A circle having radius 4 cm contains a chord of length

SOLUTION

Area of sector POQ
$=\frac{60}{360}$× $\pi 4^2$ = 8.4 $c{m}^2$

In triangle OSQ which is right angled at S, 
$O{Q}^2=S{Q}^2+O{S}^2$

$\Rightarrow 16=4+O{S}^2$
$OS=2\sqrt{3}$

Area of triangle POQ
$=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times 4\times 2\sqrt{3}$
$=6.9c{m}^2$ 

Now,
Area of segment PSQR = Area of sector POQ - Area of triangle POQ
$=8.4–6.9c{m}^2$
$=1.5c{m}^2$