Question

Each side of an equilateral triangle is 10 cm. Find

(i) the area of the triangle and (ii) the height of the triangle.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\mathrm{side}}^2 \\ =\frac{\sqrt{3}}{4}\times {10}^2 \\ =\frac{\sqrt{3}}{4}\times 100 \\ =25\sqrt{3}{\mathrm{cm}}^2 \\ \mathrm{or}25\times 1.732=43.3{\mathrm{cm}}^2 \\ \mathrm{So},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}25\sqrt{3}{\mathrm{cm}}^2\mathrm{or}43.3{\mathrm{cm}}^2. \\ \left(\mathrm{ii}\right)\mathrm{As},\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}=25\sqrt{3}{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=25\sqrt{3} \\ \Rightarrow \frac{1}{2}\times 10\times \mathrm{Height}=25\sqrt{3} \\ \Rightarrow 5\times \mathrm{Height}=25\sqrt{3} \\ \Rightarrow \mathrm{Height}=\frac{25\sqrt{3}}{5}=5\sqrt{3} \\ \mathrm{or}\mathrm{height}=5\times 1.732=8.66\mathrm{m} \\ \therefore \mathrm{The}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}5\sqrt{3}\mathrm{cm}\mathrm{or}8.66\mathrm{cm}. \end{gathered}$$