Question

Two concentric circles of radius 8cm and 5cm shown below. And sector forms an angle of 40° at

the centre O. what is the area of the shaded region?

a. 79/8n

b. 29/9

C. 39/97

d. 49/9n​

Answer 1

Answer:

Step-by-step explanation:

refer the image

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Answer 2

Answer:

The area of the shaded region is $\frac{2288}{21}\ cm^2$.

Step-by-step explanation:

Given:-

The radius of the inner circle, r = 5 cm

The radius of the outer circle, R = 8 cm

The sector forms an angle = 40°

Then,

The area of the sector OAC is,

= $\frac{40^0}{360^0}\times \pi R^2$

= $\frac{1}{9}\times \frac{22}{7}\times 8^2$

= $\frac{1408}{63}$ $cm^2$

The area of the sector OBD is,

= $\frac{40^0}{360^0}\times \pi r^2$

= $\frac{1}{9}\times \frac{22}{7}\times 5^2$

= $\frac{550}{63}$ $cm^2$

Now,

Area of the region ACDB is,

= Area of the sector OAC - Area of the sector OBD

= $\frac{1408}{63}-\frac{550}{63}$

= $\frac{858}{63}$ $cm^2$

The area of the shaded region is,

= Area of the outer circle - Area of the inner circle - Area of the region ACDB)

= $\pi R^2-\pi r^2-\frac{858}{63}$

= $\pi (R^2-r^2)-\frac{286}{21}$

= $\frac{22}{7}(8^2-5^2)-\frac{286}{21}$

= $\frac{858}{7}-\frac{286}{21}$

= $\frac{2574-286}{21}$

= $\frac{2288}{21}\ cm^2$

Therefore, the $area$ of the shaded region is $\frac{2288}{21}\ cm^2$.

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