November 23, 2019, 06:52:25 am

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#### dean.arkcoll

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##### Minor Segment of a circle
« on: May 03, 2019, 06:33:07 am »
0
Hey,
Would someone be able to help me with this question

A chord 8mm long is formed by an angle of 45 degrees subtended at the centre of a circle.
Find
a) the radius of the circle
b) the area of the minor segment cut off by the angle correct to 1 decimal place.

Thanks

#### fun_jirachi

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##### Re: Minor Segment of a circle
« Reply #1 on: May 03, 2019, 07:23:44 am »
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Hey there!

If the radii subtend an angle of 45 degrees and form a chord with length 8mm, this essentially forms an isosceles triangle with angles 45 and 67.5 and 67.5 degrees. The side of 8mm is opposite the angle of 45 degrees, so you can use the sine rule to find the radius.

For the next part, remember that the area of a sector is given by
$\frac{1}{2}r^2\theta$
(note that theta is in radians, ie. sub in pi/4) and that the area of a triangle is given by
$\frac{1}{2}ab\sin C$
Since a and b are equal to the radii, and C is the enclosed angle, use the sector - triangle ie.
$\frac{1}{2}r^2 \times (\theta - \sin \theta)$ to find the area of the minor segment.

Hope this helps
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