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Q.
In a triangle, the sum of any two sides exceed the third
side by 6 cm. Find its area (in sq. cm)
Mensuration
Solution:
Let the sides of the triangle be $a cam , b cm$, and $c cm$.
$a+b-c=a+c-b=b+c-a=6 $
$\Rightarrow a+b-c+a+c-b+b+c-a=18$
$a+b+c=18$
Area of the triangle $=$
$=\frac{1}{4} \sqrt{(18)(6)(6)(6)}=9 \sqrt{3} cm ^2 .$
