Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region Class 9

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region Class 9

Question 1.

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r. Class 9

Given:

Each circle passes through the each other's centre

To find: Area enclosed by two circles Construction: Join AC, BC, AD, BD and AB

Solution:

Let the radii of both the circle be 'r'

∵ AC = BC = AB = r

∴ ∆ABC is equilateral

∴ ∠ACB = 60°

Similarly,

∠ADB = 60°

Now, ∠CAD = ∠CAB + ∠DAB

= 60° + 60° = 120°

∴ ∠CAD = 120°

Also, ∠CBD = ∠CBA + ∠DBA = 60° + 60°

= 120°

∴ ∠CBD = 120°

∵ Area of sector (B - CAD)

$=\frac{\theta}{360} \times \pi r^2=\frac{120}{360}=\pi r^2=\frac{1}{3} \pi r^2$ ...(i)

∵ ABc is Equilateral ... [Given]

∴ A(∆ABC) = $\frac{\sqrt{3}}{4} r^2$ ...(ii)

∴ Area of segment = Area of sector - Area of triangle

$=\frac{1}{3} \pi r^2-\frac{\sqrt{3}}{4} r^2$

...[From (i) and (ii)]

∵ There are two identical segments

∴ Total area of 2 segment

=$2\left(\frac{1}{3} \pi r^2-\frac{\sqrt{3}}{4} r^2\right)$

$=\frac{2}{3} \pi r^2-\frac{\sqrt{3}}{2} r^2$

∴ Area of common region = $\left(\frac{2}{3} \pi-\frac{\sqrt{3}}{2}\right) r^2$