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Circular measures question

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Would anyone please explain the answer for me?
Thank you in advance. :heart:
Original post by maria.x
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Would anyone please explain the answer for me?
Thank you in advance. :heart:


One thing you need to do is to use the triangle to find the length of AB. Half of this length is the radius of the semicircle AYB.
Reply 2
Original post by tiny hobbit
One thing you need to do is to use the triangle to find the length of AB. Half of this length is the radius of the semicircle AYB.


Could you please write down the full answer, and I got the radius part. It's the steps after it that's confusing me.
Area of sector AOB = 1/2*r^2*2theta = r^2*theta
Area of triangle AOB = 1/2*r*r*sin(2theta)=1/2r^2*2sin(theta)*cos(theta) by the double angle formula
= r^2*sin(theta)*cos(theta)
-> Area of segment = r^2(theta-sin(theta)cos(theta)).
By the Cosine Rule, AB^2 = r^2+r^2-2(r)(r)cos(2theta) = 2r^2(1-cos(2theta))=2r^2(1-(1-2sin^2(theta)) (by the double angle formula) =4r^2sin^2(theta)
-> AB=2rsin(theta).
Thus radius of semicircle = rsin(theta), so its area = 1/2*pi*(rsin(theta))^2 = 1/2*pi*r^2sin^2(theta).
-> Shaded area = Area of semicircle - Area of segment = 1/2*pi*r^2*sin^2(theta) - r^2(theta-sin(theta)cos(theta))
= 1/2*r^2*(pi*sin^2(theta)-theta+sin(theta)cos(theta)).
Reply 4
Original post by maria.x
Could you please write down the full answer, and I got the radius part. It's the steps after it that's confusing me.


triangle + semicircle - sector
Reply 5
Original post by constellarknight
Area of sector AOB = 1/2*r^2*2theta = r^2*theta
Area of triangle AOB = 1/2*r*r*sin(2theta)=1/2r^2*2sin(theta)*cos(theta) by the double angle formula
= r^2*sin(theta)*cos(theta)
-> Area of segment = r^2(theta-sin(theta)cos(theta)).
By the Cosine Rule, AB^2 = r^2+r^2-2(r)(r)cos(2theta) = 2r^2(1-cos(2theta))=2r^2(1-(1-2sin^2(theta)) (by the double angle formula) =4r^2sin^2(theta)
-> AB=2rsin(theta).
Thus radius of semicircle = rsin(theta), so its area = 1/2*pi*(rsin(theta))^2 = 1/2*pi*r^2sin^2(theta).
-> Shaded area = Area of semicircle - Area of segment = 1/2*pi*r^2*sin^2(theta) - r^2(theta-sin(theta)cos(theta))
= 1/2*r^2*(pi*sin^2(theta)-theta+sin(theta)cos(theta)).


Original post by TeeEm
triangle + semicircle - sector


Thank you guys so much. :h:

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