrt2}{2}$

$95\sqrt3$

Correct answer:

$\frac{95\sqrt2}{2}$

Explanation:

An isoceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean theorem:

$\text{base}^2+\text{height}^2=\text{hypotenuse}^2$

Because we are working with a 45-45-90 triangle, the base and the height have the same length. We can rewrite the above equation as the following:

$\text{height}^2+\text{height}^2=\text{hypotenuse}^2$

$2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}$

Simplify.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}$

Multiply the fraction by one in the form of:

$1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}$

Substitute.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$\text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{95\sqrt2}{2}$

Report an Error

Example Question #1051 : Basic Geometry

If the hypotenuse of an isoceles right triangle is , what is the length of the height?

Possible Answers:

$13\sqrt3$

$16\sqrt3$

$26\sqrt2$

Correct answer:

$26\sqrt2$

Explanation:

An isoceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean theorem:

$\text{base}^2+\text{height}^2=\text{hypotenuse}^2$

Because we are working with a 45-45-90 triangle, the base and the height have the same length. We can rewrite the above equation as the following:

$\text{height}^2+\text{height}^2=\text{hypotenuse}^2$

$2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}$

Simplify.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}$

Multiply the fraction by one in the form of:

$1\rightarrow \frac{\sqrt{2}}{\sqrt{2}}$

Substitute.

$\text{height}=\frac{\text{hypotenuse}}{\sqrt2}\times \frac{\sqrt{2}}{\sqrt{2}}$

Solve.

$\text{height}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, substitute in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{52\sqrt2}{2}$

Simplify.

$\text{height}=26\sqrt2$

Report an Error

Example Question #71 : 45/45/90 Right Isosceles Triangles

Find the height of this triangle:

Isosceles right

Possible Answers:

4.77

12.25

1.87

Correct answer:

4.77

Explanation:

To find the height, use the Pythagorean Theorem. One of the legs is the missing side, and the other is 1.5, half of 3. The hypotenuse is 5:

(1.5)^2 + x^2 = 5^ 2

2.25 + x^2 = 25

x^2 = 22.75

$x \approx 4.77$

Report an Error

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

$What\;is\;the\;area\;of\;\Delta ABC?$

Possible Answers:

6.5

4.5

Correct answer:

4.5

Explanation:

$Area\;\Delta=\frac{1}{2}(base)(height)$

$\frac{1}{2}(3)(3)=\frac{9}{2}=4.5$

Report an Error

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Consider an isosceles triangle with a height of 24 and a base of 12. What is the area of this triangle?

Isoc._triangle

Possible Answers:

155

144

121

169

Correct answer:

144

Explanation:

The formula for the area of a trianlge is A = base * height * (1/2).

We're lucky here, because the question gives us all of the values we need. We simply need to plug them in:

A = base * height * (1/2) = 12 * 24 * (1/2) = 12 * 12 = 144

Report an Error

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Calculate the area of an isosceles right triangle who's hypotenuse is $4\sqrt{2}$ inches.

Possible Answers:

$16 \text{ in.}^2$

$8 \text{ in.}^2$

$2 \text{ in.}^2$

$4 \text{ in.}^2$

Correct answer:

$8 \text{ in.}^2$

Explanation:

The formula for the area of a triangle, right or not, is one half the base times height.

In this case, they are both $4 \text{ in.}$ Therefore, the respective values are entered, yielding:

$\frac{1}{2}(4\text{ in.})(4\text{ in.})=8 \text{ in.}^2$

Report an Error

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

The side length of the 45-45-90 right triangle is $20\sqrt{2}$ , find the area of the right triangle.

Possible Answers:

400

200

$200\sqrt{2}$

$400\sqrt{2}$

$100\sqrt{2}$

Correct answer:

400

Explanation:

The area of a triangle is:

$A=\frac{b*h}{2}$ where b=base, h=height, and A=area

$A=\frac{20\sqrt{2}*20\sqrt{2}}{2}=\frac{400\sqrt{4}}{2}=200\sqrt{4}=200*2=400$

Report an Error

Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of the triangle below.

Possible Answers:

Correct answer:

Explanation:

The key to finding the area of our triangle is to reaize that it is isosceles and therefore is a 45-45-90 triangle; therefore, we know the legs of our triangle are congruent and that each can be found by dividing the length of the hypotenuse by $\sqrt{2}$ .

$\frac{8}{\sqrt{2}}=\frac{8}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{8\sqrt{2}}{2}$

Rationalizing the denominator simplifies our result; however, we are interested in the area, not just the length of a leg; we remember that the formula for the area of a triangle is

$A = \frac{1}{2}bh$

where is the base and is the height; however, in our right triangle, the base and height are simply the two legs; therefore, we can calculate the area by substituting.

$A=\frac{1}{2}(\frac{8\sqrt2}{2})(\frac{8\sqrt2}{2})=\frac{128}{8}=16$

Report an Error

Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of a 45/45/90 triangle that has a hypotenuse of $5\sqrt3$ .

Possible Answers:

$\frac{75}{4}$

$\frac{75}{2}$

$\sqrt{\frac{75}{2}}$

37.5

$\frac{5\sqrt{6}}{2}$

Correct answer:

$\frac{75}{4}$

Explanation:

To find the area of a triangle, we must use $Area=\frac{1}{2}b\cdot h$ where b=base and h=height.

In the problem, the only information given is what type the triangle is and what its hypotenuse is.

Given the area equation, the problem hasn't given any numbers that can be substituted into the equation to solve for an area. This means that the hypotenuse value must be used to determine the height and the base.

Because this is a 45/45/90 triangle, this means that it is also isosceles. Therefore, we can logic out that the base and the height must be the same.

The missing sides can be calulated in one of two ways:

1. Using the Pythagorean Theorem a^2+b^2=c^2

2. Or using Find_the_leg_length_resolution

If we were to use the Pythagorean Theorem, since we've already determined that b=h, that mean a=b in the equation. Let's say that a=b=s .

That means the Pythagorean Theorem can be rewritten as:

s^2+s^2=c^2

${\color{Blue} 2s^2=c^2}$

Now to substitute in the value of c to solve for the height and base.

$2s^2=(5\sqrt{3})^2$

$2s^2=25\sqrt{9}$

2s^2=75

$s^2=\frac{75}{2}$

$\sqrt{s^2}=\sqrt{\frac{75}{2}}$

$s=\frac{5\sqrt{6}}{2}$

Now that we have the base and the height, we can substitute the values into the area equation and get the triangle's area.

$Area=\frac{1}{2}\cdot\frac{5\sqrt{6}}{2}\cdot\frac{5\sqrt{6}}{2}$

$Area=\frac{1}{2}\cdot\frac{(25\cdot 6)}{4}$

$Area = \frac{1}{2}\cdot\frac{150}{4}$

$Area = \frac{1}{2}\cdot \frac{75}{2} = {\color{Blue} \frac{75}{4}}$

Report an Error

View Tutors

Judith
Certified Tutor

Ohio University-Main Campus, Bachelor in Arts, Global Studies.

View Tutors

Mark
Certified Tutor

Murray State University, Bachelor of Science, Mathematics.

View Tutors

Alex
Certified Tutor

Vanderbilt University, Bachelor of Science, Psychology. Boston University, Master of Science, Speech-Language Pathology.

All Basic Geometry Resources

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

Popular Subjects

Reading Tutors in Houston, Math Tutors in Washington DC, Algebra Tutors in Dallas Fort Worth, SAT Tutors in Boston, Computer Science Tutors in Denver, Spanish Tutors in Philadelphia, Spanish Tutors in San Francisco-Bay Area, Spanish Tutors in Houston, ACT Tutors in Phoenix, GRE Tutors in New York City

Popular Courses & Classes

LSAT Courses & Classes in Philadelphia, SSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in Atlanta, GRE Courses & Classes in Philadelphia, ACT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Chicago, Spanish Courses & Classes in Houston, LSAT Courses & Classes in Chicago, SAT Courses & Classes in Atlanta

Popular Test Prep

SSAT Test Prep in Miami, ACT Test Prep in San Diego, MCAT Test Prep in Miami, GMAT Test Prep in Denver, ISEE Test Prep in Seattle, GMAT Test Prep in San Diego, LSAT Test Prep in New York City, MCAT Test Prep in Chicago, LSAT Test Prep in Philadelphia, GMAT Test Prep in Seattle

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 #71 : Triangles

Example Question #72 : Triangles

Example Question #1051 : Basic Geometry

Example Question #71 : 45/45/90 Right Isosceles Triangles

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #1 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: