Question

What is the area of an equilateral triangle whose side is 'a'?

• A. $\frac{3a^2}{2}$
• B. $\frac{\sqrt{3}{a}^2}{2}$
• C. $\frac{\sqrt{3}{a}^2}{4}$
• D. $3a^2$

Answer 1

The correct option is C $\frac{\sqrt{3}{a}^2}{4}$

In the figure $AP\perp BC$

$AB=BC=CA=a$

In $△APBand△APC$

$\angle APB=\angle APC={90}^{\circ }$

$\angle ABP=\angle ACP-{60}^{\circ }$ (ABC is equilateral triangle)

By AA similarity

$△APB\sim △APC$

$\therefore \frac{AP}{AP}=\frac{PB}{PC}=\frac{AB}{AC}$

$\Rightarrow \frac{PB}{PC}=1\Rightarrow PB=PC$

$\Rightarrow PB+PC=a$

$\Rightarrow 2PB=a$

$\Rightarrow PB=\frac{a}{2}$

In $△APB\angle APB={90}^{\circ }$

$A{B}^2=A{P}^2+P{B}^2$
$\Rightarrow a^2=A{P}^2+(\frac{a}{2}{)}^2$
$\Rightarrow A{P}^2=a^2-\frac{a^2}{4}$

$\Rightarrow A{P}^2=\frac{3a^2}{4}$

$\Rightarrow AP=\frac{\sqrt{3}a}{2}$

Hence height of the triangle is $\frac{\sqrt{3}a}{2}$

Area of the triangle$=\frac{1}{2}\times base\times height$

$\Rightarrow \frac{1}{2}\times a\times \frac{\sqrt{3}a}{2}$

$\Rightarrow Area=\frac{\sqrt{3}{a}^2}{4}$