GRE Math : How to find the area of an equilateral triangle

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Example Question #1 : Equilateral Triangles

What is the area of an equilateral triangle with a base of $\displaystyle 12$?

Possible Answers:

$36\sqrt{3}$ $\displaystyle 36\sqrt{3}$

$72\sqrt{3}$ $\displaystyle 72\sqrt{3}$

$12\sqrt{3}$ $\displaystyle 12\sqrt{3}$

$24\sqrt{3}$ $\displaystyle 24\sqrt{3}$

$144\sqrt{3}$ $\displaystyle 144\sqrt{3}$

Correct answer:

$36\sqrt{3}$ $\displaystyle 36\sqrt{3}$

Explanation:

An equilateral triangle can be considered to be 2 identical 30-60-90 triangles, giving the triangle a height of $6\sqrt{3}$ $\displaystyle 6\sqrt{3}$. From there, use the formula for the area of a triangle:

$\dpi{100} \frac{1}{2}\cdot b\cdot h=\frac{1}{2}\cdot 12\cdot 6\sqrt{3}=36\sqrt{3}$ $\displaystyle \dpi{100} \frac{1}{2}\cdot b\cdot h=\frac{1}{2}\cdot 12\cdot 6\sqrt{3}=36\sqrt{3}$

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

Gre circle

An equilateral triangle is inscribed into a circle of radius 10. What is the area of the triangle?

Possible Answers:

$100\pi$ $\displaystyle 100\pi$

$\frac{100}{3}\pi$ $\displaystyle \frac{100}{3}\pi$

The answer cannot be determined from the information given.

$150\sqrt{3}$ $\displaystyle 150\sqrt{3}$

$75\sqrt{3}$ $\displaystyle 75\sqrt{3}$

Correct answer:

$75\sqrt{3}$ $\displaystyle 75\sqrt{3}$

Explanation:

To solve this equation, first note that a line drawn from the origin to a vertex of the equilateral triangle will bisect the angle of the vertex. Furthermore, the length of this line is equal to the radius:

Gre circle solution

That this creates in turn is a 30-60-90 right triangle. Recall that the ratio of the sides of a 30-60-90 triangle is given as:

$(1:\sqrt{3}:2)$ $\displaystyle (1:\sqrt{3}:2)$

Therefore, the length of the $60^{\circ}$ $\displaystyle 60^{\circ}$ side can be found to be

$\frac{10}{2}\sqrt{3}=5\sqrt{3}$ $\displaystyle \frac{10}{2}\sqrt{3}=5\sqrt{3}$

This is also one half of the base of the triangle, so the base of the triangle can be found to be:

$b=2\cdot 5\sqrt{3}=10\sqrt{3}$ $\displaystyle b=2\cdot 5\sqrt{3}=10\sqrt{3}$

Furthermore, the length of the $30^{\circ}$ $\displaystyle 30^{\circ}$ side is:

$\frac{10}{2}=5$ $\displaystyle \frac{10}{2}=5$

The vertical section rising from the origin is the length of the radius, which when combined with the shorter section above gives the height of the triangle:

$\displaystyle h=5+10=15$

The area of a triangle is given by one half the base times the height, so we can find the answer as follows:

$Area=\frac{1}{2}bh=\frac{1}{2}(10\sqrt{3})(15)=75\sqrt{3}$ $\displaystyle Area=\frac{1}{2}bh=\frac{1}{2}(10\sqrt{3})(15)=75\sqrt{3}$

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

Find the area of an equilateral triangle when one of its sides equals 4.

Possible Answers:

4√3

2√3

Correct answer:

4√3

Explanation:

All sides of an equilateral triangle are equal, so all sides of this triangle equal 4.

Area = 1/2 base * height, so we need to calculate the height: this is easy for an equilateral triangle, since you can bisect any such triangle into two identical 30:60:90 triangles.

The ratio of lengths of a 30:60:90 triangle is 1:√3:2. The side of the equilateral triangle is 4, and we divided the base in half when we bisected the triangle, so that give us a length of 2, so our triangle must have sides of 2, 4, and 2√3; thus we have our height.

One of our 30:60:90 triangles will have a base of 2 and a height of 2√3. Half the base is 1, so 1 * 2√3 = 2√3.

We have two of these triangles, since we divided the original triangle, so the total area is 2 * 2√3 = 4√3.

You can also solve for the area of any equilateral triangle by applying the formula (s²√3)/4, where s = the length of any side.

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GRE Math : How to find the area of an equilateral triangle

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Example Question #1 : Equilateral Triangles

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

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

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