Question

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm. Also, find the height corresponding to the longest side.

Answer 1

Semiperimeter (s) = $\frac{(42+34+20)}{2}=\frac{96}{2}=48$ cm.

Using herons formula Area $=\sqrt{s(s-a)(s-b)(s-c)}$

By substituting s and a, b , c i.e the given sides.

We have $Area=\sqrt{48\times 6\times 14\times 28}=\sqrt{(4\times 6\times 2)\times 6\times (7\times 2)\times (7\times 4)}=4\times 6\times 2\times 7=336c{m}^2$

Hence area is $336c{m}^2$

Now height corresponding to longest side implies that base is 42 cm and the area remains same

So Area of triangle = $\frac{1}{2}\times base\times height$

$336=\frac{1}{2}\times 42\times H$

hence H = 16 cm

So height is 16 cm.