In the diagram below, the small circle is centered at E, the large one is centered at O. The segments FC and OD have lengths as of 2.38 and 3 cm respectively. Find the length of CD and calculate shaded area?

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1 Answer
Aug 7, 2016

#bar(CD) = 3.65273#

Explanation:

Using Thales of Mileto's famous theorem

#bar(AD)/(bar(AF)+bar(FC)) = bar(AO)/(bar(AF))# giving

#bar(AF) = (bar(AO)cdotbar(FC))/(bar(AD)-bar(AO)) = (3 xx 2.38)/(6-3)=2.38#

#bar(CD)/bar(AD) = bar(FO)/bar(AO)# so

#bar(CD)=2 bar(FO)# because #bar(AD) = 2bar(AO)#

Now using Pitagoras

#bar(FO)^2+bar(AF)^2= bar(AO)^2# because #angle(AFO) = pi/2#

giving

#bar(FO) = sqrt(3^2-2.38^2)# so

#bar(CD) = 2sqrt(3^2-2.38^2) = 3.65273#

The shaded area #S# is given by

#S = "sector"angle(FEO)+"triangle"Delta(AEF)#

#S = alphacdot bar(AE)^2+1/2bar(AF) cdot bar(AE) sin(alpha)#

where

#sin(alpha) = bar(CD)/bar(AD) = 3.65273/6 = 0.6088#

and

#alpha = arcsin( 0.6088) = 0.65453["rad"]#

giving #S = 2.5594#