Question

In the adjacent figure P is any inferior point of the equilateral triangle ABC side length of $2$ units. If $x_a,x_b$ and $x_c$ represent the distance of P from the sides BC,CA and AB respectively then $x_a+x_b+x_c$ is equal to

• A. $6$
• B. $\sqrt{3}$
• C. $\sqrt{3}/2$
• D. $2\sqrt{3}$

Answer 1

The correct option is D $2\sqrt{3}$

Area of $△ABP=\frac{1}{2}\times AB\times x_c$

Area of $△APC=\frac{1}{2}\times AC\times x_b$

Area of $△BPC==\frac{1}{2}\times BC\times x_a$

$\therefore$Area of $△ABC=$ Area of $(△ABP+△APC+△BPC)$

Formula for area of an equilateral triangle$=\frac{\sqrt{3}}{4}{a}^2$

$\Rightarrow \frac{\sqrt{3}}{4}{\left(2\right)}^2=(\frac{1}{2}\times 2\times x_c)+(\frac{1}{2}\times 2\times x_b)+(\frac{1}{2}\times 2\times x_a)$

$(\because AB=BC=AC=2)$

$\Rightarrow \sqrt{3}=x_c+x_b+x_a$