Question

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

Answer 1

Let the sides of the triangle be ​a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.
$$\begin{gathered} s=\frac{1}{2}(a+b+c) \\ s=\frac{1}{2}(20+34+42) \\ s=48\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}thetriangle\; =\sqrt{s(s-a)(s-b)(s-c)} \\ \Rightarrow \sqrt{48(48-20)(48-34)(48-42)} \\ \Rightarrow \sqrt{48\times 28\times 14\times 6} \\ \Rightarrow \sqrt{112896} \\ \Rightarrow 336{\mathrm{cm}}^2 \end{gathered}$$

Length of the longest side is 42 cm.

Area of a triangle =$\frac{1}{2}\times b\times h$
$$\begin{gathered} \Rightarrow 336=\frac{1}{2}\times 42\times h \\ \Rightarrow 672=42h \\ \Rightarrow \frac{672}{42}=h \\ \Rightarrow h=16\mathrm{cm} \end{gathered}$$

The height corresponding to the longest side is 16 cm.