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ISEE Upper Level Quantitative : Quadrilaterals

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

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All ISEE Upper Level Quantitative Resources

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Example Questions

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:

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 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.

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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.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

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 $6 \times 2 = 12$ .

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Example Question #2 : 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:

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

(a) is greater.

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.

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Example Question #1 : Rectangles

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:

(b) is greater

(a) is greater

(a) and (b) are equal

It is impossible to tell form the information given

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.

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Example Question #3 : 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.

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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.

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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

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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.

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Example Question #74 : Quadrilaterals

Rectangle

Give the area of the above rectangle in terms of .

Possible Answers:

2g + 10

$25g^{2}$

g + 5

10g

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 .

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Example Question #3 : Rectangles

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

Possible Answers:

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

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

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

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

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

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

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ISEE Upper Level Quantitative : Quadrilaterals

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 #2 : How To Find The Area Of A Rectangle

Example Question #1 : Rectangles

Example Question #3 : 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 #74 : Quadrilaterals

Example Question #3 : Rectangles

All ISEE Upper Level Quantitative Resources

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