Question

Find the length of side of an equilateral triangle inscribed in a circle of radius 4 cm. ($\sqrt{3}=1.73$)

Answer 1

Consider the figure below.


Let AB = BC = CA = a cm
We know that the median and the perpendicular bisector of an equilateral triangle coincide with each other.
$\therefore$DC = $\frac{a}{2}$ cm

$$\begin{gathered} \mathrm{As}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}△\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle},\mathrm{AD}\mathrm{is}\mathrm{the}\mathrm{median}. \\ \mathrm{Therefore},\mathrm{O}\mathrm{would}\mathrm{be}\mathrm{its}\mathrm{centroid}. \\ \mathrm{Hence},\mathrm{O}\mathrm{divides}\mathrm{the}\mathrm{median}\mathrm{in}\mathrm{the}\mathrm{ratio}2:3. \\ \mathrm{Now},\mathrm{OA}=\frac{2}{3}\times \mathrm{AD} \\ \Rightarrow 4\mathrm{cm}=\frac{2}{3}\times \mathrm{AD} \\ \Rightarrow \mathrm{AD}=6\mathrm{cm} \end{gathered}$$

In right triangle ADC, we have:

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{DC}}^2 \\ \Rightarrow a^2=6^2+{\left(\frac{a}{2}\right)}^2 \\ \Rightarrow a^2-\frac{a^2}{4}=36 \\ \Rightarrow \frac{4a^2-a^2}{4}=36 \\ \Rightarrow \frac{3a^2}{4}=36 \\ \Rightarrow 3a^2=36\times 4 \\ \Rightarrow a^2=12\times 4 \\ \Rightarrow a=4\sqrt{3} \\ \Rightarrow a=6.92 \end{gathered}$$
Thus, the side of the equilateral triangle is 6.92 cm.