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Express the area A of an equilateral triangle as a function of the height of the triangle. Thanks :)

I am not sure where to even start on how to answer this problem.

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As a follow-up to the hints below, draw a height of the triangle, and then use the Pythagorean Theorem (or trig) to express the side of the triangle in terms of the height. – user84413 Aug 24 '14 at 22:08
hint: express the base as another function of the height. – John Joy Aug 25 '14 at 13:34

Hints: (1) What is the area formula for a triangle?

(2) If you know the height of an equilateral triangle, can you find the side length? E.g., if the height were say 6 units, what would the side length be?

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Follow up: Also, draw a picture. – paw88789 Aug 24 '14 at 22:04

$$\text{ Area of an triangle }=\frac{1}{2} \cdot (\text{ base } ) \cdot (\text{ height })$$

The identity of an equilateral triangle is that all sides are equal to $x$.

enter image description here

The height is $AD$. The triangle $ABD$ is a right-angled triangle. Therefore, we can use the Pythagorean Theorem.

$$(AD)^2+(BD)^2=x^2 \Rightarrow (AD)^2+\left ( \frac{x}{2} \right )^2=x^2 \Rightarrow x^2-\frac{x^2}{4}=(AD)^2 \Rightarrow \frac{3}{4}x^2=(AD)^2 \\ \Rightarrow x=\frac{2}{\sqrt{3}}(AD) \Rightarrow x=\frac{2\sqrt{3}}{3}(AD)$$

Since all the sides are equal to $x$, the base $(BC)$ is also equal to $x$.

Therefore, $$\text{ Area of an triangle }=\frac{1}{2} \cdot (\text{ base } ) \cdot (\text{ height })=\frac{1}{2} \cdot x \cdot (AD)=\frac{1}{2} \frac{2\sqrt{3}}{3}(AD) (AD)=\frac{\sqrt{3}}{3}(AD)^2$$

So,$$A(h)=\frac{\sqrt{3}}{3}h^2$$ is the area of an equilateral triangle as a function of the height $h$ of the triangle.

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