Question

2. In the figure, O is the centre
of the circle.
m(arc PQR)= 60°
OP=10 cm.
Find he area of the shaded
region. (n = 3.14, 3=1.73)​

Answer 1

Answer:

Here is your answer. I hope this will help u

Answer 2

Answer:

Area of the shaded region is 9.08 cm².

Step-by-step-explanation:

NOTE: Kindly refer to the attachment for diagram.

We have given the radius and measure of an arc of a circle.

We have to find the area of the shaded region.

The shaded region is a segment.

Hence, we have to find the area of the segment.

$\sf\:Radius\:(\:r\:)\:=\:OP\:=\:10\:cm\\\\\sf\:m\:(\:arc\:PQR\:)\:=\:\theta\:=\:60^\circ\\\\\sf\:Segment\:of\:the\:circle\:is\:PQR\\\\\sf\:\pi\:=\:3.14\\\\\sf\:\sqrt{3}\:=\:1.73$

Now, we know that,

$\pink{\sf\:A\:(\:segment\:PQR\:)\:=\:r^2\:[\:\dfrac{\pi\:\theta}{360}\:-\:\dfrac{\sin\:\theta}{2}\:]}\sf\:\:\:-\:-\:-\:[\:Formula\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:(\:10\:)^2\:[\:\dfrac{3.14\:\times\:\cancel{60}}{\cancel{360}}\:-\:\dfrac{\sin\:60}{2}\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:[\:\dfrac{3.14}{6}\:-\:\dfrac{\sqrt{3}}{2\:\times\:2}\:\:\:-\:-\:[\:\because\:\sin\:60\:=\:\dfrac{\sqrt{3}}{2}\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:[\:\dfrac{3.14}{6}\:-\:\dfrac{1.73}{4}\:]\:\:\:-\:-\:-\:[\:\because\:\sqrt{3}\:=\:1.73\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:[\:\dfrac{3.14\:\times\:2}{6\:\times\:2}\:-\:\dfrac{1.73\:\times\:3}{4\:\times\:3}\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:[\:\dfrac{6.28}{12}\:-\:\dfrac{5.19}{12}\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:[\:\dfrac{6.28\:-\:5.19}{12}\:]\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:100\:\times\:\frac{1.09}{12}\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:\cancel{\frac{109}{12}}\\\\\implies\sf\:A\:(\:segment\:PQR\:)\:=\:9.083\\\\\implies\boxed{\red{\sf\:A\:(\:segment\:PQR\:)\:\approx\:9.08\:cm^2}}$