Question
In an equilateral triangle with side a, prove that area $=\frac{\sqrt{3}}{4}\text{a}^2.$
✓
Answer
Let ABC be the equilateral triangle with each side equal to $a$.
Let $A D$ be the altitude from $A$, meeting $B C$ at $D$.
Therefore, $D$ is the midpoint of $B C$.
Let $A D$ be $h$.
Applying Pythagoras theorem in right-angled triangle $ABD$, we have:
$A B^2=A D^2+B D^2$
$\Rightarrow\text{a}^2=\text{h}^2+\big(\frac{\text{a}}{2}\big)^2$
$\Rightarrow\text{h}^2=\text{a}^2-\frac{\text{a}}{4}=\frac{3}{4}\text{a}^2$
$\Rightarrow\text{h}=\frac{\sqrt{3}}{2}\text{a}$
Therefore,
Area of triangle $\text{ABC}=\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times\text{a}\times\frac{\sqrt{3}}{2}\text{a}$
$=\frac{\sqrt{3}}{4}\text{a}^2$
This completes the proof.
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