SOLUTION: I tried to answer this question and the correct answer according to the answer book was: 8 times r^2 times pi meters squared. Here's the problem: "In the figure shown, if the small

Algebra ->  Surface-area -> SOLUTION: I tried to answer this question and the correct answer according to the answer book was: 8 times r^2 times pi meters squared. Here's the problem: "In the figure shown, if the small      Log On


   

Question 1083421: I tried to answer this question and the correct answer according to the answer book was: 8 times r^2 times pi meters squared. Here's the problem: "In the figure shown, if the small circle inside the larger one has a radius of "r" meters and the larger circle has a diameter of "6r" meters, what is the area of the region inside the large circle and outside the small circle?" My process and answer: The area of the small circle is r^2 times pi meters squared. The area of the large circle is 3r^2 times pi meters squared. Subtract the small circle from the large circle to get 2r^2 times pi meters squared. Where did I go wrong?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's refer to the two areas as A1 and A2, where
A1 = area of small circle
A2 = area of large circle

area of small circle:
A1 = pi*(small circle radius)^2
A1 = pi*(r)^2
A1 = pi*r^2

Area of large circle
A2 = pi*(large circle radius)^2
A2 = pi*(3r)^2 ... ** see note below **
A2 = pi*(3r)*(3r)
A2 = pi*(9r^2)
A2 = 9pi*r^2

note: radius = diameter/2 = 6r/2 = 3r

Subtract the areas to get the difference D
D = A2 - A1
D = 9pi*r^2 - pi*r^2
D = (9pi - pi)*r^2
D = 8pi*r^2
which is the area of the region outside of the small circle but inside the larger circle.

8*pi*r^2 is the same as 8*r^2*pi. The order of which you multiply the terms doesn't matter.