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 #51 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

75.71

72.72

74.27

76.98

Correct answer:

72.72

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

Now, in order to find the length of the rectangle, we will first need to find the length of the hypotenuse of the triangle. The question provides us with the following relationship between the lengths of the rectangle and hypotenuse:

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=20\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{20\sqrt2}{2}$

$\text{length}=10\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=10\sqrt2 \times 9$

$\text{Area of Rectangle}=90\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{20\times 20}{2}$

$\text{Area of Triangle}=200$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=200-90\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=72.72$

Report an Error

Example Question #52 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

55.55

51.51

62.62

58.58

Correct answer:

58.58

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=20\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{20\sqrt2}{2}$

$\text{length}=10\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=10\sqrt2 \times 10$

$\text{Area of Rectangle}=100\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{20\times 20}{2}$

$\text{Area of Triangle}=200$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=200-100\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=58.58$

Report an Error

Example Question #51 : 45/45/90 Right Isosceles Triangles

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

154.29

184.58

169.21

179.35

Correct answer:

169.21

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=24\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{24\sqrt2}{2}$

$\text{length}=12\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=12\sqrt2 \times 7$

$\text{Area of Rectangle}=84\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{24\times 24}{2}$

$\text{Area of Triangle}=288$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=288-84\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=169.21$

Report an Error

Example Question #1031 : Basic Geometry

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

118.29

121.48

120.59

113.22

Correct answer:

118.29

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=24\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{24\sqrt2}{2}$

$\text{length}=12\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=12\sqrt2 \times 10$

$\text{Area of Rectangle}=120\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{24\times 24}{2}$

$\text{Area of Triangle}=288$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=288-120\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=118.29$

Report an Error

Example Question #55 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

151.29

150.86

152.24

143.26

Correct answer:

152.24

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=24\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{24\sqrt2}{2}$

$\text{length}=12\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=12\sqrt2 \times 8$

$\text{Area of Rectangle}=96\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{24\times 24}{2}$

$\text{Area of Triangle}=288$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=288-96\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=152.24$

Report an Error

Example Question #56 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

79.41

88.23

80.61

90.58

Correct answer:

80.61

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=26\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{26\sqrt2}{2}$

$\text{length}=13\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=13\sqrt2 \times 14$

$\text{Area of Rectangle}=182\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{26\times 26}{2}=338$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=338-182\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=80.61$

Report an Error

Example Question #57 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

157.84

169.59

148.20

153.02

Correct answer:

153.02

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

$\text{length}=\frac{\text{hypotenuse}}{2}$

Recall how to find the hypotenuse of a right triangle:

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

Simplify.

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

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=30\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{30\sqrt2}{2}$

$\text{length}=15\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=15\sqrt2 \times 14$

$\text{Area of Rectangle}=210\sqrt2$

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

$\text{Area of Triangle}=\frac{\text{base}\times\text{height}}{2}$

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{30\times 30}{2}$

$\text{Area of Triangle}=450$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Rectangle}$

$\text{Area of Shaded Region}=450-210\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=153.02$

Report an Error

Example Question #58 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Right triangle 7

Refer to the above figure.

True or false: $\overline{AC}$ has length $16\sqrt{2}$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

The figure shows a right triangle. The acute angles of a right triangle have measures whose sum is $90^{\circ }$ , so

$m \angle A+ m \angle C = 90^{\circ }$

Substituting $45^{\circ }$ for $m \angle C$ :

$m \angle A+ 45^{\circ } = 90^{\circ }$

$m \angle A= 45^{\circ }$

This makes $\bigtriangleup ABC$ a 45-45-90 triangle. By the 45-45-90 Triangle Theorem, the length of hypotenuse $\overline{AC}$ is equal to that of leg $\overline{BC}$ multiplied by $\sqrt{2}$ . Therefore,

$AC = AB \cdot \sqrt{2} = 16\sqrt{2}$ .

Report an Error

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

$\fn_cm In\;\Delta ABC,\;\angle BAC=45^{\circ }.$

$\angle BCA = \, ?$

Possible Answers:

$135^{\circ}$

$60^{\circ}$

$90^{\circ}$

$\fn_cm 45^{\circ}$

Correct answer:

$\fn_cm 45^{\circ}$

Explanation:

$The\;degrees\;in\;a\;triangle\;sum\;to\;180^{\circ}. \; 180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}.$

Report an Error

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

Find the height of the 45-45-90 right triangle with a hypotenuse of $15\sqrt{6}$ .

Possible Answers:

$15\sqrt{2}$

$\sqrt{6}$

$15\sqrt{3}$

7.5

Correct answer:

$15\sqrt{3}$

Explanation:

If the hypotenuse of a 45-45-90 right triangle is $15\sqrt{6}$ then:

The height and the base of the triangle will be the same length since it is a 45-45-90 triangle (isosceles).

a^2+b^2=c^2

Since both a and b will be equal let's use a=b=x and $c=15\sqrt{6}$ :

$x^2 +x^2=(15\sqrt{6})^2$

2x^2=225*6

2x^2=1350

x^2=675

$x=\sqrt{675}$

$x=\sqrt{225*3}$

$x=\sqrt{225}\, \sqrt{3}$

$x=15\sqrt{3}$

Report an Error

View Tutors

Charlynn
Certified Tutor

Florida International University, Bachelor of Science, Business Administration and Management.

View Tutors

Richard
Certified Tutor

CUNY Brooklyn College, Bachelor of Science, Chemistry. CUNY Brooklyn College, Master of Arts, Experimental Psychology.

View Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

All Basic Geometry Resources

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

Popular Subjects

Statistics Tutors in Miami, Computer Science Tutors in New York City, Math Tutors in Los Angeles, Statistics Tutors in Atlanta, Biology Tutors in Miami, Calculus Tutors in Los Angeles, Physics Tutors in Denver, SAT Tutors in Miami, ACT Tutors in Phoenix, GRE Tutors in Phoenix

Popular Courses & Classes

MCAT Courses & Classes in Washington DC, Spanish Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in San Diego, ISEE Courses & Classes in Boston, LSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Denver, MCAT Courses & Classes in Houston, SSAT Courses & Classes in Miami, LSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in New York City

Popular Test Prep

SAT Test Prep in Atlanta, MCAT Test Prep in Philadelphia, LSAT Test Prep in New York City, LSAT Test Prep in Denver, LSAT Test Prep in Seattle, ACT Test Prep in Phoenix, GMAT Test Prep in Los Angeles, ISEE Test Prep in Seattle, SAT Test Prep in Denver, GRE Test Prep in San Diego

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 #51 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Example Question #52 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Example Question #51 : 45/45/90 Right Isosceles Triangles

Example Question #1031 : Basic Geometry

Example Question #55 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Example Question #56 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Example Question #57 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

Example Question #58 : How To Find The Length Of The Hypotenuse Of A 45/45/90 Right Isosceles Triangle : Pythagorean Theorem

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

Example Question #2 : How To Find The Height 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: