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The figure shows the rectangle $\text{ABCD}$ with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

- $\left ( \sqrt{2}-1 \right )^{2}:1$
- $2\left ( \sqrt{2}-1 \right )^{2}:1$
- $\left ( \sqrt{2}-1 \right )^{2}:2$
- None of these

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Given that, the figure.

DIAGRAM

Let the radius of the semicircle be $\text{‘R’}$ and the circle be $’r’.$

The triangle $\triangle \text{OMB}$ is right-angle triangle. so we can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{(OM)}^{2} + \text{(MB)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{R}^{2} + \text{R}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{2R}^{2} $

$\Rightarrow \boxed{ \text{OB} = \sqrt{2} \; \text{R} }$

Similarly $ \boxed{ \text{O’B} = \sqrt{2} \; r }$

Now, $\text{OB} = \text{ON} + \text{NO’} + \text{O’B}$

$\Rightarrow \sqrt{2} \; \text{R} = \text{R} + r + \sqrt{2} \; r $

$\Rightarrow \sqrt{2} \; \text{R} – \text{R} = r + \sqrt{2} \; r $

$\Rightarrow \text{R}(\sqrt{2}-1) = r(\sqrt{2}+1) $

$\Rightarrow \frac{r}{\text{R}} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)}{(\sqrt{2}+1)} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{(2-1)}$

$\Rightarrow \boxed{\frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{1}}$

$\therefore$ The ratio of the area of the circle to that of the semicircle $ = \frac{\pi \; r^{2}}{\frac{\pi}{2}\; \text{R}^{2}}$

DIAGRAM

Let the radius of the semicircle be $\text{‘R’}$ and the circle be $’r’.$

The triangle $\triangle \text{OMB}$ is right-angle triangle. so we can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{(OM)}^{2} + \text{(MB)}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{R}^{2} + \text{R}^{2} $

$\Rightarrow \text{(OB)}^{2} = \text{2R}^{2} $

$\Rightarrow \boxed{ \text{OB} = \sqrt{2} \; \text{R} }$

Similarly $ \boxed{ \text{O’B} = \sqrt{2} \; r }$

Now, $\text{OB} = \text{ON} + \text{NO’} + \text{O’B}$

$\Rightarrow \sqrt{2} \; \text{R} = \text{R} + r + \sqrt{2} \; r $

$\Rightarrow \sqrt{2} \; \text{R} – \text{R} = r + \sqrt{2} \; r $

$\Rightarrow \text{R}(\sqrt{2}-1) = r(\sqrt{2}+1) $

$\Rightarrow \frac{r}{\text{R}} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)}{(\sqrt{2}+1)} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$\Rightarrow \frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{(2-1)}$

$\Rightarrow \boxed{\frac{r}{\text{R}} = \frac{(\sqrt{2}-1)^{2}}{1}}$

$\therefore$ The ratio of the area of the circle to that of the semicircle $ = \frac{\pi \; r^{2}}{\frac{\pi}{2}\; \text{R}^{2}}$