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ISEE Upper Level Quantitative : Rectangles

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Example Question #21 : Rectangles

x > 0, y > 0

Which is the greater quantity?

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

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

Possible Answers:

(a) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

Correct answer:

It is impossible to tell from the information given.

Explanation:

(a) The first rectangle has width x- 0 = x and height 4y - 0 = 4y ; its perimeter is

P = x + 4y + x + 4y = 2x + 8y .

(b) The second rectangle has width 2x- 0 = 2x and height 2y - 0 = 2y ; its perimeter is

P = 2x + 2y + 2x + 2y = 4x + 4y .

For the first rectangle to have a greater perimeter, it is necessary for

2x + 8y > 4x + 4y , or equivalently,

2x + 8y- 2x - 4y > 4x + 4y - 2x - 4y .

4y > 2x

$4y\div 4> 2x\div 4$

$y > \frac{1}{2}x$

We do not know the relative values of and , however, so we cannot compare their perimeters.

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Example Question #22 : Rectangles

A rectangle is two feet longer than it is wide; its perimeter is 11 feet. What is its area in square inches?

Possible Answers:

$4,212 \textrm{ in}^{2}$

It is impossible to determine the area from the information given

$513 \textrm{ in}^{2}$

$945 \textrm{ in}^{2}$

$3,780 \textrm{ in}^{2}$

Correct answer:

$945 \textrm{ in}^{2}$

Explanation:

The length of the rectangle is 2 feet, or 24 inches, greater than the width, so, if is the width in inches, L = W + 24 is the length in inches.

The perimeter of the rectangle is 11 feet, or $11 \times 12= 132$ inches. The perimeter, in terms of length and width, is P = 2L + 2W , so we can set up the equation:

2L + 2W = P

2 (W+24) + 2W = 132

2 W+48 + 2W = 132

4 W+48 = 132

4 W+48 -48 = 132 -48

4 W =84

$4 W \div 4 =84 \div 4$

W =21

L = W + 24 = 21 + 24 = 45

The width is 21 inches, and the length is 45 inches. The area is their product:

$21\times 45 = 945$ square inches.

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Example Question #23 : Rectangles

Rectangle

Give the perimeter of the above rectangle in terms of .

Possible Answers:

10g

$5g^{2}$

2g+10

g+5

Correct answer:

2g+10

Explanation:

Opposite sides of a rectangle are of equal length, so the two missing sidelengths are 5 (right) and (bottom). The perimeter of the rectangle is the sum of the lengths of its sides:

g + g + 5 + 5 = 2g + 10

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Example Question #24 : Rectangles

The sum of the lengths of three sides of a square is one meter. Give the perimeter of the square in millimeters.

Possible Answers:

$125\textrm{ mm}$

$750 \textrm{ mm}$

$133 \frac{1}{3} \textrm{ mm}$

$1,333 \frac{1}{3} \textrm{ mm}$

$1,250 \textrm{ mm}$

Correct answer:

$1,333 \frac{1}{3} \textrm{ mm}$

Explanation:

A square has four sides of the same length.

The sum of the lengths of three sides of a square is one meter, which is equal to 1,000 millimeters, so each side has length

$1,000 \div 3 = 333 \frac{1}{3}$ millimeters,

and the perimeter is four times this, or

$4 \times 333 \frac{1}{3} = 1,333 \frac{1}{3}$ millimeters.

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Example Question #25 : Rectangles

The area of a rectangle is 4,480 square inches. Its width is 70% of its length.

What is its perimeter?

Possible Answers:

$302\textrm{ in}$

$272 \textrm{ in}$

$284\textrm{ in}$

$224\textrm{ in}$

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

Correct answer:

$272 \textrm{ in}$

Explanation:

If the width of the rectangle is 70% of the length, then

W = 0.70 L .

The area is the product of the length and width:

LW = A

$L \cdot 0.70 L = 4,480$

$0.70 L ^{2}= 4,480$

$0.70 L ^{2} \div 0.70 = 4,480\div 0.70$

$L ^{2} = 6,400$

$L = \sqrt{6,400} = 80$

$W = 0.70 L = 0.70 \cdot 80 = 56$

The perimeter is therefore

$P = 2L + 2W = 2 \cdot 80 + 2 \cdot 56 = 160 + 112 = 272$ inches.

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Example Question #21 : Rectangles

The area of Rectangle RECT is $x^{2} - 81$ . The length of $\overline{RE}$ is x- 9 . Give the perimeter of RECT .

Possible Answers:

2x+ 18

4x-36

2x- 18

4x+ 36

Correct answer:

Explanation:

The area of the rectangle can be factored as the difference of squares:

$x^{2}-81 = x^{2} - 9^{2} = (x+9)(x-9)$

The area of a rectangle is the product of its two dimensions, one of which is x-9 ; the other dimension can be determined by dividing:

$\frac{x^{2}-81}{x-9} = \frac{ (x+9)(x-9)}{x-9} = x+9$

The perimeter is twice the sum of the two dimensions:

P = 2[(x+9)+(x-9)] = 2(x+x+9-9) = 2(2x)= 4x

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Example Question #291 : Plane Geometry

Parallelogram

Note: Figure NOT drawn to scale

The above figure shows Rhombus ABCD .

Which is the greater quantity?

(a)

(b)

Possible Answers:

It is impossible to determine which is greater from the information given

(a) is the greater quantity

(b) is the greater quantity

(a) and (b) are equal

Correct answer:

(a) and (b) are equal

Explanation:

The opposite sides of a parallelogram - a rhombus included - are congruent, so

AB = CD .

Also, Quadrilateral AXCY form a rectangle; since $\overline{AX} || \overline{CY}$ and $\overline{AY} \perp \overline{CY}$ , it follows that $\overline{AY} \perp \overline{AX}$ , and, similarly, $\overline{XC} \perp \overline{CY}$ . Therefore, AX = CY , and

AB - AX = CD - CY

BX = DY

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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 #21 : Rectangles

Example Question #22 : Rectangles

Example Question #23 : Rectangles

Example Question #24 : Rectangles

Example Question #25 : Rectangles

Example Question #21 : Rectangles

Example Question #291 : Plane Geometry

All ISEE Upper Level Quantitative Resources

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