How to find perimeter - ISEE Upper Level Quantitative Reasoning

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Question

Note: Figure NOT drawn to scale.

Which of the following is the greater quantity?

(A) The perimeter of the triangle

(B) 90

Answer

The longest side of the triangle appears opposite the angle of greatest measure. The side of length 30 appears opposite an angle of measure $180 ^{\circ } - (65 ^{\circ } + 65 ^{\circ } ) = 50 ^{\circ }$ . Therefore, the sides opposite the $65^{\circ }$ angles must have lengths greater than 30.

If we let this common length be , then

The perimeter of the triangle is therefore greater than 90.

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Question

A square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The mean of their perimeters

(B) The median of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The square, the pentagon, the hexagon, and the octagon have 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 4, 5, 6, and 8. The mean of these four perimeters is

$\left ( 4+5+6+8\right ) \div 4 = 23 \div 4 = 5.75$ units.

The median is the mean of the middle two perimeters, which are 5 and 6:

$\left ( 5 + 6\right ) \div 2 = 5.5$

The mean, (A), is greater.

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Question

An equilateral triangle, a square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The median of their perimeters

(B) The midrange of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The equilateral triangle, the square, the pentagon, the hexagon, and the octagon have 3, 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 3, 4, 5, 6, and 8.

The median of these perimeters is the middle perimeter, 5. The midrange of these perimeters is the mean of the greatest and the least perimeters:

$\left ( 3+8\right ) \div 2 = 5.5$

The midrange, (B), is greater.

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Question

The length of a side of a regular octagon is one and a half times the hypotenuse of the above right triangle. Give the perimeter of the octagon in feet.

Answer

By the Pythagorean Theorem, the hypotenuse of the right triangle is

$\sqrt{14^{2}+48^{2}} = \sqrt{196+2,304} = \sqrt{2,500} = 50$ inches.

The sidelength of the octagon is therefore

$50 \times 1\frac{1}{2} = 75$ inches,

and the perimeter of the regular octagon, which has eight sides of equal length, is

$75 \times 8 = 600$ inches,

or

$600 \div 12 = 50$ feet.

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Question

$\overline{PE}$ is a side of regular Pentagon as well as Square , which is completely outside Pentagon . $\overline{XY}$ is a side of equilateral $\bigtriangleup XYZ$ , where is a point outside Square . Which is the greater quantity?

(a) The perimeter of Pentagon

(b) The perimeter of Pentagon

Answer

The figure referenced is below:

Pentagon is regular, so all of its sides have the same length; we will examine $\overline{PE }$ in particular. The perimeter of Pentagon is the sum of the lengths of its sides, which is $5 \cdot PE$ .

Since $\overline{PE }$ is also a side of Square , it follows that ; since $\overline{XY }$ is also a side of equilateral $\bigtriangleup XYZ$ , . The perimeter of Pentagon is equal to

$= 5 \cdot PE$ ,

the same as that of Pentagon .

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Question

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB},\overline{BC},\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) Twice the perimeter of $\Delta DEF$

Answer

If segments are constructed in which the endpoints form the midpoints of the sides of a triangle, then each of the sides of the smaller triangle is half as long as the side of the larger triangle that it does not touch. Therefore:

$AB = 2 \cdot EF$

$BC = 2 \cdot DF$

$AC = 2 \cdot DE$

The perimeter of $\Delta ABC$ is:

$P = 2 \cdot EF+ 2 \cdot DF + 2 \cdot DE$

,

which is twice the perimeter of $\Delta DEF$ .

Note that the fact that the triangle is equilateral is irrelevant.

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Question

Column A Column B

The perimeter The perimeter

of a square with of an equilateral

sides of 4 cm. triangle with a side

of 9 cm.

Answer

Perimeter involves adding up all of the sides of the shape. Therefore, the square's perimeter is or 16. An equialteral shape means that all of the sides are equal. Therefore, the perimeter of the triangle is or 27. Therefore, Column B is greater.

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Question

Which is the greater quantity?

(a) The perimeter of a regular pentagon with sidelength 1 foot

(b) The perimeter of a regular hexagon with sidelength 10 inches

Answer

The sides of a regular polygon are congruent, so in each case, multiply the sidelength by the number of sides to get the perimeter.

(a) Since one foot equals twelve inches, $5 \times 12 = 60$ inches.

(b) Multiply: $6 \times 10 = 60$ inches

The two polygons have the same perimeter.

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Question

Which quantity is greater?

(a) The perimeter of the above trapezoid

(b) The perimeter of a rectangle with length and width and , respectively.

Answer

The perimeter of a rectangle is twice the sum of its length and its width:

Since the height of the trapezoid in the figure is , both of its legs must have length greater than or equal to . But for a leg to be of length, it must be perpendicular to the bases. Since perpendicularity of both legs would make the trapezoid a rectangle - which it cannot be - it follows that both legs cannot be of length . Therefore, the perimeter of the trapezoid is:

The perimeter of the trapezoid must be greater than that of the rectangle.

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Question

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 ?

Answer

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

.

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Question

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid .

Which is the greater quantity?

(a) Twice the perimeter of Trapezoid

(b) The perimeter of Trapezoid

Answer

The midsegment of a trapezoid bisects both of its legs, so

and .

For reasons that will be apparent later, we will set

Also, the length of the midsegment is half sum of the lengths of the bases:

$XY = \frac{1}{2} (12 + 36) = \frac{1}{2} \cdot 48= 24$ .

The perimeter of Trapezoid is

Twice this is

The perimeter of Trapezoid is

and , so , making (a) the greater quantity.

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Question

A rectangle has a width of 2_x_. If the length is five more than 150% of the width, what is the perimeter of the rectangle?

Answer

Given that w = 2_x_ and l = 1.5_w_ + 5, a substitution will show that l = 1.5(2_x_) + 5 = 3_x_ + 5.

P = 2_w_ + 2_l_ = 2(2_x_) + 2(3_x_ + 5) = 4_x_ + 6_x_ + 10 = 10_x_ + 10 = 10(x + 1)

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Question

A rectangle has length 72 inches and width 36 inches. What is its perimeter?

Answer

The perimeter of a rectangle is equal to twice the sum of its length and its width, which here would be, in inches,

$2 (72 + 36) = 2 \times 108 = 216 \textrm{ in}$ .

$216 \textrm{ in} = 216\div 12 = 18 \textrm{ ft} = 18 \div 3 = 6 \textrm{ yd}$

$216 \textrm{ in} = 18 \textrm{ ft} = \frac{18}{5,280} = \frac{18\div 6}{5,280\div 6}= \frac{3}{880}\textrm{ mi}$

Therefore, the correct choice is that all four measurements are equal to the perimeter.

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Question

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

Answer

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.

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Question

Which is the greater quantity?

(a) The perimeter of the rectangle on the coordinate plane with vertices

(b) The perimeter of the rectangle on the coordinate plane with vertices

Answer

(a) The first rectangle has width and height ; its perimeter is

.

(b) The second rectangle has width and height ; its perimeter is

.

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

, or equivalently,

.

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

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

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

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Question

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

Answer

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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Question

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

Answer

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.

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Question

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

What is its perimeter?

Answer

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.

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Question

Give the perimeter of the above rectangle in terms of .

Answer

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:

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Question

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

Answer

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 ; 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:

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