51
A chord of a circle of radius $10$ cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use $\pi = 3\cdot14$]
Show SolutionHide Solution↓
Area of minor segment ACB = Area of sector OACB $-$ Area of right $\triangle$OAB
Area of sector OACB = $$\begin{aligned}& \frac{90}{360} \times 3.14 \times 10 \times 10 \\ & = 78.5\end{aligned}$$ cm$^2$
Area of right $\triangle$OAB = $$\begin{aligned}& \frac{1}{2} \times 10 \times 10 \\ & = 50\end{aligned}$$ cm$^2$
Area of minor segment ACB = $$\begin{aligned}& (78.5 - 50) \\ & = 28.5\end{aligned}$$ cm$^2$
Area of sector OACB = $$\begin{aligned}& \frac{90}{360} \times 3.14 \times 10 \times 10 \\ & = 78.5\end{aligned}$$ cm$^2$
Area of right $\triangle$OAB = $$\begin{aligned}& \frac{1}{2} \times 10 \times 10 \\ & = 50\end{aligned}$$ cm$^2$
Area of minor segment ACB = $$\begin{aligned}& (78.5 - 50) \\ & = 28.5\end{aligned}$$ cm$^2$