Rectangles

Help Questions

ISEE Upper Level Quantitative Reasoning › Rectangles

Questions 1 - 10

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.

Trapezoid

Figure NOT drawn to scale.

In the above figure, $\overline{XY}$ is the midsegment of isosceles Trapezoid . Also, $TX = \frac{1}{2} XY$ .

What is the perimeter of Trapezoid ?

Explanation

The length of the midsegment of a trapezoid is half sum of the lengths of the bases, so

$XY = \frac{1}{2} (12 + 40) = \frac{1}{2} \cdot 52 = 26$ .

Also, by definition, since Trapezoid is isosceles, . The midsegment divides both legs of Trapezoid into congruent segments; combining these facts:

$TX = XP = \frac{1}{2} TP = \frac{1}{2} RA = RY = YA$

$TX = \frac{1}{2} XY = \frac{1}{2} \cdot 26 = 13$ .

, so the perimeter of Trapezoid is

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.

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

One side of a regular hexagon is 20% shorter than one side of a regular pentagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) is greater

(A) and (B) are equal

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

(B) is greater

Explanation

Let be the length of one side of the pentagon. Then its perimeter is .

Each side of the hexagon is 20% less than this length, or

The perimeter is five times this, or $6 \cdot 0.8s = 4.8s$ .

Since and is positive, , so the pentagon has greater perimeter, and (A) is greater.

The sum of the lengths of three sides of a regular pentagon is one foot. Give the perimeter of the pentagon in inches.

$20 ; \textrm{in}$

$25; \textrm{in}$

$30 ; \textrm{in}$

$15; \textrm{in}$

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

Explanation

A regular pentagon has five sides of the same length.

One foot is equal to twelve inches; since the sum of the lengths of three of the congruent sides is twelve inches, each side measures

$12 \div 3 = 4$ inches.

The perimeter is

$4 \times 5 = 20$ inches.

One side of a regular pentagon is 20% longer than one side of a regular hexagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) and (B) are equal

(B) is greater

(A) is greater

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

Explanation

Let be the length of one side of the hexagon. Then its perimeter is .

Each side of the pentagon is 20% greater than this length, or

The perimeter is five times this, or $5 \cdot 1.2s = 6s$ .

The perimeters are the same.

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 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: