Question

The sides of a triangle are $42cm$, $34cm$, and $20cm.$ Calculate its area and the length of the height on the longest side.

Answer 1

Let a $=42cm,$ b $=34cm$ and c $=20cm$

Then, s $=\frac{a+b+c}{2}$ $=\frac{(42+34+20)}{2}cm$

$=\frac{96}{2}cm=48cm$

$\therefore$ Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c}$ sq. units

$=\sqrt{48(48-42)(48-34)(48-20)}c{m}^2$

$=\sqrt{48\times 6\times 14\times 28}c{m}^2$

$=\sqrt{(6\times 8)\times 6\times 14\times (14\times 2)}c{m}^2$

$=\sqrt{\underset{–}{6\times 6}\times \underset{–}{2\times 2}\times \underset{–}{2\times 2}\times \underset{–}{14\times 14}}c{m}^2$

$=\sqrt{(2\times 2\times 6\times 14)}c{m}^2$ $=336c{m}^2$
Let the height on the side measuring $42cm$ be $hcm.$

Then, Area of the triangle $=\frac{1}{2}\times base\times height$

$\Rightarrow 336c{m}^2=\frac{1}{2}\times 42\times h$

$\Rightarrow h=\frac{336}{21}cm$ $=16cm$

$\therefore$ Area of the triangle $=336c{m}^2$

and, Height corresponding to the side measuring $42cm$ is $16cm$