Squares - ISEE Upper Level Quantitative Reasoning

Card 0 of 384

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

The length of one side of a regular octagon is 60% of that of one side of a regular pentagon. What percent of the perimeter of the pentagon is the perimeter of the octagon?

Answer

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

The length of one side of the regular octagon is 60% of , or , so its perimeter is $8 \cdot 0.6s = 4.8 s$ .The answer is therefore the percent is of , which is

$\frac{4.8s}{5s} \times 100 = 0.96 \times 100 = 96 %$

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) The perimeter of a square with sidelength 1 meter

(b) The perimeter of a regular pentagon with sidelength 75 centimeters

Answer

(a) One meter is equal to 100 centimeters; a square with this sidelength has perimeter $100 \times 4 = 400$ centimeters.

(b) A regular pentagon has five congruent sides; since the sidelength is 75 centimeters, the perimeter is $75 \times 5 = 375$ centimeters.

This makes (a) greater.

Compare your answer with the correct one above

Question

Square 1 is inscribed inside a circle. The circle is inscribed inside Square 2.

Which is the greater quantity?

(a) Twice the perimeter of Square 1

(b) The perimeter of Square 2

Answer

Let be the sidelength of Square 1. Then the length of a diagonal of this square - which is $\sqrt{2}$ times this sidelength, or $s \sqrt{2}$ by the $45^{\circ }-45^{\circ }-90^{\circ }$ Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2.

Since the perimeter of a square is four times its sidelength, Square 1 has perimeter ; Square 2 has perimeter $4 s \sqrt{2}$ , which is $\sqrt{2}$ times the perimeter of Square 1. $\sqrt{2} < 2$ , making the perimeter of Square 2 less than twice than the perimeter of Square 1.

Compare your answer with the correct one above

Question

Five squares have sidelengths one foot, two feet, three feet, four feet, and five feet.

Which is the greater quantity?

(A) The mean of their perimeters

(B) The median of their perimeters

Answer

The perimeters of the squares are

$4 \times 1 = 4$ feet

$4 \times 2 = 8$ feet

$4 \times 3 = 12$ feet

$4 \times 4 = 16$ feet

$4 \times 5 = 20$ feet

The mean of the perimeters is their sum divided by five;

$(4+8+12+16+20) \div 5 = 60 \div 5 = 12$ feet.

The median of the perimeters is the value in the middle when they are arranged in ascending order; this can be seen to also be 12 feet.

The quantities are equal.

Compare your answer with the correct one above

Question

Four squares have sidelengths one meter, one meter, 120 centimeters, and 140 centimeters. Which is the greater quantity?

(A) The mean of their perimeters

(B) The median of their perimeters

Answer

First find the perimeters of the squares:

$4 \times 100 = 400$ centimeters (one meter being 100 centimeters)

$4 \times 100 = 400$ centimeters

$4 \times 120 = 480$ centimeters

$4 \times 140 = 560$ centimeters

The mean of the perimeters is their sum divided by four:

$(400+400+480+560) \div 4 = 1,840 \div 4 = 460$ feet.

The median of the perimeters is the mean of the two values in the middle, assuming the values are in numerical order:

$(400 + 480) \div 2 = 880 \div 2 = 440$

The mean, (A), is greater.

Compare your answer with the correct one above

Question

The perimeters of six squares form an arithmetic sequence. The smallest square has area 9; the second smallest square has area 25. Give the perimeter of the largest of the six squares.

Answer

The two smallest squares have areas 9 and 25, so their sidelengths are the square roots of these, or, respectively, 3 and 5. Their perimeters are the sidelengths multiplied by four, or, respectively, 12 and 20. Therefore, in the arithmetic sequence,

$P_{1} = 12$

$P_{2} = 20$

and the common difference is .

The perimeter of the th smallest square is

$P_{n}=P_{1}+ (n-1)d$

Setting $P_{1} = 12, n = 6, d=8$ , the perimeter of the largest (or sixth-smallest) square is

$P_{6}=P_{1}+ (6-1)8 = 12+ 5 \cdot 8 = 12 + 40 = 52$ .

Compare your answer with the correct one above

Question

The area of a square is $36t ^{2} + 84t + 49$ .

Give the perimeter of the square.

Answer

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

$36t ^{2} + 84t + 49$

$=( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}$

$= (6t+7)^{2}$

Therefore, the square root of the area is

$\sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|$ ,

which is the length of one side.

The perimeter of the square is four times this length, or

.

Compare your answer with the correct one above

Question

The perimeters of six squares form an arithmetic sequence. The second-smallest square has sides that are two inches longer than those of the smallest square.

Which, if either, is the greater quantity?

(a) The perimeter of the third-smallest square

(b) The length of one side of the largest square

Answer

Let the length of one side of the first square be $s_{1}$ . Then the length of one side of the second-smallest square is $s_{2}= s_{1}+ 2$ , and the perimeters of the squares are

$P_{1}= 4s_{1}$

and

$P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8$

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the th-smallest square is

$P_{n} = P_{1} + (n-1)d$

Since , this becomes

$P_{n} = P_{1} + 8 (n-1)$

The perimeter of the third-smallest square is

$P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16$

The perimeter of the largest, or sixth-smallest, square is

$P_{6} = P_{1} + 8 (6-1) = P_{1}+ 8 \cdot 5= P_{1} + 40$

The length of one side of this square is one fourth of this, or

$s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10$

Therefore, we are comparing $P_{1} + 16$ and $\frac{1}{4} P_{1} + 10$ .

Since a perimeter must be positive,

$\frac{1}{4} P_{1} < P_{1}$ ;

also, .

Therefore, regardless of the value of $P_{1}$ ,

$P_{1} + 16 > \frac{1}{4} P_{1} + 10$ ,

and

$P_{3}> s_{6}$ ,

making (a) the greater quantity.

Compare your answer with the correct one above

Question

The sidelength of a square is $2^{x}$ . Give its perimeter in terms of .

Answer

The perimeter of a square is four times the length of a side, which here is $2 ^{x }$ :

$P = 4 \cdot 2^{x} = 2 ^{2} \cdot 2^{x} =2 ^{x+2}$

Compare your answer with the correct one above

Question

A diagonal of a square has length $2^{x}$ . Give its perimeter.

Answer

The length of a side of a square can be determined by dividing the length of a diagonal by $\sqrt{2}$ - that is, $2 ^{\frac{1}{2}}$ . A diagonal has length $2^{x}$ , so the sidelength is

$2^{x} \div 2 ^{\frac{1}{2}}= 2 ^{x-\frac{1}{2}}$

Multiply this by four to get the perimeter:

$4 \cdot 2 ^{x-\frac{1}{2}} = 2^{2} \cdot 2 ^{x-\frac{1}{2}} = 2 ^{x-\frac{1}{2}+2} = 2 ^{x+\frac{3}{2}}$

Compare your answer with the correct one above

Question

The perimeter of a square is one yard. Which is the greater quantity?

(a) The area of the square

(b) $\frac{1}{2}$ square foot

Answer

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\frac{3}{4}$ feet. The area of the square is $\frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\frac{9}{16} > \frac{1}{2}$ , making (a) greater.

Compare your answer with the correct one above

Question

Square 1 is inscribed inside a circle. The circle is inscribed inside Square 2.

Which is the greater quantity?

(a) Twice the area of Square 1

(b) The area of Square 2

Answer

Let be the sidelength of Square 1. Then the length of a diagonal of this square - which is $\sqrt{2}$ times this sidelength, or $s \sqrt{2}$ , by the $45^{\circ }-45^{\circ }-90^{\circ }$ Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2.

Therefore, Square 1 has area $A = s^{2}$ , and Square 2 has area $A = \left ( s \sqrt{2 } \right )^{2} = 2s^{2}$ , twice that of Square 1.

Compare your answer with the correct one above

Tap the card to reveal the answer