Question

in figure segment PQ is the diameter of semicircle P and Q the centre of p m q is equals to 10 cm and angle poq equal to 60 degree find the shaded portion​

Answer 1

Answer:

wt is the shaded portion???

Answer 2

The shaded portion is 30.25 $cm^2$.

Given

To find the shaded portion.

Radius ( r ) = 10 cm.

              $\begin{equation}\theta\end$ = 60°

From the picture,

It consist of segment PMQ,

Area of segment PMQ = $r^2$ × [ (π $\begin{equation}\theta\end$ / 360) - (sin $\begin{equation}\theta\end$ / 2) ]

                      = $10^2$ × [ (3.14 × 60°) - (sin 60° / 2) ]

                      = 100 × [ (0.52) - (0.43) ]

                      = 100 × [ 0.52 - 0.43 ]

                      = 100 × 0.09

                      = 9.

            Area of segment PMQ = 9 $cm^2$.

In ΔOPQ, segment OP ≅ segment OQ

                ∠OPQ = m∠OPQ = x

                m∠OPQ + m∠OQP + m∠POQ = 180°

                           x + x + 60° = 180°

                                         2x = 180° - 60°

                                         2x = 120°

                                           x = 60°

Hence,        m∠OPQ = m∠OQP = m∠POQ = 60°

                   ΔOPQ is an equilateral triangle

                   OP = OQ = PQ = 10 cm.

Area of Semi-circle = $\frac{1}{2}$ × π$r^2$

Here, Diameter PQ = 10 cm.

          Radius ( r ) = 10 / 2 = 5 cm.

                         r = 5 cm  

Area of Semi-circle = $\frac{1}{2}$ × π$r^2$

                                =  $\frac{1}{2}$ × 3.14 × 5 × 5

                                = 39.25 $cm^2$.

                         Area of semi-circle = 39.25 $cm^2$.

Area of shaded portion = Area of semi-circle - Area of segment PMQ

                                       = 39.25 - 9        

                                       = 30.25 $cm^2$

Therefore, the shaded portion is 30.25 $cm^2$.

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