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# Where am I going wrong? Watch

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1. I have been stuck on this question for hours...
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The way I was going to work this out was to find the area of the triangle in figure 1 and then double it and take away the area 'S' seen in figure 4. So,

For the area of the sector, from figure two you can see that the angle is .
So

In order to work out the triangle in figure one, I worked out 'G' from figure 2.
The hypotenuse 'OP' has already been worked out from a previous part of the question where the X-Coord of P:

Area of the triangle in figure one is g/2. Multiplied by 2 gives the whole ares including the sector (figure 2).

Which is not the right answer.
2. (Original post by KloppOClock)
...
Sorry, I can't see all the attachments, so I don't know what your working is refering to.

However.

Let C be the centre of the circle. And drop a perpendicular from C to OP meeting it at M, say.

Then, agree area of sector CQM = pi/3

Area of quadrilateral CQPM is twice area of triangle CPM.

CPM is a 30,60,90 triangle and since CM=1, it follows MP = root(3)

Area CPM is thus root(3) / 2, and area of quadrilateral is root(3).

Hence....
3. (Original post by ghostwalker)
Sorry, I can't see all the attachments, so I don't know what your working is refering to.

However.

Let C be the centre of the circle. And drop a perpendicular from C to OP meeting it at M, say.

Then, agree area of sector CQM = pi/3

Area of quadrilateral CQPM is twice area of triangle CPM.

CPM is a 30,60,90 triangle and since CM=1, it follows MP = root(3)

Area CPM is thus root(3) / 2, and area of quadrilateral is root(3).

Hence....
4. (Original post by KloppOClock)
...
See my previous post for a working.

You've used diagram 3, when working out G, using OP as the hypoteneuse.

However, diagram 3 is incorrect. The perpendicular to RP at Q, does not go through O. It would only go through O if the angle theta was 45 degrees. There's your problem.
5. (Original post by ghostwalker)
See my previous post for a working.

You've used diagram 3, when working out G, using OP as the hypoteneuse.

However, diagram 3 is incorrect. The perpendicular to RP at Q, does not go through O. It would only go through O if the angle theta was 45 degrees. There's your problem.
Thank you! I was wondering what I was doing wrong for ages!

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Updated: October 10, 2016
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