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ISEE Upper Level Math : Geometry

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Example Questions

Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

Which of the following could be the lengths of the three sides of a scalene triangle?

Possible Answers:

$6 \textrm{ ft, }84 \textrm{ in, }5\textrm{ ft}$

$7 \textrm{ ft, }84 \textrm{ in, }72\textrm{ in}$

$5 \textrm{ ft, }60 \textrm{ in, }4\textrm{ ft}$

$5 \textrm{ ft, }66 \textrm{ in, }5 \frac{1}{2}\textrm{ ft}$

$3 \textrm{ ft, }36 \textrm{ in, }1\textrm{ yd}$

Correct answer:

$6 \textrm{ ft, }84 \textrm{ in, }5\textrm{ ft}$

Explanation:

A scalene triangle, by definition, has three sides of different lengths. We can identify the scalene triangle by converting measures to the same unit. We will convert to inches in this case.

$5 \textrm{ ft, }60 \textrm{ in, }4\textrm{ ft}$

5 feet = $5 \times 12 = 60$ inches. This triangle can be eliminated, since two sides have length 60 inches.

$7 \textrm{ ft, }84 \textrm{ in, }72\textrm{ in}$

7 feet = $7 \times 12 = 84$ inches. This triangle can be eliminated, since two sides have length 84 inches.

$3 \textrm{ ft, }36 \textrm{ in, }1\textrm{ yd}$

These three measures are equal. This is an equilateral triangle, and it can be eliminated.

$5 \textrm{ ft, }66 \textrm{ in, }5 \frac{1}{2}\textrm{ ft}$

$5 \frac{1}{2}$ feet = $5 \frac{1}{2} \times 12 = 66$ inches. This triangle can be eliminated, since two sides have length 66 inches.

$6 \textrm{ ft, }84 \textrm{ in, }5\textrm{ ft}$

5 feet = 60 inches.

6 feet = $6\times 12 = 72$ inches.

The sides of this triangle measure 60, 72, and 84 inches, so the triangle is scalene.

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Example Question #1 : Equilateral Triangles

For an equilateral triangle, Side A measures 2x+2 and Side B measures 4x-4 . What is the length of Side A?

Possible Answers:

Correct answer:

Explanation:

First you need to recognize that for an equilateral triangle, all 3 sides have equal lengths.

This means you can set the two values for Side A and Side B equal to one another, since they measure the same length, to solve for .

4x-4 = 2x+2

2x-4 = 2

2x = 6

x = 3

You now know that x = 3 , but this is not your answer. The question asked for the length of Side A, so you need to plug 3 into that equation.

2x+2

=2(3)+2

So the length of Side A (and Side B for that matter) is 8.

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Example Question #101 : Geometry

One angle of an equilateral triangle is 60 degrees. One side of that triangle is 12 centimeters. What are the measures of the two other angles and two other sides?

Possible Answers:

55 degrees, 65 degrees, 12 centimeters, 13 centimeters

50 degrees, 70 degrees, 5 centimeters, 12 centimeters

45 degrees, 75 degrees, 12 centimeters, 13 centimeters

90 degrees, 30 degrees, 12 centimeters, 12 centimeters

60 degrees, 60 degrees, 12 centimeters, 12 centimeters

Correct answer:

60 degrees, 60 degrees, 12 centimeters, 12 centimeters

Explanation:

An equilateral triangle is one in which all three sides are congruent. It also has the property that all three interior angles are equal. In other words, all three angles of an equilateral triangle are always 60°. Since all sides are congruent, the other two sides both measure 12 centimeters.

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Example Question #3 : Equilateral Triangles

If an equilateral triangle has a perimeter of 18in, what is the length of one side?

Possible Answers:

$3\text{in}$

$8\text{in}$

$4\text{in}$

$7\text{in}$

$6\text{in}$

Correct answer:

$6\text{in}$

Explanation:

To find the perimeter of an equilateral triangle, we will use the following formula:

P = 3a

where a is the length of any side of the triangle. Because an equilateral triangle has 3 equal sides, we can use any side in the formula. To find the length of one side of the triangle, we will solve for a.

Now, we know the perimeter of the equilateral triangle is 18in. So, we will substitute.

$18\text{in} = 3a$

$\frac{18\text{in}}{3} = \frac{3a}{3}$

$6\text{in} = a$

$a=6\text{in}$

Therefore, the length of one side of the equilateral triangle is 6in.

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Example Question #102 : Plane Geometry

An equilateral triangle has a perimeter of 39in. Find the length of one side.

Possible Answers:

$9\text{in}$

$13\text{in}$

$16\text{in}$

$14\text{in}$

$12\text{in}$

Correct answer:

$13\text{in}$

Explanation:

An equilateral triangle has 3 equal sides. So, to find the length of one side, we will use what we know. We know the perimeter of the equilateral triangle is 39in. So, we will look at the formula for perimeter. We get

P = 3a

where a is the length of one side of the triangle. So, to find the length of one side, we will solve for a. Now, as stated before, we know the perimeter of the triangle is 39in. So, we will substitute and solve for a. We get

$39\text{in} = 3a$

$\frac{39\text{in}}{3} = \frac{3a}{3}$

$13\text{in} = a$

$a = 13\text{in}$

Therefore, the length of one side of the equilateral triangle is 13in.

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Example Question #1 : Equilateral Triangles

The length of one side of an equilateral triangle is 6 inches. Give the area of the triangle.

Possible Answers:

$9\ in^2$

$6\sqrt{3}\ in^2$

$6\ in^2$

$9\sqrt{3}\ in^2$

$9\sqrt{2}\ in^2$

Correct answer:

$9\sqrt{3}\ in^2$

Explanation:

$Area=\frac{1}{2}absinC$ ,

where and are the lengths of two sides of the triangle and is the angle measure.

In an equilateral triangle, all of the sides have the same length, and all three angles are always $60^{\circ}$ .

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 6\times 6\times sin60^{\circ}$

$sin60^{\circ}=\frac{\sqrt{3}}{2}\Rightarrow Area=\frac{1}{2}\times 6\times 6\times \frac{\sqrt{3}}{2}=9\sqrt{3}\ in^2$

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Example Question #3 : Equilateral Triangles

The perimeter of an equilateral triangle is $30\sqrt[4]{3}$ . Give its area.

Possible Answers:

$50\sqrt{3}$

150

$25\sqrt{6}$

$25\sqrt{3}$

Correct answer:

Explanation:

An equilateral triangle with perimeter $30\sqrt[4]{3}$ has three congruent sides of length

$s = 30\sqrt[4]{3} \div 3 = 10\sqrt[4]{3}$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{ \left ( 10\sqrt[4]{3} \right )^{2}\sqrt{3}}{4} = \frac{100 \sqrt[4]{9} \cdot \sqrt{3}}{4}$

$\sqrt[4]{9} = \sqrt[4]{3^{2}} = 3^{2/4} = 3^{1/2} = \sqrt{3}$ , so

$A = \frac{100 \sqrt{3} \cdot \sqrt{3}}{4} = \frac{100 \cdot 3 }{4}= 75$

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Example Question #3 : Equilateral Triangles

The perimeter of an equilateral triangle is $18\sqrt{3}$ . Give its area.

Possible Answers:

The correct answer is not among the other four choices.

$27\sqrt{3}$

$36\sqrt{3}$

Correct answer:

$27\sqrt{3}$

Explanation:

An equilateral triangle with perimeter $18\sqrt{3}$ has three congruent sides of length

$s = 18\sqrt{3} \div 3 = 6\sqrt{3}$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{ \left ( 6\sqrt{3} \right )^{2}\sqrt{3}}{4} = \frac{36 \cdot 3 \cdot \sqrt{3}}{4} = 27\sqrt{3}$

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Example Question #3 : Equilateral Triangles

The perimeter of an equilateral triangle is . Give its area in terms of .

Possible Answers:

$\frac{n^{2} }{12}$

$\frac{n^{2} \sqrt{3}}{18}$

$\frac{n^{2} \sqrt{3}}{36}$

$\frac{n^{2} }{6}$

$\frac{n^{2} \sqrt{3}}{12}$

Correct answer:

$\frac{n^{2} \sqrt{3}}{36}$

Explanation:

An equilateral triangle with perimeter has three congruent sides of length $\frac{n}{3}$ . Substitute this for in the following area formula:

$A =s^{2} \cdot \frac{\sqrt{3}}{4}$

$A = \left ( \frac{n}{3} \right )^{2} \cdot \frac{\sqrt{3}}{4}$

$A = \frac{n^{2} }{9} \cdot \frac{\sqrt{3}}{4}$

$A = \frac{n^{2} \sqrt{3}}{36}$

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Example Question #4 : Equilateral Triangles

An equilateral triangle is inscribed inside a circle of radius 8. Give its area.

Possible Answers:

$48\sqrt{3}$

$64\sqrt{3}$

$48\sqrt{2}$

Correct answer:

$48\sqrt{3}$

Explanation:

The trick is to know that the circumscribed circle, or the circumcircle, has as its center the intersection of the three altitudes of the triangle, and that this center, or circumcenter, divides each altitude into two segments, one twice the length of the other - the longer one being a radius. Because of this, we can construct the following:

Circumcircle

Each of the six smaller triangles is a 30-60-90 triangle, and all six are congruent.

We will find the area of $\Delta COY$ , and multiply it by 6.

By the 30-60-90 Theorem, $CY = OY \cdot \sqrt{3} = 4\sqrt{3}$ , so the area of $\Delta COY$ is

$A = \frac{1}{2}\cdot OY \cdot CY = \frac{1}{2} \cdot 4\cdot 4\sqrt{3} = 8 \sqrt{3}$ .

Six times this - $6 \cdot 8 \sqrt{3} = 48 \sqrt{3}$ - is the area of $\Delta ABC$ .

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ISEE Upper Level Math : Geometry

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

Example Question #1 : Equilateral Triangles

Example Question #101 : Geometry

Example Question #3 : Equilateral Triangles

Example Question #102 : Plane Geometry

Example Question #1 : Equilateral Triangles

Example Question #3 : Equilateral Triangles

Example Question #3 : Equilateral Triangles

Example Question #3 : Equilateral Triangles

Example Question #4 : Equilateral Triangles

All ISEE Upper Level Math Resources

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