A.Maths (radian)

圖片參考：http://i263.photobucket.com/albums/ii157/mathmate/math/belt1.jpg
figure 1

Question:
1. Find the length of the straight part of the belt
2. Find the area enclosed by the belt.
Separate the area into four, separated by the radii perpendicular to the tangents, and the line joining the centres (OO'). Denote these areas, from left to right, by A1, A2, A3 and A4.
Let α be the angle between OO' and the 17 cm radius perpendicular to the upper tangent.
Let l be the length of each of the tangents
= sqrt(41^2-9^2)=40
1. length of straight part of the belt=40
2. Area enclosed by the belt, A
The area sought, A = A1 + A2 + A3 + A4
α = cos-1(9/41)=1.34948.. radians
A1=π*8^2*(2α/2π)=86.3668...
A2=A3=l*(8+17)/2=40*12.5=500
A4=π*17^2*((2π-2α)/2π)=517.9200..
Total enclosed by the belt, A
=A1+A2+A3+A4
=86.3668+500+500+517.9200
=1604.29

Part 2.

圖片參考：http://i65.photobucket.com/albums/h226/fung0828/0003.jpg
Figure 2

Angle ACB=90 degrees (angle at circumference subtended by diameter)
Angle CBA=60 degrees (equilateral triangle)
Angle BAC=30 degrees (sum of angles of triangle)
Angle DAC=BAE-BAC=60-30=30 degrees
Angle DOC=2*30=60 degrees (angle subtended by chord at centre)
Therefore triangle DOC is equilateral, with sides equal to 5 cm, as annotated.
Also, triangle EDC is also equilateral, with sides equal to 5 cm.

Area EDC (shaded)
=Area of triangle EDC - area of segment DC
=5^2*(sqrt(3)/2)/2-(π *5^2 /6 - 5^2*(sqrt(3)/2)/2)
=2*(5^2*(sqrt(3)/2)/2) -π *5^2 /6
=2*10.825317 - 13.089969
=8.56 cm^2