Question

The equation to the base of an equilateral triangle is $(\sqrt{3}+1)x+(\sqrt{3}-1)y+2\sqrt{3}=0$ and the opposite vertex is $A(1,1)$, then the area of the triangle (in sq. units) is

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

Answer 1

Answer: (C)

$2\sqrt{3}$

$d=\left|\frac{4\sqrt{3}}{\sqrt{4+2\sqrt{3}+4-2\sqrt{3}}}\right|=2\sqrt{\frac{3}{2}}$

From figure,

$a\cos {30}^{\circ }=d$

$a=\frac{\sqrt{\frac{3}{2}}\times 2\times 2}{\sqrt{3}}=2\sqrt{2}$

$\therefore$ Area of equilateral $\Delta =\frac{1}{2}a\times d$

$=\frac{1}{2}\times 2\sqrt{2}\times 2\times \frac{\sqrt{3}}{\sqrt{2}}$

$=2\sqrt{3}$