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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Answer 1 · 2016-08-07T12:21:28

$bar(CD) = 3.65273$

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 $S = 2.5594$