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ISEE Upper Level Quantitative : Other Polygons

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Example Questions

Example Question #1 : Other Polygons

Right_triangle

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.

Possible Answers:

$50 \textrm{ ft}$

$16 \frac{2}{3} \textrm{ ft}$

$12 \frac{1}{2} \textrm{ ft}$

$31 \frac{1}{4} \textrm{ ft}$

$37 \frac{1}{2} \textrm{ ft}$

Correct answer:

$50 \textrm{ ft}$

Explanation:

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,

$600 \div 12 = 50$ feet.

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Example Question #1 : Other Polygons

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

Possible Answers:

(A) and (B) are equal

(A) is greater

(B) is greater

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

Correct answer:

(B) is greater

Explanation:

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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Example Question #2 : Other Polygons

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

Possible Answers:

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

(A) and (B) are equal

(A) is greater

(B) is greater

Correct answer:

(A) is greater

Explanation:

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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Example Question #1 : Other Polygons

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

(a) The perimeter of Pentagon PEATN

(b) The perimeter of Pentagon PEXZY

Possible Answers:

(a) and (b) are equal

(b) is the greater quantity

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

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The figure referenced is below:

Thingy

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

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

PE + EX + XZ + ZY + YP

= PE + PE + PE + PE + PE

$= 5 \cdot PE$ ,

the same as that of Pentagon PEATN .

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Example Question #4 : Other Polygons

Right_triangle

A regular decagon has the same perimeter as the above right triangle. Give the length of one side.

Possible Answers:

$14 \textrm{ in}$

$6 \frac{1}{5 }\textrm{ in}$

$11 \frac{1}{5} \textrm{ in}$

$7 \frac{3}{4} \textrm{ in}$

$9 \frac{1}{3} \textrm{ in}$

Correct answer:

$11 \frac{1}{5} \textrm{ in}$

Explanation:

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, making its perimeter

14 + 48 + 50 =112 inches.

A regular decagon has ten sides of equal length, so each side measures

$112 \div 10 = 11 \frac{1}{5}$ inches.

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Example Question #3 : Other Polygons

A regular octagon has perimeter one meter. Which is the greater quantity?

(A) The length of one side

(B) 125 millimeters

Possible Answers:

(B) is greater

(A) is greater

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

(A) and (B) are equal

Correct answer:

(A) and (B) are equal

Explanation:

A regular octagon has eight sides of equal length. The perimeter of this octagon is one meter, which is equal to 1,000 millimeters; each side, therefore, has length

$1,000 \div 8 = 125$ millimeters

making the quantities equal.

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Example Question #4 : Other Polygons

A regular pentagon has sidelength 72; the perimeter of a regular hexagon is 80% of that of the pentagon. Which is the greater quantity?

(A) The length of one side of the hexagon

(B) 50

Possible Answers:

(A) is greater

(B) is greater

(A) and (B) are equal

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

Correct answer:

(B) is greater

Explanation:

A regular pentagon has five sides of equal length; since one side is 72 units long, its perimeter is

$72 \times 5 = 360$ .

80% of this is

$0.8 \times 360 = 288$ ,

so this is the length of the hexagon, and, since all six sides are of equal length, one side measures

$288 \div 6 = 48$

(B) is greater.

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Example Question #102 : Geometry

A regular octagon has twice the perimeter of a regular pentagon. What is the ratio of the sidelength of the octagon to that of the pentagon?

Possible Answers:

$5\textrm{ to }8$

$5\textrm{ to }2$

$8\textrm{ to }5$

$5\textrm{ to }4$

$4\textrm{ to } 5$

Correct answer:

$5\textrm{ to }4$

Explanation:

The solution is independent of the actual lengths, so we assume the pentagon has sidelength 1. Its perimeter is therefore 5. Subsequently, the octagon's perimeter is twice this, or 10, and its sidelength is one-eighth of this, or

$\frac{10}{8} = \frac{5}{4}$ .

The ratio of the sidelength of the octagon to that of the pentagon is

$\frac{5}{4} : 1$ or 5 to 4.

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Example Question #5 : Other Polygons

A regular octagon has perimeter one mile.Which is the greater quantity?

(a) The length of one side

(b) 880 feet

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

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

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

A regular octagon has eight sides of equal length. The perimeter, which is the sum of the lengths of these sides, is one mile, which is equal to 5,280 feet. Therefore, the length of one side is

$5, 280 \textrm{ ft} \div 8 = 660 \textrm{ ft}$ . This makes the length of a side less than 880 feet.

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Example Question #1 : How To Find Area

Rectangle 1

The above rectangle, which has perimeter 360, is divided into squares of equal size. Give the area of the shaded portion.

Possible Answers:

2,800

700

5, 600

1,400

Correct answer:

2,800

Explanation:

The sides of the rectangle, in total, are divided into 18 segments of equal measure, as indicated below:

Rectangle 2

The rectangle has a total perimeter of 360, so each segment - one side of a square - measures $36 0 \div 18 = 20$ . Each square has area $20 ^{2} = 400$ , so the shaded portion of the rectangle, which comprises seven squares, has area

$400 \times 7 = 2,800$ .

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ISEE Upper Level Quantitative : Other Polygons

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #1 : Other Polygons

Example Question #1 : Other Polygons

Example Question #2 : Other Polygons

Example Question #1 : Other Polygons

Example Question #4 : Other Polygons

Example Question #3 : Other Polygons

Example Question #4 : Other Polygons

Example Question #102 : Geometry

Example Question #5 : Other Polygons

Example Question #1 : How To Find Area

All ISEE Upper Level Quantitative Resources

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