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Answer 1

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Answer 2

$\huge{\red{\underline{\underline{\sf{Question:}}}}}$

30] The shaded region ABCD shows the space enclosed by two concentric circles with center 'O' and angle at the center is 75°. If the radii of the circles are 21 cm and 42 cm, find the area of the shaded region and perimeter of ABCD.

$\huge{\red{\underline{\underline{\tt{Answer:}}}}}$

$\blue{\boxed{\sf{Given:-}}}$

Center 'O' [Having two concentric circles]

angle = 75° = $\theta$

Radii

→ radius = r = 21 cm

→ Radius = R = 42 cm

$\blue{\boxed{\tt{To \:\:be\:\:found:-}}}$

(i) Area of the shaded region (ABCD)

(ii) Perimeter of ABCD

$\huge{\boxed{\boxed{\sf{Solution}}}}$

(i) Area of the shaded region (ABCD)

The formula for finding the area of sector = $\pink{\boxed{\boxed{ \sf \dfrac{\theta}{360}\times \pi r^{2} }}}}$

We can find the area of the shaded region by subtraction area of small sector OAD from big sector OBC.

Method 1

We will find the area of the two sectors differently.

So, the area of sector OAD

$\bf = \frac{\theta}{360}\times \pi r^{2} \\ \\ = \frac{75}{360} \times \frac{22}{7} \times (21)^{2}\\ \\ = 288.75 cm^{2}$

$\rule{150}2$

Area of sector OBC

$\bf = \frac{\theta}{360}\times \pi R^{2} \\ \\ = \frac{75}{360} \times \frac{22}{7} \times (42)^{2}\\ \\ = 1155 cm^{2}$

Now,

Area of the shaded region

= $\red{\boxed{\sf{Area\:of\: (sector\: OBC) - Area \:of \:sector\:OAD}}}$

= 1155 - 288.75 = $\green{\sf{866.25cm^{2} }}$

Method 2

$\red{\boxed{\sf{Area\:of\: (sector\: OBC) - Area \:of \:sector\:OAD}}}$

$= \bigg( \frac{\theta}{360} \times \pi R^2 \bigg) - \bigg(\frac{\theta}{360}\times \pi r^2 \bigg)$

$\red{\sf{ = \bigg( \frac{ \theta }{360} \times \pi \bigg) \bigg( R^2 - r^2 \bigg) }}$

$\red{\sf{= \bigg( \frac{75}{360} \times \pi \bigg) \bigg( (42)^2 - (21)^2 \bigg) }}$

$\red{\sf{ = \bigg( \frac{5}{24} \times \frac{22}{7} \bigg) \bigg( 1764 - 441 \bigg) }}$

$\red{\sf = \bigg( \frac{55}{84} \bigg) \bigg( 1323 \bigg) }$

$\red{= \boxed{\sf 866.25 cm^2 }}$

$\rule{200}2$

Now,

(ii) The Perimeter of ABCD.

To get it we need the length of AD arc, BC arc, length of AB, and DC.

Now,

Formula to find the of arc =  $\pink{\boxed{\boxed{\sf{\frac{\theta}{360}\times2 \pi r}}}}$

So,

arc AD = $\pink{\sf{\frac{\theta}{360}\times2 \pi r}}$

=  $\bf \frac{75}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{55}{2}=\boxed{\sf 27.5cm}$

arc BC =  $\pink{\sf{\frac{\theta}{360}\times2 \pi R}}$

= $\bf \frac{75}{360} \times 2 \times \frac{22}{7} \times 42 =\boxed{\sf 55 cm}$

Now length of AB

AB = OB - OA

AB = 42 - 21 = $\boxed{\bf{21cm}}$

And length of DC

DC = OC - OD

DC = 42 - 21 = $\boxed{\bf{21cm}}$

Therefore,

Perimeter of ABCD

= length of arc AD + length of arc BC + length of AB and DC

= 27.5 + 55 + 21 + 21

= $\red{\sf{124.5cm}}$

$\rule{200}2$

Hence,

Area of the shaded region $\red{= \boxed{\sf 866.25 cm^2 }}$

and,

Perimeter of ABCD = $\red{\sf{124.5cm}}$