Ans: D
Let $O$ be the centre of the circle. $A$ and $B$ be the end points of the chord. In $\Delta ABO$,
$\begin{array}{rcl}
\cos \theta & = & \dfrac{5^2+5^2-8^2}{2(5)(5)} \\
\theta & = & 106.260\ 204\ 7^\circ
\end{array}$
Therefore, the area of the major segment
$\begin{array}{cl}
= & \text{area of $\Delta ABO$} +\text{area of major sector $OAB$} \\
= & \dfrac{1}{2} (5)(5) \sin 106.260\ 204\ 7^\circ + \pi (5)^2 \dfrac{360^\circ -106.260\ 204\ 7^\circ}{360^\circ} \\
= & 67.357\ 435\ 89 \\
\approx & 67 \text{ cm}^2
\end{array}$
