Question

find area of triangle whose sides are 11 cm , 8 cm , 13cm . fIND THE ALTITUDE using longest side as base .

Answer 1

$$\begin{gathered} \mathrm{Take}\mathrm{a}=11\mathrm{cm},\mathrm{b}=8\mathrm{cm},\mathrm{c}=13\mathrm{cm} \\ \mathrm{The}\mathrm{semi}\mathrm{perimeter}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\ =\frac{11\mathrm{cm}+8\mathrm{cm}+13\mathrm{cm}}{2} \\ =\frac{32}{2}\mathrm{cm} \\ =16\mathrm{cm} \\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{traingle}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula}, \\ \mathrm{Area}\left(△\right)=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)} \\ =\sqrt{16\left(16-11\right)\left(16-8\right)\left(16-13\right)} \\ =\sqrt{16\left(5\right)\left(8\right)\left(3\right)} \\ =\sqrt{\left(2\times 2\times 2\times 2\right)\left(5\right)\left(2\times 2\times 2\right)\left(3\right)} \\ =2\times 2\times 2\sqrt{2\times 3\times 5} \\ =\mathbf{8}\sqrt{\mathbf{30}}\mathbf{}{\mathbf{cm}}^{\mathbf{2}} \\ \mathrm{Note}\mathrm{that}\mathrm{the}\mathrm{area}\mathrm{of}\left(△\mathrm{ABC}\right)\mathrm{is}\mathrm{also}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{formula}, \\ \mathrm{area}\left(△\right)=\frac{1}{2}\times \mathrm{base}\times \mathrm{altitude} \\ \mathrm{area}\left(△\right)=\frac{1}{2}\times \mathrm{c}\times \mathrm{altitude} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{to}\mathrm{get}, \\ 8\sqrt{30}=\frac{1}{2}\times 13\times \mathrm{altitude} \\ \mathrm{altitude}=\frac{2\times 8\sqrt{30}}{13} \\ =\frac{16\sqrt{30}}{13} \\ =\mathbf{6}\mathbf{.}\mathbf{74}\mathbf{}\mathbf{cm}\left(\mathrm{approx}\right) \end{gathered}$$