ISEE Upper Level Quantitative Reasoning - How to find the height of an eqilateral triangle

The length of one side of an equilateral triangle is 4 centimeters. Give the height (altitude) of the triangle.

$2\sqrt{3}\ cm$

$4\sqrt{3}\ cm$

$3\sqrt{3}\ cm$

$4\ cm$

$3\ cm$

Explanation

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

where and are the lengths of two sides and is the corresponding angle.

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

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 4\times 4\times sin60^{\circ}$

$sin60^{\circ}=\frac{\sqrt{3}}{2}\Rightarrow Area=\frac{1}{2}\times 4\times 4\times \frac{\sqrt{3}}{2}=4\sqrt{3}\ cm^2$

Now plug this area into the alternate formula for the area of the triangle and solve for the height:

$Area =\frac{bh}2{}\Rightarrow 4\sqrt{3}=\frac{4\times h}{2}\Rightarrow h=2\sqrt{3}\ cm$

The height of the equilateral triangle below is 5 centimeters. Give the perimeter of the triangle.

$\frac{8\sqrt{3}}{3}\ cm$

$\frac{8}{3}\ cm$

$8\ cm$

$4\ cm$

$\frac{4\sqrt{3}}{3}\ cm$

Explanation

In an equilateral triangle all of the sides have the same length, so all we need to do is find the length of one side.

We know that, in an equilateral triangle, all three angles measure $60^{\circ}$ .

$sin\angle A=\frac{h}{AB}=\frac{h}{c}$

$sin\angle A=\frac{h}{c}\Rightarrow sin60^{\circ}=\frac{4}{c}\Rightarrow \frac{\sqrt{3}}{2}=\frac{4}{c}\Rightarrow c=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3} \ cm$

How to find the height of an eqilateral triangle

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