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#

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#
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Answer 2

To find the length of CD, we use the Pythagorean theorem because CD is the hypotenuse of right triangle COD.

Using the Pythagorean theorem: (CD^2 = OD^2 - OC^2)

Substituting the given values: (CD^2 = (3 , \text{cm})^2 - (2.38 , \text{cm})^2)

Solve for CD: (CD^2 = 9 , \text{cm}^2 - 5.6644 , \text{cm}^2) (CD^2 = 3.3356 , \text{cm}^2) (CD \approx \sqrt{3.3356} , \text{cm}) (CD \approx 1.8276 , \text{cm})

Now, to calculate the shaded area, we first find the areas of the two circles and then subtract the area of the smaller circle from the area of the larger circle.

The area of a circle is given by the formula: (A = \pi r^2)

For the larger circle: (A_{\text{large}} = \pi (OD/2)^2) (A_{\text{large}} = \pi (3/2)^2) (A_{\text{large}} = \pi \cdot 2.25 , \text{cm}^2)

For the smaller circle: (A_{\text{small}} = \pi (FC/2)^2) (A_{\text{small}} = \pi (2.38/2)^2) (A_{\text{small}} = \pi \cdot 2.2409 , \text{cm}^2)

Now, subtract the area of the smaller circle from the area of the larger circle to find the shaded area: (A_{\text{shaded}} = A_{\text{large}} - A_{\text{small}}) (A_{\text{shaded}} = 2.25\pi - 2.2409\pi) (A_{\text{shaded}} = 0.0091\pi , \text{cm}^2) (A_{\text{shaded}} \approx 0.0286 , \text{cm}^2)

Therefore, the length of CD is approximately 1.83 cm and the shaded area is approximately 0.0286 square centimeters.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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