# help on circles

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• December 16th 2008, 03:14 AM
helloying
help on circles
help me solve part (b)
• December 16th 2008, 05:13 AM
Grandad
Area
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
• December 17th 2008, 04:34 PM
helloying
Quote:

Originally Posted by Grandad
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

Hello Grandad

Thanks I got it now. However i got stuck in part (c) . Can u teach me please?
• December 17th 2008, 09:41 PM
Grandad
Area
Hi -

In triangle OBT we know OB = 6, angle BOT = $\frac{\pi}{3}$ and angle OBT = $90^o$. 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