Question

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=91\mathrm{m},b=98\mathrm{m}\mathrm{and}c=105\mathrm{m} \\ \therefore s=\frac{a+b+c}{2}=\frac{91+98+105}{2}=147\mathrm{m} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{147(147-91)(147-98)(147-105)} \\ =\sqrt{147\times 56\times 49\times 42} \\ =\sqrt{7\times 3\times 7\times 2\times 2\times 2\times 7\times 7\times 7\times 7\times 3\times 2} \\ =7\times 7\times 7\times 2\times 3\times 2 \\ =4116{\mathrm{m}}^2 \end{gathered}$$

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=4116{\mathrm{m}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=4116 \\ \Rightarrow \mathrm{Height}=\frac{4116\times 2}{105}=78.4\mathrm{m} \end{gathered}$$