Question

In a triangle, the average of any two sides is $6cm$ more than half of the third side. Find area of the triangle. (in sq. $cm$ )

• A. $64\sqrt{3}$
• B. $48\sqrt{3}$
• C. $72\sqrt{3}$
• D. $36\sqrt{3}$

Answer 1

Answer: (D)

Let the sides of the triangle be $a,b$, and $l$.
Given, $\frac{a+b}{2}=6+\frac{c}{2},\frac{a+c}{2}=6+\frac{b}{2}$ and $\frac{b+c}{2}$
$=6+\frac{a}{2}.$
$\Rightarrow a+b=12+c$(1)
$b+c=12+a$(2)
$c+a=12+b$(3)
By solving, we get:
$a=b=c=12$
$\Rightarrow \text{ Area of }△ABC=\frac{\sqrt{3}}{2}\times 12^2$
$=36\sqrt{3}sq.cm$.