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Square ABCD lies inside circle O. Square ABCD has been given side lengths to be 2 by 2. Square ABCD is NOT shaded. The remaining space in circle O is shaded. Find area of the shaded region.
Let me see.
Let A_r = area of shaded region.
A_r = area of circle O - area of Square ABCD.
Correct?
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Yes,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yes,
Bob
This is correct but the other posts are not. Am I right? I am doing my best to describe the textbook pictures.
Bob, I have an idea. My questions come from Michael Sullivan's College Algebra Edition 9 book, which I can email to you or you can freely download yourself. It is a better idea for you to have the same book I have here. I can then post the page number for each set of questions.
You say?
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Yes,
Bob
Problem 40; page 37.
Let me see.
Let A_r = area of shaded region.
A_r = area of circle O - area of Square ABCD.
From Problem 39, I know that r = sqrt{2}.
A_r = pi(sqrt{2})^2 - 2^2
A_r = 2pi - 4
You say?
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Yes correct. On the way to this answer you have done Q39 correctly.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yes correct. On the way to this answer you have done Q39 correctly.
Bob
Thanks but I did 39 over because you said I got it wrong. Is my original work for 39 correct?
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