Question

The height of an equilateral triangle is 6 cm. Find its area. [Take $\sqrt{3}$ = 1.73]

Answer 1

Let the side of the equilateral triangle be x cm.

$$\begin{gathered} \mathrm{As},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2=\frac{x^2\sqrt{3}}{4} \\ \mathrm{Also},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=\frac{1}{2}\times x\times 6=3x \\ \mathrm{So},\frac{x^2\sqrt{3}}{4}=3x \\ \Rightarrow \frac{x\sqrt{3}}{4}=3 \\ \Rightarrow x=\frac{12}{\sqrt{3}} \\ \Rightarrow x=\frac{12}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \Rightarrow x=\frac{12\sqrt{3}}{3} \\ \Rightarrow x=4\sqrt{3}\mathrm{cm} \\ \mathrm{Now},\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}=3x \\ =3\times 4\sqrt{3} \\ =12\sqrt{3} \\ =12\times 1.73 \\ =20.76{\mathrm{cm}}^2 \end{gathered}$$