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    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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    (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!

    Name:  ImageUploadedByStudent Room1364766743.503828.jpg
Views: 81
Size:  123.4 KB


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    Work out the area of the quadrilateral - area of the sector
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    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.
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    (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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    (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!

    Name:  ImageUploadedByStudent Room1364766743.503828.jpg
Views: 81
Size:  123.4 KB


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    A=A_{OBCA}-A_{OAB}
    A =area
    The area of OBCA quadrilateral is sum of area of two triangles (OBC+ABC)
    A_{OBC \Delta} =3.6^2\cdot \sin \frac{2\pi}{3}
    Assuming that AB and AC is tangent to the circle so OBA and OCA angle
    is right angle. From this follows BOA angle is \frac{\pi}{3}
    So from the OBA right angled triangle you can get AB.
    The BAC angle is \pi-\frac{2\pi}{3}=\frac{\pi}{3}
    So You can get the area of ABC triangle.
    Generally the area of a sector is
    a=\frac{1}{2}r^2\cdot \alpha
    where r is the radius and \alpha is the angle in radians.
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    (Original post by ztibor)
    A=A_{OBCA}-A_{OAB}
    A =area
    The area of OBCA rectangle is sum of area of two triangles (OBC+ABC)
    A_{OBC \Delta} =3.6^2\cdot \sin \frac{2\pi}{3}
    Assuming that AB and AC is tangent to the circle so OBA and OCA angle
    is right angle. From this follows BOA angle is \frac{\pi}{3}
    So from the OBA right angled triangle you can get AB.
    The BAC angle is \pi-\frac{2\pi}{3}=\frac{\pi}{3}
    So You can get the area of ABC triangle.
    Generally the area of a sector is
    a=\frac{1}{2}r^2\cdot \alpha
    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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    (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.
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    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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    (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.
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    (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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    (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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