Question

Let $f\left(x\right)$ be a real values function such that the area of an equilateral triangle with two of its vertices at $(0,0)$ and $(x,f(x\left)\right)$ is $\frac{\sqrt{3}}{4}$ square units. Then perimeter of the equilateral triangle is

• A. $1$
• B. $3$
• C. $6$
• D. $3\sqrt{3}$

Answer 1

The correct option is B $3$

Assume the side length of equilateral triangle is $a$,

$\therefore$ Area of equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\to a^2=1\to a=1$

Perimeter will be$=a+a+a=3a=3\times 1=3$

Hence, $\left(B\right)$