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SECTION-D
This section comprises 4 Long Answer (LA) type questions of 5 marks each.
(A) In the adjoining figure,
$\triangle OAB$ is an equilateral
triangle and the area of the
shaded region is $750 \pi$ cm$^2$.
Find the perimeter of the
shaded region.
OR
(B) O and O' are the centres of the circles of radius $r$ as shown in figures
(i) and (ii) respectively.
Find the ratio of area of
shaded region in figure (i) to
that of area of shaded region
in figure (ii).
This section comprises 4 Long Answer (LA) type questions of 5 marks each.
(A) In the adjoining figure,
$\triangle OAB$ is an equilateral
triangle and the area of the
shaded region is $750 \pi$ cm$^2$.
Find the perimeter of the
shaded region.
OR
(B) O and O' are the centres of the circles of radius $r$ as shown in figures
(i) and (ii) respectively.
Find the ratio of area of
shaded region in figure (i) to
that of area of shaded region
in figure (ii).
Show SolutionHide Solution↓
Ans.
(A) $\angle AOB = 60^\circ$ (1/2 Mark)
Thus, angle of sector corresponding to shaded region = $300^\circ$
So, $\frac{300}{360} \pi r^2 = 750\pi$ (1 Mark)
$r^2 = 900$ (1 Mark)
$r = 30$ cm (1 Mark)
Perimeter of shaded region = $\frac{300}{360} \times 2\pi \times 30 + 2.30$ (1 Mark)
$= 50\pi + 60 = \frac{1520}{7}$ cm or 217.14 cm (1/2 Mark)
OR
(B) Area of unshaded segments in figure (i) = $2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2)$ (1 Mark)
Getting length of rectangle = $\sqrt{3}r$ and width = $r$ (1 Mark)
Area of rectangle = $\sqrt{3}r^2$ (1 Mark)
Area of shaded region in figure (i) = $\pi r^2 - 2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2)$ (1 Mark)
$= \frac{2}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2$ (1 Mark)
Area of shaded region in figure (ii) = $2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2) = \frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2$ (1/2 Mark)
Required ratio = $\frac{\frac{2}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2}{\frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2} = \frac{4\pi-3\sqrt{3}}{2\pi-3\sqrt{3}}$ (1/2 Mark)
or $(88 - 21\sqrt{3}): (44 - 21\sqrt{3})$
(A) $\angle AOB = 60^\circ$ (1/2 Mark)
Thus, angle of sector corresponding to shaded region = $300^\circ$
So, $\frac{300}{360} \pi r^2 = 750\pi$ (1 Mark)
$r^2 = 900$ (1 Mark)
$r = 30$ cm (1 Mark)
Perimeter of shaded region = $\frac{300}{360} \times 2\pi \times 30 + 2.30$ (1 Mark)
$= 50\pi + 60 = \frac{1520}{7}$ cm or 217.14 cm (1/2 Mark)
OR
(B) Area of unshaded segments in figure (i) = $2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2)$ (1 Mark)
Getting length of rectangle = $\sqrt{3}r$ and width = $r$ (1 Mark)
Area of rectangle = $\sqrt{3}r^2$ (1 Mark)
Area of shaded region in figure (i) = $\pi r^2 - 2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2)$ (1 Mark)
$= \frac{2}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2$ (1 Mark)
Area of shaded region in figure (ii) = $2 (\frac{60}{360} \pi r^2 - \frac{\sqrt{3}}{4} r^2) = \frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2$ (1/2 Mark)
Required ratio = $\frac{\frac{2}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2}{\frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{2} r^2} = \frac{4\pi-3\sqrt{3}}{2\pi-3\sqrt{3}}$ (1/2 Mark)
or $(88 - 21\sqrt{3}): (44 - 21\sqrt{3})$