Considering the three circles are tangent. Find the area of the shaded region in terms of π or to the nearest 0.1 cm2.
(Source: Discovering Geometry Workbook by Michael Serra, 2008)
Hit Counts
- 41,975 visits since Sept. 7, 2011
-
Problems
- Problem#L1
- Problem#L2
- Problem#L3
- Problem#L4
- Problem#L5
- Problem#L6
- Problem#L7
- Problem#L8
- Problem#K1
- Problem#K2
- Problem#K3
- Problem#K4
- Problem#K5
- Problem#K6
- Problem#K7
- Problem#K8
- Problem#J1
- Problem#J2
- Problem#J3
- Problem#J4
- Problem#J5
- Problem#J6
- Problem#J7
- Problem#J8
- Problem#I1
- Problem#I2
- Problem#I3
- Problem#I4
- Problem#I5
- Problem#I6
- Problem#I7
- Problem#I8
- Problem#H1
- Problem#H2
- Problem#H3
- Problem#H4
- Problem#H5
- Problem#H6
- Problem#H7
- Problem#H8
- Problem#G1
- Problem#G2
- Problem#G3
- Problem#G4
- Problem#G5
- Problem#G6
- Problem#G7
- Problem#G8
- Problem#F1
- Problem#F2
- Problem#F3
- Problem#F4
- Problem#F5
- Problem#F6
- Problem#F7
- Problem#F8
- Problem#E1
- Problem#E2
- Problem#E3
- Problem#E4
- Problem#E5
- Problem#E6
- Problem#E7
- Problem#E8
- Problem#D1
- Problem#D2
- Problem#D3
- Problem#D4
- Problem#D5
- Problem#D6
- Problem#D7
- Problem#D8
- Problem#C1
- Problem#C2
- Problem#C3
- Problem#C4
- Problem#C5
- Problem#C6
- Problem#C7
- Problem#C8
- Problem#B1
- Problem#B2
- Problem#B3
- Problem#B4
- Problem#B5
- Problem#B6
- Problem#B7
- Problem#B8
- Problem#A1
- Problem#A2
- Problem#A3
- Problem#A4
- Problem#A5
- Problem#A6
- Problem#A7
- Problem#A8
Calendar
EDSC 353: Teaching Secondary Geometry 
The way I would solve this problem is by taking the area of the big circle, subtract the combined area of the two smaller circles, and then divide the answer by 2. Because the radius of the smaller circle is 5, the radius of the big circle is 10. So, the area of the big circle is pi(r)^2 = 100pi.
The are of the little circle is pi(r)^2 = 25pi. Double that and you have 50pi.
100pi – 50pi = 50pi. Divide 50pi in half and you get 25pi. That’s my answer!
Well done!
I just have a few comments. You stated that you would divide your answer by 2, but you didn’t state the reason for doing so. I completely agree with you, and would have solved it in pretty much the same manner. However, I looked at the problem in terms of semicircles. Thus, the area of the shaded region would be (1/2) the area of the big circle minus 2* (1/2) the area of the small circle (or rather, the area of the small circle).
So, the shaded region is (100/2 – 25)π = 50 – 25π = 25π (thus obtaining the same exact value)