2022-II-16

Let $O$ be the centre of the circle. $A$ and $B$ be the end points of the chord. In $\mathrm{\Delta }ABO$,

$\begin{array}{rcl}\cos \theta & =& {\displaystyle \frac{5^2+5^2-8^2}{2(5)(5)}}\\ \theta & =& 106.2602047^{\circ }\end{array}$

Therefore, the area of the major segment

$\begin{array}{cl}=& \text{area of }\mathrm{\Delta }ABO+\text{area of major sector }OAB\\ =& {\displaystyle \frac{1}{2}}(5)(5)\sin 106.2602047^{\circ }+\pi (5{)}^2{\displaystyle \frac{{360}^{\circ }-106.2602047^{\circ }}{{360}^{\circ }}}\\ =& 67.35743589\\ \approx & 67{\text{ cm}}^2\end{array}$