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Q.
In a triangle, the average of any two sides is $6 cm$ more than half of the third side. Find area of the triangle. (in sq. $cm$ )
Mensuration
Solution:
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:
$a=b=c=12$
$\Rightarrow \text { Area of } \triangle A B C =\frac{\sqrt{3}}{2} \times 12^2$
$ =36 \sqrt{3} sq . cm $.