How to find perimeter

Help Questions

ISEE Upper Level Quantitative Reasoning › How to find perimeter

Questions 1 - 10

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.

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.

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.

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

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.

$\Delta ABC$ and $\Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ , respectively.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) The perimeter of $\Delta DEF$

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

Explanation

No information is given about the legs of either triangle; therefore, no information about their perimeters can be deduced.

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.

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.

Right_triangle

Note: Figure NOT drawn to scale

Refer to the above triangle. Starting at point A, an insect walks clockwise along the sides of the triangle until he has walked 75% of the length of $\overline{CB}$ . What percent of the perimeter of the triangle has the insect walked?

$73 \frac{1}{3} %$

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

Explanation

By the Pythagorean Theorem, the distance from B to C, which we will call , is equal to

$D = \sqrt{13^{2}- 5^{2}} = \sqrt{169- 25 }= \sqrt{144} = 12$ .

The perimeter of the triangle is

The insect traveled the entirety of the hypotenuse, which is 13 units long, and 75% of the longer leg, which adds 75% of 12, or $0.75 \times 12 = 9$ units. Therefore, the insect has traveled 22 out of the 30 units perimeter, or

$\frac{22}{30} \times 100 = 73 \frac{1}{3} %$ of the perimeter.

Parallelogram

Figure NOT drawn to scale.

The above figure depicts Rhombus with and .

Give the perimeter of Rhombus .

Explanation

All four sides of a rhombus have the same length, so we can find the perimeter of Rhombus by taking the length of one side and multiplying it by four. Since , the perimeter is four times this, or $4 \cdot AB = 4 \cdot 32 = 128$ .

Note that the length of $\overline{AY}$ is actually irrelevant to the problem.

Page 1 of 4

Return to subject