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Question Number 193093 by Mingma last updated on 04/Jun/23

	Answered by Subhi last updated on 04/Jun/23
	$(1/2).π.r^2 = 3π r = (√6) AB = AD = BE = 2(√6) (CD)^2 = CE.BC=CE.(CE+2(√6)) (CD)^2 =(CE)^2 +2(√6).CE →(1) or {(CD)^2 +6=(CE+(√6))^2 =(CE)^2 +2(√6).CE+6} ((CD)/(sin(90−x)))=((√6)/(sin(x))) (sin law) ((sin(x))/(cos(x)))=((√6)/(CD)) ⇛ tan(x)=((√6)/(CD)) (i) in the Δ ABC ((2(√6))/(sin(x)))=((2(√6)+CE)/(sin(90−x))) tan(x) = ((2(√6))/(2(√6)+CE)) (ii) ((2(√6))/(2(√6)+CE))=((√6)/(CD)) CD = (√6)+((CE)/2) → (2) substitute in (1) 6+(√6).CE+(((CE)^2 )/4)=(CE)^2 +2(√6).CE (3/4)(CE)^2 +(√6).CE−6 = 0 CE = ((2(√6))/3) , CD = ((4(√6))/3) [ΔABC] = (1/2)×2(√6)×(2(√6)+((2(√6))/3))=16 the shaded area = 16−3π$
	$\frac{1}{2} . π . r^{2} = 3 π$ $r = \sqrt{6}$ $A B = A D = B E = 2 \sqrt{6}$ ${(C D)}^{2} = C E . B C = C E . (C E + 2 \sqrt{6})$ ${(C D)}^{2} = {(C E)}^{2} + 2 \sqrt{6} . C E \to (1)$ $o r$ ${{(C D)}^{2} + 6 = {(C E + \sqrt{6})}^{2} = {(C E)}^{2} + 2 \sqrt{6} . C E + 6}$ $\frac{C D}{s i n (90 - x)} = \frac{\sqrt{6}}{s i n (x)} (s i n l a w)$ $\frac{s i n (x)}{c o s (x)} = \frac{\sqrt{6}}{C D} ⇛ t a n (x) = \frac{\sqrt{6}}{C D} (i)$ $i n t h e Δ A B C$ $\frac{2 \sqrt{6}}{s i n (x)} = \frac{2 \sqrt{6} + C E}{s i n (90 - x)}$ $t a n (x) = \frac{2 \sqrt{6}}{2 \sqrt{6} + C E} (i i)$ $\frac{2 \sqrt{6}}{2 \sqrt{6} + C E} = \frac{\sqrt{6}}{C D}$ $C D = \sqrt{6} + \frac{C E}{2} \to (2)$ $s u b s t i t u t e i n (1)$ $6 + \sqrt{6} . C E + \frac{{(C E)}^{2}}{4} = {(C E)}^{2} + 2 \sqrt{6} . C E$ $\frac{3}{4} {(C E)}^{2} + \sqrt{6} . C E - 6 = 0$ $C E = \frac{2 \sqrt{6}}{3}, C D = \frac{4 \sqrt{6}}{3}$ $[Δ A B C] = \frac{1}{2} \times 2 \sqrt{6} \times (2 \sqrt{6} + \frac{2 \sqrt{6}}{3}) = 16$ $t h e s h a d e d a r e a = 16 - 3 π$
	Commented by Subhi last updated on 04/Jun/23

	$Y o u c a n a l s o u s e t h e s i m i l a r i t y$ $Δ D C O i s s i m i l a r t o Δ B C A$ $\frac{D C}{C E + 2 \sqrt{6}} = \frac{D O}{B A} = \frac{\sqrt{6}}{2 \sqrt{6}} = \frac{1}{2}$ $2 D C = C E + 2 \sqrt{6}$
	Commented by Subhi last updated on 04/Jun/23

	Commented by Mingma last updated on 04/Jun/23
	Perfect ��
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