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SSAT Upper Level Math : Equilateral Triangles

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Example Questions

Example Question #733 : Ssat Upper Level Quantitative (Math)

The perimeter of an equilateral triangle is $30\sqrt{6}$ . Give its area.

Possible Answers:

$75\sqrt{3}$

$150\sqrt{3}$

$150\sqrt{6}$

$75\sqrt{6}$

$150\sqrt{2}$

Correct answer:

$150\sqrt{3}$

Explanation:

An equilateral triangle with perimeter $30\sqrt{6}$ has three congruent sides of length

$s = 30\sqrt{6}\div 3 = 10\sqrt{6}$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{ \left ( 10\sqrt{6} \right )^{2}\sqrt{3}}{4} = \frac{100 \cdot 6 \cdot \sqrt{3}}{4} = 150\sqrt{3}$

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Example Question #1 : How To Find The Area Of An Equilateral Triangle

Hexagon ABCDEF is regular and has perimeter 72. $\Delta ACE$ is constructed. What is its area?

Possible Answers:

$144\sqrt{3}$

$54\sqrt{3}$

$72 \sqrt{3}$

$36\sqrt{3}$

$108 \sqrt{3}$

Correct answer:

$108 \sqrt{3}$

Explanation:

Since the perimeter of the (six-congruent-sided) regular hexagon is 72, each side has length one sixth this, or 12.

The figure described is given below, with a perpendicular segment drawn from to side $\overline{AC}$ :

Hexagon

Each angle of a regular hexagon measures $120^{\circ}$ . Therefore, $m \angle ABM = 60^{\circ }$ , and $\Delta ABM$ is a 30-60-90 triangle.

By the 30-60-90 Theorem,

$BM = \frac{1}{2}AB = \frac{1}{2} \cdot 12 = 6$

and

$AM = BM \sqrt{3} = 6 \sqrt{3}$

$AC = 2 \cdot AM = 2 \cdot 6 \sqrt{3} =12 \sqrt{3}$ .

$\Delta ACE$ is equilateral, and $12 \sqrt{3}$ is its sidelength, making its area

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{\left (12 \sqrt{3} \right )^{2}\sqrt{3}}{4} = \frac{144 \cdot 3 \cdot \sqrt{3}}{4} =108 \sqrt{3}$

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Example Question #2 : How To Find The Area Of An Equilateral Triangle

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

Possible Answers:

$81 \sqrt{3}$

$108\sqrt{2}$

$81 \sqrt{2}$

$108\sqrt{3}$

108

Correct answer:

$81 \sqrt{3}$

Explanation:

An equilateral triangle with perimeter 54 has three congruent sides of length

$s = 51 \div 3 = 18$

The area of this triangle is

$A = \frac{s^{2}\sqrt{3}}{4} = \frac{18^{2}\sqrt{3}}{4} = \frac{324 \sqrt{3}}{4}= 81 \sqrt{3}$

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Example Question #71 : Properties Of Triangles

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

Possible Answers:

$\frac{3,969\sqrt{3}}{4}$

$441\sqrt{3}$

$\frac{1,323 \sqrt{3}}{4}$

$2,646\sqrt{3}$

$1,323 \sqrt{3}$

Correct answer:

$1,323 \sqrt{3}$

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:

Equilateral

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, $CY = OY \cdot \sqrt{3} = 21\sqrt{3}$ , so the area of $\Delta COY$ is

$A = \frac{1}{2}\cdot OY \cdot CY = \frac{1}{2} \cdot 21\cdot 21\sqrt{3} =\frac{ 441 \sqrt{3}}{2}$ .

Six times this - $6 \cdot \frac{ 441 \sqrt{3}}{2} = 1,323 \sqrt{3}$ - is the area of $\Delta ABC$ .

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Example Question #2 : How To Find The Area Of An Equilateral Triangle

An equilateral triangle has side lengths of . What is the area of this triangle?

Possible Answers:

$4\sqrt3$

$x^2\sqrt3$

$\sqrt3$

$4x^2\sqrt3$

Correct answer:

$x^2\sqrt3$

Explanation:

The area of an equilateral triangle can be found using this formula:

$\frac{\sqrt3}{4}(\text{side length})^2$

Using $\text{side length}=2x$ , we can find the area of the equilateral triangle.

$\text{Area}=\frac{\sqrt3}{4}4x^2=x^2\sqrt3$

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Example Question #1 : How To Find The Height Of A Triangle

The side of an equilateral triangle, in feet, is . What is the height of this triangle?

Possible Answers:

$18\sqrt3$

$9\sqrt3$

Correct answer:

$9\sqrt3$

Explanation:

The triangle in question looks like this:

Triangle

The height of a triangle will always cut one of the sides of an equilateral triangle in half. Now, to find the length of the height is just a matter of using the Pythagorean Theorem.

$9^2+\text{height}^2=18^2$

$\text{height}^2=243$

$\text{height}=9\sqrt3$

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Example Question #1 : How To Find The Height Of A Triangle

The length of a side of an equilateral triangle is feet. In feet, what is the height of this triangle?

Possible Answers:

$7\sqrt3$

$14\sqrt3$

Correct answer:

$7\sqrt3$

Explanation:

The given equilateral should look similar:

Triangle

Because the height of an equilateral triangle always cuts a side length in half, figuring out the height becomes a matter of applying the Pythagorean Theorem.

$\text{height}^2+7^2=14^2$

$\text{height}^2=147$

$\text{height}=7\sqrt3$

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Example Question #3 : How To Find The Height Of A Triangle

The length of a side of an equilateral triangle is centimeters. In centimeters, what is the length of the height of this triangle?

Possible Answers:

$5\sqrt3$

$10\sqrt3$

Correct answer:

$5\sqrt3$

Explanation:

Draw out the equilateral triangle:

Triangle

Since the height of an equilateral triangle will always cut one of the sides in half, find the height using the Pythagorean Theorem.

$\text{height}^2+5^2=10^2$

$\text{height}^2=75$

$\text{height}=5\sqrt3$

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Example Question #74 : Properties Of Triangles

The length of a side of an equilateral triangle is . Find the length of the height of this triangle.

Possible Answers:

$\sqrt3$

$\text{Cannot determine with the given information}$

$x\sqrt3$

Correct answer:

$x\sqrt3$

Explanation:

Draw out and label the triangle.

Triangle

Even though you are given exponents for the lengths of this triangle, use the Pythagorean Theorem to solve it. The height of an equilateral triangle will always cut one of its bases in half.

$\text{height}^2+x^2=(2x)^2$

$\text{height}^2=3x^2$

$\text{height}=x\sqrt3$

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Example Question #74 : Properties Of Triangles

Find the height of an equilateral triangle that has side lengths of 12r .

Possible Answers:

$6\sqrt3$

$r\sqrt3$

$6r\sqrt3$

$12r\sqrt3$

Correct answer:

$6r\sqrt3$

Explanation:

Draw out and label the triangle.

Triangle

Since the height of an equilateral triangle will always cut its base in half, use the Pythagorean Theorem to find the height.

$\text{height}^2+(6r)^2=(12r)^2$

$\text{height}^2=108r^2$

$\text{height}=6r\sqrt3$

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SSAT Upper Level Math : Equilateral Triangles

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #733 : Ssat Upper Level Quantitative (Math)

Example Question #1 : How To Find The Area Of An Equilateral Triangle

Example Question #2 : How To Find The Area Of An Equilateral Triangle

Example Question #71 : Properties Of Triangles

Example Question #2 : How To Find The Area Of An Equilateral Triangle

Example Question #1 : How To Find The Height Of A Triangle

Example Question #1 : How To Find The Height Of A Triangle

Example Question #3 : How To Find The Height Of A Triangle

Example Question #74 : Properties Of Triangles

Example Question #74 : Properties Of Triangles

All SSAT Upper Level Math Resources

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