1. Two circles with radius 2R and $$\sqrt{2R}$$ 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.
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By: anil on 05 May 2019 03.05 pm
Let us draw the diagram according to the given info, We can see that AD = AO*cos60° = 2R*$$dfrac{1}{2}$$ = R In triangle, ACD $$Rightarrow$$ $$sinACD=dfrac{AC}{AC}$$ $$Rightarrow$$ $$sinACD=dfrac{R}{sqrt{2}*R}$$ = $$dfrac{1}{sqrt{2}}$$
$$Rightarrow$$ $$angle$$ ACD = 45°
By symmetry we can say that $$angle$$ BCD = 45° Therefore we can say that $$angle$$ ACB = 90° Hence, the area colored by green color = $$dfrac{270°}{360°}*pi*(sqrt{2}R)^2$$ = $$dfrac{3}{2}*pi*R^2$$ ... (1) Area of triangle ACB = $$dfrac{1}{2}*R*2R$$ = $$R^2$$ ... (2) Area shown in blue color = $$dfrac{60°}{360°}*pi*(2R)^2-dfrac{sqrt{3}}{4}*(2R)^2$$ = $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ ... (3) By adding (1) + (2) + (3) Therefore, the area of the common region between two circles = $$dfrac{3}{2}*pi*R^2$$ + $$R^2$$ + $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ $$Rightarrow$$ $$(dfrac{13pi}{6}+1-sqrt{3})R^2$$ Hence, option C is the correct answer.
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$$Rightarrow$$ $$angle$$ ACD = 45°
By symmetry we can say that $$angle$$ BCD = 45° Therefore we can say that $$angle$$ ACB = 90° Hence, the area colored by green color = $$dfrac{270°}{360°}*pi*(sqrt{2}R)^2$$ = $$dfrac{3}{2}*pi*R^2$$ ... (1) Area of triangle ACB = $$dfrac{1}{2}*R*2R$$ = $$R^2$$ ... (2) Area shown in blue color = $$dfrac{60°}{360°}*pi*(2R)^2-dfrac{sqrt{3}}{4}*(2R)^2$$ = $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ ... (3) By adding (1) + (2) + (3) Therefore, the area of the common region between two circles = $$dfrac{3}{2}*pi*R^2$$ + $$R^2$$ + $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ $$Rightarrow$$ $$(dfrac{13pi}{6}+1-sqrt{3})R^2$$ Hence, option C is the correct answer.