Question
In an equilateral triangle with side a, prove that area $=\frac{\sqrt{3}}{4}\text{a}^2.$
✓
Answer
Let $\triangle\text{ABC}$ be an equilateral triangle with side a.
Then, AB = AC = BC = a.
Draw $\text{AD}\perp\text{BC}.$
In $\triangle\text{ADB}$ and $\triangle\text{A},$ we have
$\text{AB}=\text{AC}(\text{given}),\angle\text{B}=\angle\text{C}=60^\circ$
and $\angle\text{ADB}=\angle\text{ADC}=90^\circ$
$\therefore\triangle\text{ADB}\cong\triangle\text{ADC}$
$\therefore\text{BD}=\text{DC}=\frac{\text{a}}{2}$
From right $\triangle\text{ADB},$ we have
$\text{AB}^2=\text{AD}^2+\text{BD}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AD}=\sqrt{\text{AB}^2-\text{AB}^2}$
$\Rightarrow\text{AD}=\sqrt{\text{a}^2-\Big(\frac{\text{a}}{2}\Big)^2}$
$\Rightarrow\text{AD}=\sqrt{\text{a}^2-\frac{\text{a}}{4}^2}$
$\Rightarrow\text{AD}=\sqrt{\frac{3\text{a}}{4}^2}$
$\Rightarrow\text{AD}=\frac{\sqrt{3}\text{a}}{2}$
So, the altitude is $\frac{\sqrt{3\text{a}}}{2},$
Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{BC}\times\text{AD}$
$=\frac{1}{2}\times\text{a}\times\frac{\sqrt{3\text{a}}}{2}$
$=\frac{\sqrt{3}\text{a}^2}{4}$
Hence proved.
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