Question

Two circles with radius 2r and √2r intersect each other at points a and



b. The centers of both the circles are on the same side of ab. O is the center of the bigger circle and ∠aob is 60°. Find the area of the common region between two circles.

Answer 1

Answer:

$r^{2} (\frac{13{\pi}}{6}+1-\sqrt{3})$

Step-by-step explanation:

We are given two circles with radius 2r and $\sqrt{2r}$ intersect each other at points a and b. From the diagram, we have,  AD= AO cos60°

⇒AD=$AO$×$\frac{1}{2}$

⇒AD=r

Noe, in ΔACD, $\frac{AD}{AC} =sinACD$

⇒$\frac{r}{\sqrt{2r} } =sinACD$

⇒sinACD[/tex]=$\frac{1}{\sqrt{2} }$

⇒∠ACD=45°

Therefore by similarity, ∠BCD=45°⇒∠ACB=90°

Now,area of portion (1) will be: $\frac{60^{\circ}}{360^{\circ}} {\pi}r^{2} -\frac{1}{2} r^{2} sin60^{\circ}$

⇒$\frac{60^{\circ}}{360^{\circ}} {\pi}(2r)^{2} -\frac{1}{2} (2r)^{2} sin60^{\circ}$

⇒$\frac{2{\pi}r^{2} }{3} -\sqrt{3} r^{2}$                             (A)

Now, area of portion (2) will be: $\frac{270^{\circ}}{360^{\circ}} {\pi}(\sqrt{2} r^{2})$

⇒$\frac{3}{2}{\pi}r^{2}$                                                     (B)

Also,area triangle ACB= $\frac{1}{2}$×$r$×$2r$= $r^{2}$                             (C)

Adding (A), (B) and (C),we have

Area of the shaded portion=$\frac{2{\pi}r^{2} }{3} -\sqrt{3} r^{2} +\frac{3{\pi}r^{2} }{2} +r^{2}$

= $r^{2} (\frac{13{\pi}}{6}+1-\sqrt{3})$