Call Now to Set Up Tutoring:

(888) 888-0446

ISEE Upper Level Quantitative : Rectangles

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

Create An Account

All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

Example Question #1 : How To Find The Area Of A Rectangle

x > 0, y > 0

Which is the greater quantity?

(a) The area of the rectangle on the coordinate plane with vertices (0,0), (x,0 ), (0,4y), (x, 4y)

(b) The area of the rectangle on the coordinate plane with vertices (0,0), (2x,0 ), (0,2y), (2x, 2y)

Possible Answers:

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

Explanation:

(a) The first rectangle has width x- 0 = x and height 4y - 0 = 4y ; its area is $A = x \cdot 4y = 4xy$ .

(b) The second rectangle has width 2x- 0 = 2x and height 2y - 0 = 2y ; its area is $A = 2x \cdot 2y = 4xy$ .

The areas of the rectangle are the same.

Report an Error

Example Question #1 : How To Find The Area Of A Rectangle

A rectangle on the coordinate plane has its vertices at the points (3,4), (-6,4), (-6,-2), (3,-2) .

Which is the greater quantity?

(a) The area of the portion of the rectangle in Quadrant I

(b) The area of the portion of the rectangle in Quadrant III

Possible Answers:

It is impossible to tell from the information given.

(b) is greater.

(a) is greater.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

Explanation:

(a) The portion of the rectangle in Quadrant I is a rectangle with vertices (3,4), (0,4), (0,0), (3,0) , so its area is $3 \times 4 = 12$ .

(a) The portion of the rectangle in Quadrant III is a rectangle with vertices (0,0), (-6,0), (-6,-2), (0,-2) , so its area is $6 \times 2 = 12$ .

Report an Error

Example Question #1 : How To Find The Area Of A Rectangle

A >0 . Which is the greater quantity?

(a) The area of the square on the coordinate plane with vertices (0,0), (A,0 ), (0,A), (A,A)

(b) The area of the rectangle on the coordinate plane with vertices (0,0), (A+1,0 ), (0,A-1), (A+1,A-1)

Possible Answers:

(a) is greater.

(a) and (b) are equal.

(b) is greater.

It is impossible to tell from the information given.

Correct answer:

(a) is greater.

Explanation:

(a) The square has sidelength A - 0 = A , and therefore has area $A ^{2}$ .

(b) The rectangle has width A + 1 - 0 = A + 1 and height A - 1 - 0 = A - 1 , and therefore has area $(A + 1)(A - 1) = A^{2} - 1$ .

Since $A^{2} > A^{2} - 1$ , the square in (a) has the greater area.

Report an Error

Example Question #4 : How To Find The Area Of A Rectangle

The perimeter of a rectangle is one yard. The rectangle is three times as long as it is wide. Which is the greater quantity?

(a) The area of the rectangle

(b) 60 square inches

Possible Answers:

(a) is greater

(b) is greater

It is impossible to tell form the information given

(a) and (b) are equal

Correct answer:

(a) is greater

Explanation:

Let be the width of the rectangle. Then its length is , and its perimeter is

P = 2 (3W + W) = 8W

Since the perimeter is one yard, or 36 inches,

$8W\div 8 = 36 \div 8$

$W = 4 \frac{1}{2}$ inches is the wiidth, and $3W = 3\cdot 4 \frac{1}{2}= 13 \frac{1}{2}$ inches is the length, so the area is

$4 \frac{1}{2} \cdot 13 \frac{1}{2} = 60 \frac{3}{4}$ square inches. (a) is the greater quantity.

Report an Error

Example Question #1 : How To Find The Area Of A Rectangle

If one rectangular park measures $45ft\times 20ft$ and another rectangular park measures $30ft\times 90ft$ , how many times greater is the area of the second park than the area of the first park?

Possible Answers:

2.5

Correct answer:

Explanation:

First, you must compute the area of both parks. The area of a rectangle is length times width. Therefore, the area of park one is $45\cdot 20$ , which is 900 . The area of park two is $30\cdot 90$ , which is 2700. Then, divide the area of the second park by the area of the first park ( $2700\div 900$ ). This yields 3 as the answer.

Report an Error

Example Question #6 : How To Find The Area Of A Rectangle

The sum of the lengths of three sides of a rectangle is 572 inches; the width of the rectangle is 60% of its length. Give its area in square inches.

Possible Answers:

It is impossible to determine the area from the given information.

$29,040 \textrm{ in}^{2}$

$40,560 \textrm{ in}^{2}$

$67,600\textrm{ in}^{2}$

$48,400\textrm{ in}^{2}$

Correct answer:

It is impossible to determine the area from the given information.

Explanation:

Since the width of the rectangle is 60% of its length, we can write W = 0.6L .

However, it is not clear from the problem which three sides - two lengths and a width or two widths and a length - we are choosing to have sum 572 inches. Depending on the three sides chosen, we can either set up

2L + W = 572

2L + 0.6L = 572

2.6L = 572

$2.6L \div 2.6 = 572 \div 2.6$

L = 220

L + 2W = 572

L + 2 (0.6L) = 572

L +1.2L = 572

2.2L = 572

$2.2L \div 2.2 = 260$

Since the length cannot be determined with certainty, neither can the width, and, subsequently, neither can the area.

Report an Error

Example Question #7 : How To Find The Area Of A Rectangle

Five rectangles each have the same length, which we will call . The widths of the five rectangles are 7, 5, 8, 10, and 12. Which of the following expressions is equal to the mean of their areas?

Possible Answers:

16.8L

2L + 16.8

2L + 16

8.4L

Correct answer:

8.4L

Explanation:

The area of a rectangle is the product of its width and its length. The areas of the five rectangles, therefore, are 7L, 5L, 8L, 10L, 12L . The mean of these five areas is their sum divided by 5, or

$\left (7L+5L+ 8L+10L+ 12L \right ) \div 5$

$= \left (7+5+ 8+10+ 12 \right ) L \div 5$

$=42L \div 5$

= 8.4 L

Report an Error

Example Question #8 : How To Find The Area Of A Rectangle

Two rectangles, A and B, each have perimeter 32 feet. Rectangle A has length 12 feet; Rectangle B has length 10 feet. The area of Rectangle A is what percent of the area of Rectangle B?

Possible Answers:

$75 \%$

$83 \frac{1}{3} \%$

$80 \%$

$60 \%$

$87 \frac{1}{2} \%$

Correct answer:

$80 \%$

Explanation:

The perimeter of a rectangle can be given by the formula

P = 2(L+W)

Since for both rectangles, 30 is the perimeter, this becomes

2(L+W) = 32 , and subsequently

L+W = 16

w = 16 - W .

Rectangle A has length 12 feet and, subsequently. width 4 feet, making its area

$12 \times 4 = 48$ square feet

Rectangle B has length 10 feet and, subsequently. width 6 feet, making its area

$10 \times 6 = 60$ square feet

The area of Rectangle A is

$\frac{48}{60} \times 100 = 80 \%$

of the area of Rectangle B.

Report an Error

Example Question #9 : How To Find The Area Of A Rectangle

Rectangle

Give the area of the above rectangle in terms of .

Possible Answers:

g + 5

10g

$25g^{2}$

2g + 10

Correct answer:

Explanation:

The area of a rectangle is equal to the product of its length and height, which here are 5 and . This product is .

Report an Error

Example Question #10 : How To Find The Area Of A Rectangle

A rectangle is two feet shorter than twice its width; its perimeter is six yards. Give its area in square inches.

Possible Answers:

$1,386 \textrm{ in}^{2}$

$7,128 \textrm{ in}^{2}$

$2,772 \textrm{ in}^{2}$

$2,816 \textrm{ in}^{2}$

$2,475 \textrm{ in}^{2}$

Correct answer:

$2,816 \textrm{ in}^{2}$

Explanation:

The length of the rectangle is two feet, or 24 inches, shorter than twice the width, so, if is the width in inches, the length in inches is

L = 2W - 24

Six yards, the perimeter of the rectangle, is equal to $36 \times 6 = 216$ inches. The perimeter, in terms of length and width, is P = 2L + 2W , so we can set up the equation:

2L + 2W = P

2 (2W-24) + 2W = 216

4W-48 + 2W= 216

6W-48 = 216

6W-48+ 48 = 216 + 48

6W= 264

$6W \div 6 = 264 \div 6$

W = 44

L = 2W - 24 = 2(44)-24 = 88 - 24 = 64

The length and width are 64 inches and 44 inches; the area is their product, which is

$44 \times 64 = 2,816$ square inches

Report an Error

← Previous 1 2 3 Next →

View Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Tutors

Kaitlyn
Certified Tutor

Fairfield University, Bachelor of Science, Biology, General. NUI Galway Ireland, Master of Science, Neuroscience.

View Tutors

Pushpendra
Certified Tutor

university of allahabad, Bachelor of Science, Mathematics.

All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Miami, Chemistry Tutors in Dallas Fort Worth, ACT Tutors in San Diego, Calculus Tutors in Dallas Fort Worth, Statistics Tutors in Seattle, English Tutors in Los Angeles, Chemistry Tutors in Miami, Calculus Tutors in Boston, Physics Tutors in Houston, Statistics Tutors in Atlanta

Popular Courses & Classes

SSAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Philadelphia, GRE Courses & Classes in Seattle, MCAT Courses & Classes in Chicago, GMAT Courses & Classes in Boston, LSAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Miami, ISEE Courses & Classes in Miami, SSAT Courses & Classes in Miami, GRE Courses & Classes in Chicago

Popular Test Prep

GRE Test Prep in San Diego, GRE Test Prep in Seattle, GRE Test Prep in Chicago, GRE Test Prep in Philadelphia, LSAT Test Prep in Houston, LSAT Test Prep in New York City, SSAT Test Prep in Los Angeles, ACT Test Prep in Washington DC, GMAT Test Prep in Los Angeles, MCAT Test Prep in Chicago

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

ISEE Upper Level Quantitative : Rectangles

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #1 : How To Find The Area Of A Rectangle

Example Question #1 : How To Find The Area Of A Rectangle

Example Question #1 : How To Find The Area Of A Rectangle

Example Question #4 : How To Find The Area Of A Rectangle

Example Question #1 : How To Find The Area Of A Rectangle

Example Question #6 : How To Find The Area Of A Rectangle

Example Question #7 : How To Find The Area Of A Rectangle

Example Question #8 : How To Find The Area Of A Rectangle

Example Question #9 : How To Find The Area Of A Rectangle

Example Question #10 : How To Find The Area Of A Rectangle

All ISEE Upper Level Quantitative Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: