Posted by hii on Sunday, September 16, 2007 at 5:43pm.
draw a perpendicular from O to AB. Let x be the intersection. You know xb, and bo lengths, so find the sine of obx, thence the measure of obx. Double it to get boa.
The rest come from that.
Iwill be happy to critique your thinking.
This splits the triangle into two right triangles, each with
hypotenuse 2, and adjacent side = 1.5
cos = adjacent / hypotenuse = (1.5) / 2 = 0.75
= arccos (0.75) 0.723 radians
Angle(AOB) = - 2 - 1.446 1.696 radians.
Area of one of the small right triangles is (1/2)bh
Need to find h (length of segment from 0 to midpt AB)
sin = opposite / hypotenuse
opposite = (hypotenuse) * sin = 2 * sin (0.723 radians)
1.323 cm
So area of one of the small right triangles
(1/2) (1.5) (1.323) 0.992 cm^2
Area of triangle AOB is two times this amount or 1.984 cm^2
for v) area oc circle , pie r ^2 = pie 2^2 = 12.56
so 100 / 12.56 x 1.984 = 15.796 which is 15%
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