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Find the area of the segment shown in Fig 2, if the radius of the circle is 21 cm and $\angle \mathrm{AOB} = 120\degree$ . $\left(\text{Use }\pi =\frac{22}{7} \right )$

Answers (1)

In a given circle,

Radius (r) = 21 cm

and $\angle \mathrm{AOB} = 120\degree$

Area of segment AYB = Area of sector OAYB - Area of $\Delta$ OAB

$\begin{align*} \text{Area of sector OAYB} & = \frac{120}{360}\times \pi r^2 & = \frac{120}{360}\times \frac{22}{7}\times (21)^2 \\ & = 462 \mathrm{cm}^2 \end{align*}$

Finding the area of $\Delta$ AOB

$\begin{align*} \mathrm{Area of \Delta OAB} & = \frac{1}{2}\mathrm {Base\times Height} \end{align*}$

We draw ON $\perp$ AB

$\therefore\angle \text{OMB} = \angle \text{ONA} = 90\degree$

In $\Delta$ OMA and $\Delta$ OMB

$\angle OMA = \angle OMB\quad (Both \ 90\degree)$

$OA = OB \quad (\text{Radius of circle})$

$OM = OM \quad (\text{Common side})$

$\therefore \Delta OMA \cong \Delta OMB\quad (\text{By R.H.S congruence})$

$\Rightarrow \angle AOM = \angle BOM\quad (\text{CPCT})$

$\Rightarrow \angle AOM = \angle BOM = \frac{1}{2}\angle BOA = 60\degree$

$\Delta$ OMB

$\frac{OM}{OA} = \cos 60\degree$

$\frac{OM}{21} = \frac{1}{2}$

$\Rightarrow OM = \frac{21}{2}cm$

Now, $\frac{AM}{OA} = \sin 60\degree = \frac{\sqrt{3}}{2}$

$AM = \frac{21\sqrt{3}}{2}cm$

Therefore, $AB = 2AM = 2\times \frac{21\sqrt{3}}{2}cm = 21\sqrt{3} cm$

So, Area of $\Delta$ OAB $= \frac{1}{2}AB\times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441}{4}\sqrt{3}cm^2$

Therefore area of the segment AYB

$\\ = \left[462 - \frac{441}{4} \sqrt{3}\mathrm{ \ cm}^2\right]$

Posted by

Safeer PP

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