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 #145 : Triangles

A triangle is placed so that it shares a base with a parallelogram.

If the height of the triangle is one-third that of the parallelogram, find the area of the shaded region.

Possible Answers:

2.5

10.5

8.5

12.5

Correct answer:

12.5

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=3\times 5=15$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{3}$

$\text{Height of Triangle}=\frac{3}{3}=1$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(5\times 1)=\frac{5}{2}=2.5$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=15-2.5=12.5$

Report an Error

Example Question #146 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=14\times8=112$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{8}{2}=4$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(14 \times 4)=28$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=112-28=84$

Report an Error

Example Question #591 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half the height of the parallelogram, find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=10 \times 6 = 60$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{6}{2}=3$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(10 \times 3)=15$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=60-15=45$

Report an Error

Example Question #592 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half the height of the parallelogram, find the area of the shaded region.

Possible Answers:

508

496

522

532

Correct answer:

522

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=29\times 24 = 696$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{24}{2}=12$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(29\times 12)=174$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=696-174=522$

Report an Error

Example Question #591 : Intermediate Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-fifth that of the parallelogram, find the area of the shaded region.

Possible Answers:

650

675

700

625

Correct answer:

675

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=30 \times 25 = 750$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{5}$

$\text{Height of Triangle}=\frac{25}{5}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(30 \times 5)=75$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=750-75=675$

Report an Error

Example Question #593 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

795

775

755

765

Correct answer:

765

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 30 =1020$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{30}{2}=15$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(34 \times 15)=255$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=1020-255=765$

Report an Error

Example Question #155 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-ninth that of the parallelogram, find the area of the shaded region.

Possible Answers:

2100

2150

2175

2125

Correct answer:

2125

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=50 \times45 = 2250$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{9}$

$\text{Height of Triangle}=\frac{45}{9}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(50 \times 5)=125$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=2250-125=2125$

Report an Error

Example Question #156 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

1700

1900

1812

1716

Correct answer:

1716

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=52\times 44 = 2288$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{44}{2}=22$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(52\times 22)=572$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=2288-572=1716$

Report an Error

Example Question #157 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

510

490

480

500

Correct answer:

510

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 20 = 680$

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

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

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{20}{2}=10$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(340 \times 10)=170$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=680-170=510$

Report an Error

Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

In ΔABC: a = 10, c = 15, and B = 50°.

Find b (to the nearest tenth).

Possible Answers:

9.7

14.3

12.1

11.9

11.5

Correct answer:

11.5

Explanation:

This problem requires us to use either the Law of Sines or the Law of Cosines. To figure out which one we should use, let's write down all the information we have in this format:

A = ? a = 10

B = 50° b = ?

C = ? c = 15

Now we can easily see that we do not have any complete pairs (such as A and a, or B and b) but we do have one term from each pair (a, B, and c). This tells us that we should use the Law of Cosines. (We use the Law of Sines when we have one complete pair, such as B and b).

Law of Cosines:
$a^2 = b^2 + c^2 - 2bc\cos(A)$
$b^2 = a^2 + c^2 - 2ac\cos(B)$
$c^2 = a^2 + b^2 - 2ab\cos(C)$

Since we are solving for b, let's use the second version of the Law of Cosines.
This gives us:

$b^2 = 10^2 + 15^2 - 2(10)(15)\cos(50)$
b^2 = 100 + 225 - 192.836
b = 11.5

Report an Error

View Tutors

Michele
Certified Tutor

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

View Tutors

Sophia
Certified Tutor

Universidad del Azuay, Bachelor in Arts, International Relations.

View Tutors

Waren
Certified Tutor

Laguna State Polytechnic University, Master of Science, Science Technology.

All Intermediate Geometry Resources

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

Popular Subjects

ISEE Tutors in Los Angeles, Chemistry Tutors in Phoenix, GMAT Tutors in Seattle, GRE Tutors in Chicago, Spanish Tutors in New York City, ISEE Tutors in Miami, Chemistry Tutors in Boston, ISEE Tutors in Boston, MCAT Tutors in Boston, Math Tutors in Los Angeles

Popular Courses & Classes

SAT Courses & Classes in New York City, GRE Courses & Classes in Atlanta, MCAT Courses & Classes in Chicago, MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Denver, ISEE Courses & Classes in Seattle, ACT Courses & Classes in Philadelphia, MCAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Phoenix, GMAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

LSAT Test Prep in Philadelphia, ACT Test Prep in Philadelphia, SSAT Test Prep in Washington DC, SSAT Test Prep in Miami, SAT Test Prep in Philadelphia, ISEE Test Prep in Dallas Fort Worth, LSAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Phoenix, LSAT Test Prep in Miami, ISEE 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

Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #145 : Triangles

Example Question #146 : Triangles

Example Question #591 : Plane Geometry

Example Question #592 : Plane Geometry

Example Question #591 : Intermediate Geometry

Example Question #593 : Plane Geometry

Example Question #155 : Triangles

Example Question #156 : Triangles

Example Question #157 : Triangles

Example Question #1 : How To Find The Length Of The Side 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: