We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Parallelograms

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 #81 : Parallelograms

In the figure, the area of the parallelogram is 150 . 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)=150

Expand this equation.

x^2+5x=150

x^2+5x-150=0

Now factor this equation.

(x-10)(x+15)=0

Solve for .

x=10, x=-5

Since lengths of bases and heights can only be positive, x=10 .

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.

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

$\text{Length of Base}=10+5$

$\text{Length of Base}=15$

Report an Error

Example Question #421 : Intermediate Geometry

In the figure, the area of the parallelogram is 500 . 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)=500

Expand this equation.

x^2+5x=500

x^2+5x-500=0

Now factor this equation.

(x-20)(x+25)=0

Solve for .

x=20, x=-25

Since lengths of bases and heights can only be positive, x=20 .

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.

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

$\text{Length of Base}=20+5$

$\text{Length of Base}=25$

Report an Error

Example Question #40 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 164 . 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+37)=164

Expand this equation.

x^2+37x=164

x^2+37x-164=0

Now factor this equation.

(x-4)(x+41)=0

Solve for .

x=4, x=-41

Since lengths of bases and heights can only be positive, x=4 .

Notice that the length of the base is given by the expression x+37 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+37$

$\text{Length of Base}=4+37$

$\text{Length of Base}=41$

Report an Error

Example Question #81 : Parallelograms

In the figure, the area of the parallelogram is 256 . 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+24)=256

Expand this equation.

x^2+24x=256

x^2+24x-256=0

Now factor this equation.

(x-8)(x+32)=0

Solve for .

x=8, x=-32

Since lengths of bases and heights can only be positive, x=8 .

Notice that the length of the base is given by the expression x+24 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+24$

$\text{Length of Base}=8+24$

$\text{Length of Base}=32$

Report an Error

Example Question #82 : Parallelograms

In the figure, the area of the parallelogram is 128 . 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-8)=128

Expand this equation.

x^2-8x=128

x^2-8x-128=0

Now factor this equation.

(x-16)(x+8)=0

Solve for .

x=16, x=-8

Since lengths of bases and heights can only be positive, x=16 .

Notice that the length of the base is given by the expression x-8 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x-8$

$\text{Length of Base}=16-8$

$\text{Length of Base}=8$

Report an Error

Example Question #421 : Plane Geometry

Find the area of the parallelogram.

Possible Answers:

169.48

184.09

199.23

186.23

Correct answer:

186.23

Explanation:

Recall how to find the area of a parallelogram:

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

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{13^2-7^2}=\sqrt{169-49}=\sqrt{120}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=17\times\sqrt{120}=186.23$

Remember to round to places after the decimal.

Report an Error

Example Question #252 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

54.99

60.31

42.19

58.29

Correct answer:

54.99

Explanation:

Recall how to find the area of a parallelogram:

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

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=12\times\sqrt{21}=54.99$

Remember to round to places after the decimal.

Report an Error

Example Question #253 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

256

240

189

213

Correct answer:

240

Explanation:

Recall how to find the area of a parallelogram:

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

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{15^2-9^2}=\sqrt{225-81}=12$

Now, use the height to find the area of the parallelogram.

$\text{Area}=12\times20=240$

Remember to round to places after the decimal.

Report an Error

Example Question #254 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

62.61

69.22

60.06

78.45

Correct answer:

62.61

Explanation:

Recall how to find the area of a parallelogram:

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

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{6^2-4^2}=\sqrt{36-16}=\sqrt{20}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=16\times\sqrt{20}=62.61$

Remember to round to places after the decimal.

Report an Error

Example Question #255 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

93.67

90.88

82.50

95.61

Correct answer:

93.67

Explanation:

Recall how to find the area of a parallelogram:

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

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{8^2-5^2}=\sqrt{64-25}=\sqrt{39}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=15\times\sqrt{39}=93.67$

Remember to round to places after the decimal.

Report an Error

View Tutors

Adair
Certified Tutor

Marymount College, Bachelor of Science, Chemistry.

View Tutors

Paarth
Certified Tutor

New Jersey Institute of Technology, Bachelor of Science, Computer Science.

View Tutors

Joshua
Certified Tutor

Murray State University, Bachelor in Arts, Spanish. Lipscomb University, Bachelor of Science, Teacher Assistant.

All Intermediate Geometry Resources

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

Popular Subjects

LSAT Tutors in Miami, SAT Tutors in Denver, Algebra Tutors in Boston, Math Tutors in Philadelphia, Physics Tutors in San Francisco-Bay Area, MCAT Tutors in Washington DC, Chemistry Tutors in Miami, French Tutors in Boston, Statistics Tutors in Los Angeles, Biology Tutors in San Diego

Popular Courses & Classes

LSAT Courses & Classes in Houston, GMAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Phoenix, GMAT Courses & Classes in Houston, SAT Courses & Classes in New York City, Spanish Courses & Classes in San Diego, LSAT Courses & Classes in New York City, Spanish Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Boston, GRE Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Seattle, GRE Test Prep in Denver, SSAT Test Prep in Miami, GRE Test Prep in Miami, GMAT Test Prep in Miami, GMAT Test Prep in Philadelphia, LSAT Test Prep in Miami, SAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Atlanta, SSAT Test Prep in Los Angeles

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 : Parallelograms

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #81 : Parallelograms

Example Question #421 : Intermediate Geometry

Example Question #40 : How To Find The Area Of A Parallelogram

Example Question #81 : Parallelograms

Example Question #82 : Parallelograms

Example Question #421 : Plane Geometry

Example Question #252 : Quadrilaterals

Example Question #253 : Quadrilaterals

Example Question #254 : Quadrilaterals

Example Question #255 : Quadrilaterals

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: