Question

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?

Answer 1

`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`