Question

$a$ and $b$ are real numbers between $0$ and $1.$ $A\equiv (a,1),B\equiv (1,b)$ and $C\equiv (0,0)$ are the vertices of a $△ABC$. If the $△ABC$ is equilateral, its area is equal to

• A. $\frac{7\sqrt{3}+12}{4}$
• B. $\frac{7\sqrt{3}-12}{4}$
• C. $\frac{\sqrt{3}}{4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{7\sqrt{3}-12}{4}$

$a^2+1=b^2+1={(a-1)}^2+{(b-1)}^2\Rightarrow a=b[\because 0<a,b<1]$
And

$2(a^2-2a+1)=a^2+1\Rightarrow a^2-4a+1=0\Rightarrow a=2-\sqrt{3}$

Hence area of triangle is $=\frac{\sqrt{3}}{4}{(2-\sqrt{3})}^2=\frac{7\sqrt{3}-12}{4}$