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 #101 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is $22\sqrt2$ .

Possible Answers:

144

$3\sqrt{111}$

484

121

242

Correct answer:

242

Explanation:

Notice that the given triangle is a right isosceles triangle. The hypotenuse of the triangle is the same as the diameter of the circle; therefore, we can use the Pythagorean theorem to find the length of the legs of this triangle.

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

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

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

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

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

$\text{side}=\frac{22\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=22$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{22\times 22}{2}$

Solve.

$\text{Area}=242$

Report an Error

Example Question #102 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is .

Possible Answers:

$4\sqrt{3}$

Correct answer:

Explanation:

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

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

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

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

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

$\text{side}=\frac{8(\sqrt2)}{2}$

Simplify.

$\text{side}=4\sqrt2$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{4\sqrt2\times 4\sqrt2}{2}$

Solve.

$\text{Area}=16$

Report an Error

Example Question #103 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is $10\sqrt2$ .

Possible Answers:

110

125

$2\sqrt{45}$

100

Correct answer:

Explanation:

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

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

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

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

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

$\text{side}=\frac{10\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=10$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{10\times 10}{2}$

Solve.

$\text{Area}=50$

Report an Error

Example Question #104 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

$3\sqrt3$

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 3$

Simplify.

$\text{Hypotenuse}=6$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{6^2}{4}$

Solve.

$\text{Area}=18$

Report an Error

Example Question #105 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is

Possible Answers:

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 4$

Simplify.

$\text{Hypotenuse}=8$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{8^2}{4}$

Solve.

$\text{Area}=16$

Report an Error

Example Question #106 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 2$

Simplify.

$\text{Hypotenuse}=4$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{4^2}{4}$

Solve.

$\text{Area}=4$

Report an Error

Example Question #107 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 1$

Simplify.

$\text{Hypotenuse}=2$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{2^2}{4}$

Solve.

$\text{Area}=1$

Report an Error

Example Question #108 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

100

125

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 5$

Simplify.

$\text{Hypotenuse}=10$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{10^2}{4}$

Solve.

$\text{Area}=25$

Report an Error

Example Question #109 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

144

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2 \times6$

Simplify.

$\text{Hypotenuse}=12$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{12^2}{4}$

Solve.

$\text{Area}=36$

Report an Error

Example Question #110 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 7$

Simplify.

$\text{Hypotenuse}=14$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{14^2}{4}$

Solve.

$\text{Area}=49$

Report an Error

View Tutors

Hirdeshwar
Certified Tutor

Indian Institute of Technology, Powai, Bombay, India, Doctor of Philosophy, Mathematics.

View Tutors

Megha
Certified Tutor

Navrachana University,, Bachelor of Technology, Mechanical Engineering. Lakehead University, Master of Science, Mechanical En...

View Tutors

William
Certified Tutor

CUNY Graduate School and University Center, Doctor of Philosophy, Theater Literature, History, and Criticism.

All Basic Geometry Resources

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

Popular Subjects

Computer Science Tutors in San Francisco-Bay Area, French Tutors in Atlanta, Biology Tutors in Seattle, Calculus Tutors in Denver, Spanish Tutors in Dallas Fort Worth, Reading Tutors in Chicago, SAT Tutors in San Francisco-Bay Area, MCAT Tutors in Atlanta, SSAT Tutors in Phoenix, ACT Tutors in Philadelphia

Popular Courses & Classes

GMAT Courses & Classes in Washington DC, GRE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Philadelphia, ISEE Courses & Classes in Los Angeles, ACT Courses & Classes in Phoenix, MCAT Courses & Classes in San Diego, Spanish Courses & Classes in Boston, GRE Courses & Classes in Seattle, ISEE Courses & Classes in San Diego, GMAT Courses & Classes in Seattle

Popular Test Prep

SSAT Test Prep in Washington DC, ACT Test Prep in New York City, ACT Test Prep in Phoenix, MCAT Test Prep in Seattle, GMAT Test Prep in Atlanta, ISEE Test Prep in San Diego, SSAT Test Prep in New York City, GMAT Test Prep in Seattle, MCAT Test Prep in Dallas Fort Worth, ISEE 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 #101 : 45/45/90 Right Isosceles Triangles

Example Question #102 : 45/45/90 Right Isosceles Triangles

Example Question #103 : 45/45/90 Right Isosceles Triangles

Example Question #104 : 45/45/90 Right Isosceles Triangles

Example Question #105 : 45/45/90 Right Isosceles Triangles

Example Question #106 : 45/45/90 Right Isosceles Triangles

Example Question #107 : 45/45/90 Right Isosceles Triangles

Example Question #108 : 45/45/90 Right Isosceles Triangles

Example Question #109 : 45/45/90 Right Isosceles Triangles

Example Question #110 : 45/45/90 Right Isosceles Triangles

All Basic Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: