Find the area of the shaded portion intersecting between the two circles.

Mathematics

2 answers:

defon2 years ago

8 0

Answer: A=16/3 pi - 8Г3

Step-by-step explanation:

Zolol [24]2 years ago

3 0

Look at the picture.

ΔABC and ΔBDC are <span>equilateral triangles.

Area of sector of a circle:
$\dfrac{60^o}{360^o}=\dfrac{1}{6}\\\\area\ of\ a\ circle\\A_O=\pi r^2;\ r=4\\A_O=\pi\cdot4^2=16\pi\\\\A_s=\dfrac{1}{6}A_O\to A_s=\dfrac{1}{6}\cdot16\pi=\boxed{\dfrac{8}{3}\pi}$

Area of the triangle ABC:
$A_\Delta=\dfrac{a^2\sqrt3}{4};\ a=4\\\\A_\Delta=\dfrac{4^2\sqrt3}{4}=\dfrac{16\sqrt3}{4}=\boxed{4\sqrt3}$

Area of the shaded portion:
$A=2(A_s-A_\Delta)\\\\\boxed{A=2\left(\dfrac{8}{3}\pi-4\sqrt3\right)\approx2.89}$
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$\bf \begin{array}{clclll}
-6&+&6\sqrt{3}\ i\\
\uparrow &&\uparrow \\
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-------------------------------\\\\$

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\\\\\\
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\\\\\\
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