Rectangles

Help Questions

ISEE Upper Level Quantitative Reasoning › Rectangles

Questions 1 - 10

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

What is its perimeter?

$272 \textrm{ in}$

$224\textrm{ in}$

$284\textrm{ in}$

$302\textrm{ in}$

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

Explanation

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

The area is the product of the length and width:

$L \cdot 0.70 L = 4,480$

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

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

$L ^{2} = 6,400$

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

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

The perimeter is therefore

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

A rectangle has perimeter 140 inches and area 1,200 square inches. Which is the greater quantity?

(A) The length of a diagonal of the rectangle.

(B) 4 feet

(A) is greater

(B) is greater

(A) and (B) are equal

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

Explanation

Let and be the dimensions of the rectangle. Then

and, subsequently,

Since the product of the length and width is the area, we are looking for two numbers whose sum is 70 and whose product is 1,200; through trial and error, they are found to be 30 and 40. We can assign either to be and the other to be since the result is the same.

The length of a diagonal of the rectangle can be found by applying the Pythagorean Theorem:

$D = \sqrt{L^{2}+ W^{2}}$

$D = \sqrt{30^{2}+ 40^{2}} = \sqrt{900+1,600} = \sqrt{2,500} = 50$

A diagonal is 50 inches long; since 4 feet are equivalent to 48 inches, (A) is the greater quantity.

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.

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

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

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

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

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

Explanation

Since the width of the rectangle is 60% of its length, we can write .

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

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

$2.2L \div 2.2 = 260$

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

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

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

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

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

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

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

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

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

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

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

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

Parallelogram

Note: Figure NOT drawn to scale

The above figure shows Rhombus .

Which is the greater quantity?

(a)

(b)

(a) and (b) are equal

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

(b) is the greater quantity

(a) is the greater quantity

Explanation

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

Also, Quadrilateral 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, , and

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.

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

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

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

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

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

Explanation

Since the width of the rectangle is 60% of its length, we can write .

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

$2.2L \div 2.2 = 260$

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

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

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

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

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

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

It is impossible to determine the area from the information given

Explanation

The length of the rectangle is 2 feet, or 24 inches, greater than the width, so, if is the width in inches, 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 , so we can set up the equation:

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

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

$21\times 45 = 945$ square inches.

Which quantity is greater?

(a) The perimeter of a square with area 10,000 square centimeters

(b) The perimeter of a rectangle with area 8,000 square centimeters

It is impossible to tell from the information given

(a) and (b) are equal

(b) is greater

(a) is greater

Explanation

A square with area 10,000 square centimeters has sidelength $\sqrt{10,000} = 100$ centimeters, and perimeter $4 \times 100 = 400$ centimeters.

Not enough information is given about the rectangle with area 8,000 square centimeters to determine its perimeter. For example, if its dimensions are 100 centimeters by 80 centimeters, its perimeter is centimeters. If the dimensions are 200 centimeters by 40 centimeters, its perimeter is centimeters. Both cases are consistent with the conditions of the problem, yet one makes (a) greater and one makes (b) greater.