# Circles

• Aug 12th 2012, 05:12 AM
Arulmozhi
Circles
Could someone help solve this problem, please. Thanks.
• Aug 12th 2012, 05:30 AM
HallsofIvy
Re: Circles
Can you find the area of each quarter of the square? Can you find the area of each quarter of a circle? Do you see how to get the black area from that?
• Aug 12th 2012, 05:39 AM
Plato
Re: Circles
Quote:

Originally Posted by Arulmozhi
Could someone help solve this problem, please.

The question is "can you solve this problem?" After all it is your problem.
What do you know about the problem? What have you even tried?
• Aug 12th 2012, 09:28 AM
Deveno
Re: Circles
first of all, what you want to do is look at each 1/4 of "the big square" (the one with sides of length 12). each of these "quarter-squares" has sides of length 6.

contribution form lower-left 1/4 = ?
contribution from lower-right 1/4 = ?
contribution from upper-left 1/4 = ?
contribution from upper-right 1/4 = ?

2 of these questions should be "easy" to answer. how can you use symmetry to reduce the computation involved in the last 2?

at some point you will have to subtract the area of a "quarter-circle" from one of the "quarter-squares". do you know how to compute the area of a circle, if you know its radius?

there are a lot of lines and curves in that picture. not all of them are relevant to the solution. what *is* relevant is that you are told some of the curves are "semi-circles" (1/2 a circle), and that they are all the same size.
• Aug 12th 2012, 06:33 PM
bjhopper
Re: Circles
Finding shaded area in the lower left square is the difficult part of problem.Label square from top left to lower left clockwise ABCD, Intersection of circles centered at A and B meet at P. P is on the perpendicular bisector of AB and DC. DPC is not shaded but is equal to the shaded. Draw a line parallel to DC thru P meeting AD at R and BC at S
Circle centered at A (0,0) X^2 + y^2 = 36 X =3 y = 3 rad3 M is midpoint of AB and N midpoint of DC. MP = 3rad3 NP = 6-3rad3.
Area of DPC = 6*(6-3rad3) - segment of 6cm circle whose height is 6-3rad3, chord 6 sector angle 60 deg
• Aug 12th 2012, 08:35 PM
Deveno
Re: Circles
Quote:

Originally Posted by bjhopper
Finding shaded area in the lower left square is the difficult part of problem.Label square from top left to lower left clockwise ABCD, Intersection of circles centered at A and B meet at P. P is on the perpendicular bisector of AB and DC. DPC is not shaded but is equal to the shaded. Draw a line parallel to DC thru P meeting AD at R and BC at S
Circle centered at A (0,0) X^2 + y^2 = 36 X =3 y = 3 rad3 M is midpoint of AB and N midpoint of DC. MP = 3rad3 NP = 6-3rad3.
Area of DPC = 6*(6-3rad3) - segment of 6cm circle whose height is 6-3rad3, chord 6 sector angle 60 deg

there is a MUCH easier way