We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

Create An Account

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #9 : How To Find The Area Of An Acute / Obtuse Triangle

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$68.3\:units^2$

$75.1\:units^2$

$153.5\:units^2$

$182.4\:units^2$

Correct answer:

$153.5\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(24^{\circ})=\frac{height}{14}$

Now, solve for the height.

$\text{Height}=14\cos(24^{\circ})=12.79$

Now you can find the area.

$\text{Area}=\frac{24\times 12.79}{2}=153.5\:units^2$

Report an Error

Example Question #10 : How To Find The Area Of An Acute / Obtuse Triangle

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$124.2\:units^2$

$44.5\:units^2$

$81.9\:units^2$

$68.8\:units^2$

Correct answer:

$68.8\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(55^{\circ})=\frac{height}{12}$

Now, solve for the height.

$\text{Height}=12\sin(55^{\circ})=9.83$

Now you can find the area.

$\text{Area}=\frac{14\times 9.83}{2}=68.8\:units^2$

Report an Error

Example Question #71 : Acute / Obtuse Triangles

Find the area of the triangle. Round to the nearest tenths.

Possible Answers:

$35.4\:units^2$

$22.9\:units^2$

$42.5\:units^2$

$65.1\:units^2$

Correct answer:

$35.4\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(41^{\circ})=\frac{height}{9}$

Now, solve for the height.

$\text{Height}=9\sin(41^{\circ})=5.90$

Now you can find the area.

$\text{Area}=\frac{12\times 5.90}{2}=35.4\:units^2$

Report an Error

Example Question #72 : Acute / Obtuse Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$24.8\:units^2$

$8.2\:units^2$

$14.7\:units^2$

$16.5\:units^2$

Correct answer:

$16.5\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(36^{\circ})=\frac{height}{4}$

Now, solve for the height.

$\text{Height}=4\sin(36^{\circ})=2.35$

Now you can find the area.

$\text{Area}=\frac{14\times 2.35}{2}=16.5\:units^2$

Report an Error

Example Question #133 : Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$145.2\:units^2$

$97.9\:units^2$

$80.3\:units^2$

$79.8\:units^2$

Correct answer:

$80.3\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(42^{\circ})=\frac{height}{8}$

Now, solve for the height.

$\text{Height}=8\sin(42^{\circ})=5.35$

Now you can find the area.

$\text{Area}=\frac{30\times 5.35}{2}=80.3\:units^2$

Report an Error

Example Question #73 : Acute / Obtuse Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$116.0\:units^2$

$111.4\:units^2$

$104.4\:units^2$

$92.8\:units^2$

Correct answer:

$116.0\:units^2$

Explanation:

The formula used to find the area of the triangle is

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

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(48^{\circ})=\frac{height}{12}$

Now, solve for the height.

$\text{Height}=12\sin(48^{\circ})=8.92$

Now you can find the area.

$\text{Area}=\frac{26\times 8.92}{2}=116.0\:units^2$

Report an Error

Example Question #131 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

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

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=6 \times 12$

$\text{Area of Rectangle}= 72$

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

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

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

$\text{height}=\frac{6}{2}$

$\text{height}=3$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 3$

$\text{Area of Triangle}=18$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=72-18$

Solve.

$\text{Area of Shaded Region}=54$

Report an Error

Example Question #131 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

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

Substitute in the given length and width to find the area.

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

$\text{Area of Rectangle}= 96$

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

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

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

$\text{height}=\frac{8}{2}$

$\text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 4$

$\text{Area of Triangle}=24$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=96-24$

Solve.

$\text{Area of Shaded Region}=72$

Report an Error

Example Question #17 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

168

154

132

170

Correct answer:

168

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

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

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=14 \times 16$

$\text{Area of Rectangle}= 224$

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

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

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

$\text{height}=\frac{14}{2}$

$\text{height}=7$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 16\times 7$

$\text{Area of Triangle}=56$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=224-56$

Solve.

$\text{Area of Shaded Region}=168$

Report an Error

Example Question #18 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{5}{2}$

$\frac{7}{2}$

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

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

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=1 \times 4$

$\text{Area of Rectangle}= 4$

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

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

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

$\text{height}=\frac{1}{2}$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 4\times \frac{1}{2}$

$\text{Area of Triangle}=1$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=4-1$

Solve.

$\text{Area of Shaded Region}=3$

Report an Error

View Tutors

Manikya
Certified Tutor

Osmania University, Bachelor of Science, Biology Teacher Education. Osmania University, Master of Science, Biology, General.

View Tutors

Clay
Certified Tutor

Grand Canyon University, Bachelor in Arts, History.

View Tutors

Michele
Certified Tutor

Georgian Court University, Bachelor in Arts, English. Georgian Court University, Master of Arts, Elementary and Middle School...

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Atlanta, French Tutors in Philadelphia, SSAT Tutors in Houston, Physics Tutors in Boston, Chemistry Tutors in Denver, Computer Science Tutors in Dallas Fort Worth, GMAT Tutors in Miami, SSAT Tutors in Philadelphia, MCAT Tutors in Houston, ISEE Tutors in Chicago

Popular Courses & Classes

SAT Courses & Classes in Philadelphia, GRE Courses & Classes in Los Angeles, MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Houston, GMAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Washington DC, SAT Courses & Classes in Miami, SAT Courses & Classes in New York City, Spanish Courses & Classes in Denver, SAT Courses & Classes in Los Angeles

Popular Test Prep

ACT Test Prep in Denver, SAT Test Prep in Boston, MCAT Test Prep in Dallas Fort Worth, SAT Test Prep in New York City, ACT Test Prep in Miami, SSAT Test Prep in New York City, ACT Test Prep in Atlanta, SSAT Test Prep in Los Angeles, MCAT Test Prep in Phoenix, GMAT Test Prep in Dallas Fort Worth

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

Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #9 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #10 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #71 : Acute / Obtuse Triangles

Example Question #72 : Acute / Obtuse Triangles

Example Question #133 : Triangles

Example Question #73 : Acute / Obtuse Triangles

Example Question #131 : Triangles

Example Question #131 : Triangles

Example Question #17 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #18 : How To Find The Area Of An Acute / Obtuse Triangle

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: