Expert Answer to "To find:The area of region common to both..."

Brennan Flores

Answered question

2021-08-13

To find:The area of region common to both circles.

Clara Reese

Skilled2021-08-14Added 120 answers

Explanation of solution

2021-12-24

Now taking sector OAPB

$\sin O B X = \frac{3}{6}$

$\sin O B X = 0.5$

$m ∠ O B X = 30$ and

$\cos 30 = \frac{X B}{6}$

$X B = 3 \sqrt{3}$

$A X = X B = 3 \sqrt{3}$ and

$A X = X B = \frac{1}{2} A B$

AB=2 AX

$A B = (2) 3 \sqrt{3}$

$A B = 6 \sqrt{3}$

Now in triangle AOB

$∠ A O B + ∠ O B X + ∠ O A X = 180$

$∠ A O B + 60 = 180$

$∠ A O B = 120$

so $m \overset{⌢}{A B} = 120$

let $\overset{⌢}{A B} = x$

Now area of sector

$A_{A O B} = \frac{x}{360} π r^{2}$

$A_{A O B} = \frac{120}{360} π (6)^{2}$

$A_{A O B} = 12 π$

Now area of shaded region in OAPB is

$A = 12 π - 9 \sqrt{3} c m^{3}$

Similarly in the right circle

Shaded region is also $A = 12 π - 9 \sqrt{3} c m^{2}$

Thus, common shaded area

$A = (12 π - 9 \sqrt{3}) + (12 π - 9 \sqrt{3}) c m^{2}$

$A = 24 π - 18 \sqrt{3} c m^{2}$

Hence, $A = 24 π - 18 \sqrt{3} c m^{2}$

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