Equilateral Triangles - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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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

$\Delta ABC$ 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 $\Delta DEF$

Answer

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

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Question

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?

Answer

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.

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Question

Refer to the above diagram, in which $\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Which is the greater quantity?

(a) Four times the perimeter of $\bigtriangleup AXB$

(b) Three times the perimeter of $\bigtriangleup CXB$

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of $\bigtriangleup CXB$ to that of $\bigtriangleup AXB$ (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the perimeters of two similar triangles is the same as their similarity ratio; therefore, if is the perimeter of $\bigtriangleup AXB$ and is the perimeter of $\bigtriangleup CXB$ , it follows that

$\frac{Y}{X} = \frac{4}{3}$

$\frac{Y}{X} \cdot X = \frac{4}{3} \cdot X$

$Y = \frac{4}{3} X$

Multiply both sides by 3:

$3 \cdot Y =3 \cdot \frac{4}{3} X$

Three times the perimeter of $\bigtriangleup CXB$ is therefore equal to four times that of $\bigtriangleup AXB$ .

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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

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

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

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

$\overline{PE}$ 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

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.

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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

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

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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.

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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.

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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.

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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.

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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.

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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.

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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

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

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