need help from a math wiz!!!

[reefburnaby]

Hello,

Here we go....

Were R is the radius, AC is perpendicular to BO. We are trying to find the area in semicircle ABCD.
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First, find the area of the pie ABCO.

R = AO = BO = CO = 1.25"
BD = 0.5"

DO = BO - BD = 1.25" - 0.5" = 0.75"

Angle DOC = inv cos (DO / CO)
Angle DOC = 0.9273 rad or 53.13 degrees.

Since AD = DC, then angle DOC = angle DOA.

Therefore Angle AOC = 2 * angle DOC = 106.26 deg.

A circle has 360 degrees, so the pie ABCO is 106.26 / 360 = 29.52% of the circle.

Area of ABCO pie = pi * R * R * pie's percentage of the circle.
Area of ABCO pie = 1.4489 square inches.

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Next, find area of triangle ADCO.

DC = AD = sqrt (CO * CO - DO * DO)
DC = AD = 1"

Area of ADCO = AC * DO / 2 = 0.75 square inches.

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Area of the semicircle.

Semicircle ABCD = Pie ABCO - Triangle ADCO
Semicircle ABCD = 0.69889 square inches.

- Victor.

[ 23 October 2001: Message edited by: reefburnaby ]