Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Quadrilaterals

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 #28 : How To Find The Area Of A Parallelogram

In the figure, the height of the parallelogram is half the height of the equilateral triangle. Find the area of the shaded region.

Possible Answers:

39.71

49.62

40.08

43.30

Correct answer:

43.30

Explanation:

First, we will need to find the height of the equilateral triangle.

Recall that the height of an equilateral triangle will cut the equilateral triangle into two congruent $30^\circ-60^\circ-90^\circ$ triangles that have side lengths in the the following ratio:

$1:\sqrt3:2$

Use the given side length of the equilateral triangle in order to find the length of the height.

$\frac{\text{side length}}{2}=\frac{\text{height}}{\sqrt3}$

$\text{height}=(\sqrt3)\times \frac{\text{side length}}{2}$

$\text{height}=\frac{20(\sqrt3)}{2}$

$\text{height}=10\sqrt3$

Now, find the area of the equilateral triangle.

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

$\text{Area of Equilateral Triangle}=\frac{1}{2}(20\times10\sqrt3)$

$\text{Area of Equilateral Triangle}=100\sqrt3$

Now, use the height of the equilateral triangle to find the height of the parallelogram.

$\text{height of parallelogram}=\frac{\text{height of triangle}}{2}$

$\text{height of parallelogram}=\frac{10\sqrt3}{2}$

$\text{height of parallelogram}=5\sqrt3$

Next, find the area of the parallelogram.

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

$\text{Area of Parallelogram}=15\times 5\sqrt3$

$\text{Area of Parallelogram}=75\sqrt3$

Now, from the figure, we can see that we will need to subtract the area of the parallelogram from the area of the triangle in order to find the area of the shaded region.

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

$\text{Area of Shaded Region}=100\sqrt3-75\sqrt3$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=43.30$

Report an Error

Example Question #29 : How To Find The Area Of A Parallelogram

In the figure, the height of the parallelogram is half the height of the equilateral triangle. Find the area of the shaded region.

Possible Answers:

82.00

90.07

95.26

85.63

Correct answer:

90.07

Explanation:

First, we will need to find the height of the equilateral triangle.

Recall that the height of an equilateral triangle will cut the equilateral triangle into two congruent $30^\circ-60^\circ-90^\circ$ triangles that have side lengths in the the following ratio:

$1:\sqrt3:2$

Use the given side length of the equilateral triangle in order to find the length of the height.

$\frac{\text{side length}}{2}=\frac{\text{height}}{\sqrt3}$

$\text{height}=(\sqrt3)\times \frac{\text{side length}}{2}$

$\text{height}=\frac{26(\sqrt3)}{2}$

$\text{height}=13\sqrt3$

Now, find the area of the equilateral triangle.

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

$\text{Area of Equilateral Triangle}=\frac{1}{2}(26\times13\sqrt3)$

$\text{Area of Equilateral Triangle}=169\sqrt3$

Now, use the height of the equilateral triangle to find the height of the parallelogram.

$\text{height of parallelogram}=\frac{\text{height of triangle}}{2}$

$\text{height of parallelogram}=\frac{13\sqrt3}{2}$

Next, find the area of the parallelogram.

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

$\text{Area of Parallelogram}=18\times \frac{13\sqrt3}{2}$

$\text{Area of Parallelogram}=117\sqrt3$

Now, from the figure, we can see that we will need to subtract the area of the parallelogram from the area of the triangle in order to find the area of the shaded region.

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

$\text{Area of Shaded Region}=169\sqrt3-117\sqrt3$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=90.07$

Report an Error

Example Question #30 : How To Find The Area Of A Parallelogram

In the figure, the height of the parallelogram is half the height of the equilateral triangle. Find the area of the shaded region.

Possible Answers:

174.06

162.39

173.21

181.29

Correct answer:

173.21

Explanation:

First, we will need to find the height of the equilateral triangle.

Recall that the height of an equilateral triangle will cut the equilateral triangle into two congruent $30^\circ-60^\circ-90^\circ$ triangles that have side lengths in the the following ratio:

$1:\sqrt3:2$

Use the given side length of the equilateral triangle in order to find the length of the height.

$\frac{\text{side length}}{2}=\frac{\text{height}}{\sqrt3}$

$\text{height}=(\sqrt3)\times \frac{\text{side length}}{2}$

$\text{height}=\frac{40(\sqrt3)}{2}$

$\text{height}=20\sqrt3$

Now, find the area of the equilateral triangle.

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

$\text{Area of Equilateral Triangle}=\frac{1}{2}(40\times20\sqrt3)$

$\text{Area of Equilateral Triangle}=400\sqrt3$

Now, use the height of the equilateral triangle to find the height of the parallelogram.

$\text{height of parallelogram}=\frac{\text{height of triangle}}{2}$

$\text{height of parallelogram}=\frac{20\sqrt3}{2}$

$\text{height of parallelogram}=10\sqrt3$

Next, find the area of the parallelogram.

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

$\text{Area of Parallelogram}=30\times 10\sqrt3$

$\text{Area of Parallelogram}=300\sqrt3$

Now, from the figure, we can see that we will need to subtract the area of the parallelogram from the area of the triangle in order to find the area of the shaded region.

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

$\text{Area of Shaded Region}=400\sqrt3-300\sqrt3$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=173.21$

Report an Error

Example Question #71 : Parallelograms

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Cannot be determined

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+1)=6

Expand this equation.

x^2+x=6

x^2+x-6=0

Now factor this equation.

(x-2)(x+3)=6

Solve for .

x=2, x=-3

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #32 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 100 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-48)=100

Expand this equation.

x^2-48x=100

x^2-48x-100=0

Now factor this equation.

(x-50)(x+2)=0

Solve for .

x=50, x=-2

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #31 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 120 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-14)=120

Expand this equation.

x^2-14x=120

x^2-14x-120=0

Now factor this equation.

(x-20)(x+6)=0

Solve for .

x=20, x=-6

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #31 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 240 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+34)=240

Expand this equation.

x^2+34x=240

x^2+34x-240=0

Now factor this equation.

(x-6)(x+40)=0

Solve for .

x=6, x=-40

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #411 : Intermediate Geometry

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+11)=80

Expand this equation.

x^2+11x=80

x^2+11x-80=0

Now factor this equation.

(x-5)(x+16)=0

Solve for .

x=5, x=-16

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #412 : Intermediate Geometry

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+13)=90

Expand this equation.

x^2+13x=90

x^2+13x-90=0

Now factor this equation.

(x-5)(x+18)=0

Solve for .

x=5, x=-18

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #71 : Parallelograms

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-5)=84

Expand this equation.

x^2-5x=84

x^2-5x-84=0

Now factor this equation.

(x-12)(x+7)=0

Solve for .

x=12, x=-7

Since lengths of bases and heights can only be positive, x=12 .

Notice that the length of the base is given by the expression x-5 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x-5$

$\text{Length of Base}=12-5$

$\text{Length of Base}=7$

Report an Error

View Tutors

Jacob
Certified Tutor

Ohio State University-Main Campus, Bachelor of Science, Political Science and Government.

View Tutors

Brandon
Certified Tutor

Flager College, Bachelor, Elementary Education .

View Tutors

Alexander
Certified Tutor

Manhattan College, Bachelor of Science, Mathematics. New York University, Doctor of Philosophy, Mathematics Teacher Education.

All Intermediate Geometry Resources

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

Popular Subjects

GRE Tutors in San Francisco-Bay Area, Algebra Tutors in Chicago, LSAT Tutors in Denver, Calculus Tutors in New York City, Math Tutors in Houston, Computer Science Tutors in Chicago, SSAT Tutors in Washington DC, Calculus Tutors in Boston, English Tutors in Dallas Fort Worth, Chemistry Tutors in Seattle

Popular Courses & Classes

GMAT Courses & Classes in Washington DC, SSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Seattle, MCAT Courses & Classes in San Diego, SSAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in New York City, MCAT Courses & Classes in Atlanta, Spanish Courses & Classes in Los Angeles, SSAT Courses & Classes in Los Angeles

Popular Test Prep

SAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Seattle, SAT Test Prep in Houston, ACT Test Prep in Dallas Fort Worth, SAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Seattle, SSAT Test Prep in Boston, ISEE Test Prep in Atlanta, SSAT Test Prep in Miami

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 : Quadrilaterals

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #28 : How To Find The Area Of A Parallelogram

Example Question #29 : How To Find The Area Of A Parallelogram

Example Question #30 : How To Find The Area Of A Parallelogram

Example Question #71 : Parallelograms

Example Question #32 : How To Find The Area Of A Parallelogram

Example Question #31 : How To Find The Area Of A Parallelogram

Example Question #31 : How To Find The Area Of A Parallelogram

Example Question #411 : Intermediate Geometry

Example Question #412 : Intermediate Geometry

Example Question #71 : Parallelograms

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: