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1. I don't know how to go about finding the area of the shaded region in this diagram :/ Any hints? It's a really dodgy quality picture, sorry! So the radius is 3.6cm and the angle is 2/3 pi. Thanks!

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2. (Original post by JodieW)
I don't know how to go about finding the area of the shaded region in this diagram :/ Any hints? It's a really dodgy quality picture, sorry! So the radius is 3.6cm and the angle is 2/3 pi. Thanks!

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Work out the area of the quadrilateral - area of the sector
3. I presume the lines AB and AC are tangents?

Draw a line AO and work out the area of triangle OBA. You should make use of a circle theorem to do this.

Now find the area of the quadrilateral ABOC.

Subtract the area of the sector.
4. (Original post by Mr M)
I presume the lines AB and AC are tangents?

Draw a line AO and work out the area of triangle OBA. You should make use of a circle theorem to do this.

Now find the area of the quadrilateral ABOC.

Subtract the area of the sector.
I don't know how to find the area of ABOC, that's the thing. I've worked out the area of the sector though

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5. (Original post by JodieW)
I don't know how to go about finding the area of the shaded region in this diagram :/ Any hints? It's a really dodgy quality picture, sorry! So the radius is 3.6cm and the angle is 2/3 pi. Thanks!

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A =area
The area of OBCA quadrilateral is sum of area of two triangles (OBC+ABC)

Assuming that AB and AC is tangent to the circle so OBA and OCA angle
is right angle. From this follows BOA angle is
So from the OBA right angled triangle you can get AB.
The BAC angle is
So You can get the area of ABC triangle.
Generally the area of a sector is

where r is the radius and \alpha is the angle in radians.
6. (Original post by ztibor)

A =area
The area of OBCA rectangle is sum of area of two triangles (OBC+ABC)

Assuming that AB and AC is tangent to the circle so OBA and OCA angle
is right angle. From this follows BOA angle is
So from the OBA right angled triangle you can get AB.
The BAC angle is
So You can get the area of ABC triangle.
Generally the area of a sector is

where r is the radius and \alpha is the angle in radians.
Isn't the area BCO 1/2 x 3.6^2 x sin2/3pi or have I got it wrong?

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7. (Original post by JodieW)
I don't know how to find the area of ABOC, that's the thing. I've worked out the area of the sector though

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You have all the angles and one side length of a right-angled triangle. Consequently you can easily find the other sides and the area.

8. Could someone please help me do this? AB and AC are tangents to the circle. The angle in the diagram is 2/3 pi and the radius is 3.6 cm, and I don't know how to find the area of the shaded area. I know what the area of the sector is, but how do I find the area of the entire quadrilateral? I know I should split it to two triangles but from there I'm lost. Thanks!

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9. (Original post by JodieW)
...
Don't spam the forum with multiple threads. You have been told how to answer the question. Finding the side lengths of a right-angled triangle by using trigonometry is a grade B skill at GCSE and finding the area of a right-angled triangle is a grade E skill. If you just think for a couple of minutes you will get the answer. If not, you need to revise SOHCAHTOA.
10. (Original post by Mr M)
Don't spam the forum with multiple threads. You have been told how to answer the question. Finding the side lengths of a right-angled triangle by using trigonometry is a grade B skill at GCSE and finding the area of a right-angled triangle is a grade E skill. If you just think for a couple of minutes you will get the answer. If not, you need to revise SOHCAHTOA.
Sorry, I was just really stressed but I've done it now, I got the right answer. Thanks people!

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11. (Original post by JodieW)
Isn't the area BCO 1/2 x 3.6^2 x sin2/3pi or have I got it wrong?

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You are right.

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