Circles

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by sk818020 » Sun May 16, 2010 11:48 am
The radius of the circle is 9.

r = 9

Thus, the circumference is;

c = 2r(pi) = 18(pi)

Which means the circumference of the top half of this circle is;

c/2 = [18(pi)]/2 = 9(pi)

To figure out the length of arc PQ;

PQ = 9 - QR - PO, but,

QR = PO, so;

PQ= 9(pi) - 2(PO)

We can figure out PO by multiplying the proportion of the top half of the circle that PO takes up by the circumference of the top half the circle.

2*PO is (because we want to figure out two PO to subtract from 9(pi)), therefore;

2PO = 2(35/180)*9(pi) = (7/18) * 9(pi) = (21/6)pi = 3.5(pi)

We can now solve for PQ;

PQ = 9(pi) - 3.5(pi) = 4.5pi = 9(pi)/2

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by harshavardhanc » Sun May 16, 2010 1:06 pm
dkumar.83 wrote:Image
The answer will be 2 PI.

Solution :

as PQ|| OR, <RPQ will be 35 degrees.

Now the property to be used here:
angle subtended by an arc at the center is twice the angle subtended by the arc on the circumference.
therefore, minor arcs OP and QR will subtend 70 degrees each at the center. Hence, a total of 180 - 70 * 2 remain in the semi-circle, which will be subtended by minor arc PQ at the center.

Therefore, the measure of minor arc will be 40/360 * 2 PI * 9 = 2 PI .
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by dkumar.83 » Tue May 18, 2010 8:11 am
Thanks Harsh for the simple reply.