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
Cesareo R.
Aug 7, 2016

bar(CD) = 3.65273¯¯¯¯¯¯CD=3.65273

Explanation:

Using Thales of Mileto's famous theorem

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

bar(AF) = (bar(AO)cdotbar(FC))/(bar(AD)-bar(AO)) = (3 xx 2.38)/(6-3)=2.38¯¯¯¯¯¯AF=¯¯¯¯¯¯AO⋅¯¯¯¯¯¯FC¯¯¯¯¯¯AD−¯¯¯¯¯¯AO=3×2.386−3=2.38

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

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

Now using Pitagoras

bar(FO)^2+bar(AF)^2= bar(AO)^2¯¯¯¯¯¯FO2+¯¯¯¯¯¯AF2=¯¯¯¯¯¯AO2 because angle(AFO) = pi/2∠(AFO)=π2

giving

bar(FO) = sqrt(3^2-2.38^2)¯¯¯¯¯¯FO=√32−2.382 so

bar(CD) = 2sqrt(3^2-2.38^2) = 3.65273¯¯¯¯¯¯CD=2√32−2.382=3.65273

The shaded area SS 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

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