Question

Find the area of a triangle whose sides. are $18$ $cm$ , $10$ $cm$ and $14$ $cm$.

Answer 1

Given sides are $18cm,10cm$ and $14cm$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow a=18cm,b=10cm$ and $c=14cm$

$\Rightarrow s=\frac{a+b+c}{2}=\frac{18+10+14}{2}=\frac{42}{2}=21cm$

Area $=\sqrt{21(21-18)(21-10)(21-14)}$

$=\sqrt{21\times 3\times 11\times 7}$

$=21\sqrt{11}sqcm$

Hence, the answer is $21\sqrt{11}sqcm.$