Find angle, area and perimeter of sectors given two similar sectors with the same center

Question Sample Titled 'Find angle, area and perimeter of sectors given two similar sectors with the same center'

In the figure, ${O}{A}{B}$ and ${O}{C}{B}$ are sectors with centre ${O}$, where ${O}{A}={82}$ cm and ${O}{C}={98}$ cm. The area of the shaded region ${A}{B}{D}{C}$ is ${496}\pi$ $\text{cm}^{{2}}$. Which of the following is/are true?

 I. The angle of the sector ${O}{A}{B}$ is ${60}^{\circ}$ . II. The area of the sector ${O}{A}{B}$ is $\dfrac{{1271}}{{45}}\pi$ $\text{cm}^{{2}}$ . III. The perimeter of the sector ${O}{C}{D}$ is ${\left({196}+\dfrac{{1519}}{{45}}\pi\right)}$ cm .

${O}$${D}$${C}$${B}$${A}$
A
III only
B
I only
C
I and III only
D
II and III only

${O}$${D}$${C}$${B}$${A}$$\theta$${82}$${16}$

 I. Let the angle of the sector ${O}{A}{B}$ be $\theta$ . $\dfrac{\theta}{{360}^{\circ}}\pi{\left({98}\right)}^{{2}}-\dfrac{\theta}{{360}^{\circ}}\pi{\left({82}\right)}^{{2}}$ $={496}\pi$ area of sector $=\dfrac{\theta}{{360}^{\circ}}\times\pi{r}^{{2}}$ $\theta$ $={62}^{\circ}$ ∴   The angle of the sector ${O}{A}{B}$ is ${62}^{\circ}$ . I is false.  II. The area of the sector ${O}{A}{B}$ $=\dfrac{{62}^{\circ}}{{360}^{\circ}}\pi{\left({82}\right)}^{{2}}$ area of sector $=\dfrac{\theta}{{360}^{\circ}}\times\pi{r}^{{2}}$ $=\dfrac{{52111}}{{45}}\pi$ $\text{cm}^{{2}}$ II is false.  III. The perimeter of the sector ${O}{C}{D}$ $={O}{D}+{O}{C}+$ CD⏜$\stackrel{⏜}{CD}$$\overparen{CD}$ $={98}+{98}+\dfrac{{62}^{\circ}}{{360}^{\circ}}{2}\pi{\left({98}\right)}$ arc length of sector $=\dfrac{\theta}{{360}^{\circ}}\times{2}\pi{r}$ $={\left({196}+\dfrac{{1519}}{{45}}\pi\right)}$ $\text{cm}$ III is true.

 II. Let the area of the sector ${O}{A}{B}$ be ${x}$ $\text{cm}^{{2}}$ . $\dfrac{{x}}{{{x}+{496}\pi}}$ $={\left(\dfrac{{82}}{{98}}\right)}^{{2}}$ $\dfrac{{A}_{{2}}}{{A}_{{1}}}={\left(\dfrac{{l}_{{2}}}{{l}_{{1}}}\right)}^{{2}}$ for similar figures ${x}$ $={\left(\dfrac{{41}}{{49}}\right)}^{{2}}{x}+{\left(\dfrac{{41}}{{49}}\right)}^{{2}}\cdot{496}\pi$ ${\left({1}-{\left(\dfrac{{41}}{{49}}\right)}^{{2}}\right)}{x}$ $={\left(\dfrac{{41}}{{49}}\right)}^{{2}}\cdot{496}\pi$ ${x}$ $=\dfrac{{{\left(\dfrac{{41}}{{49}}\right)}^{{2}}\cdot{496}\pi}}{{{1}-{\left(\dfrac{{41}}{{49}}\right)}^{{2}}}}$ ${x}$ $=\dfrac{{52111}}{{45}}\pi$ ∴   The area of the sector ${O}{A}{B}=\dfrac{{52111}}{{45}}\pi$ $\text{cm}^{{2}}$

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