Question

Find the area of shaded region​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

OQSR is a rectangle such that, RT = 1 cm and OQ = 3 cm

Now,

Let assume that radius of semi-circle, OT = r cm

So, OS = r cm

Now, OR = OT - RT = r - 1 cm

Now, In right-angle triangle ORS

We have,

OR = r - 1 cm

OS = r cm

RS = OQ = 3 cm

By using Pythagoras Theorem, we have

$\rm \: {OS}^{2} = {OR}^{2} + {RS}^{2}$

So, on substituting the values, we get

$\rm \: {r}^{2} = {(r - 1)}^{2} + {3}^{2}$

$\rm \: {r}^{2} = {r}^{2} + 1 - 2r + 9$

$\rm \: 0 = - 2r + 10$

$\rm \: 2r \: = \: 10$

$\rm\implies \:r \: = \: 5 \: cm \:$

Now,

Dimensions of rectangle OQSR

OQ = 3 cm

OR = OT - RT = 5 - 1 = 4 cm

$\rm \: Area_{(rectangle)}\: OQSR \: = \: 3 \times 4 = 12 \: {cm}^{2}$

Now,

$\rm \: Area_{(quadrant)}\:OTSP = \frac{1}{4} \: {\pi \: r}^{2}$

$\rm \: = \: \frac{1}{4} \times \pi \times {5}^{2}$

$\rm \: = \: \frac{25\pi}{4}$

$\rm \: = \: 6.25 \: \pi \: {cm}^{2}$

$\rm\implies \: Area_{(quadrant)}\:OTSP \: = \: 6.25 \: \pi \: {cm}^{2} \:$

So,

Required area of shaded region is

$\rm \: = \: Area_{(quadrant)}\:OTSP \: - \: Area_{(rectangle)}\: OQSR$

$\rm \: = \: (6.25 \: \pi \: - \: 12) \: {cm}^{2}$

Hence, Option (ii) is correct.

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Answer:

$\color{red} \[ \begin{array}{l} \text{Given that,} \\ \tt R T=1 cm \\ \tt O Q=3 cm \end{array} \]$

$\color{navy}\[ \begin{array}{l} \text{Here,} \\ \text { let } \tt O R=x \\ \tt \: O S=O R+R T( \because \text { radius }) \\ \tt \Rightarrow r=O R+1=x+1 \qquad \ldots \text { \tt (1) } \end{array} \]$

$\color{darkviolet} \begin{aligned} & \tt \: r^{2}=3^{2}+O R^{2}(\text { pythagoras theorem }) \\ \\ \Rightarrow & \tt(x+1)^{2}=9+x^{2} \\ \\ \tt\Rightarrow & \tt \: x^{2}+1+2 x=9+x^{2} \\ \\ \Rightarrow & \tt 2 x=8 \\ \\ \Rightarrow & \tt x=4 \end{aligned}$

$\color{green} \text{Hence, area of \( \tt O Q S R=3 \times 4=12 \)} \\ \text{ radius \( \tt=O R+R T=x+1=4+1=5 \)}$

$\color{brown} \text{Area of shaded region }= \dfrac{\tt(Area\:\: of \:\:circle) }{4}- (\tt Area\:\: of \:\: OQ S R )$

$\color{olive}\[ \begin{array}{l} \tt=\dfrac{\pi \times 5 \times 5}{4}-12 \qquad\left(\because A=\pi r^{2}\right) \\ \\ = \tt\dfrac{25 \pi}{4}-12 \\ \\ \boxed{\boxed{ \tt =6.25 \pi-12}} \end{array} \]$