help me solve part (b)

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- Dec 16th 2008, 03:14 AMhelloyinghelp on circles
help me solve part (b)

- Dec 16th 2008, 05:13 AMGrandadArea
Hello -

Area of APBC = 2 x Area of triangle OBC + area of sector OAPB, found in part (a)

Area of triangle OBC = 0.5 x base x height, where base = OC and height = perpendicular distance from B to OP.

Height = OB x sin (angle BOP), and angle BOP = 2 x angle BCP (angle at centre = twice angle at circumference.

I hope you can do it now.

Grandad - Dec 17th 2008, 04:34 PMhelloying
- Dec 17th 2008, 09:41 PMGrandadArea
Hi -

In triangle OBT we know OB = 6, angle BOT = $\displaystyle \frac{\pi}{3}$ and angle OBT = $\displaystyle 90^o$. So you can work out the length of BT and hence the area of the triangle.

Then subtract the area of the sector OAPB (found in b) from area of quadrilateral OABT.

Grandad