## diamondboy one year ago quick simple question

1. diamondboy

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2. diamondboy

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3. diamondboy

4. diamondboy

5. diamondboy

6. Nnesha

tag jim :P :D first)gtg in few mints 2nd) not good at geometry :P

7. diamondboy

ok @jim_thompson5910 help

8. Nnesha

or ganeshie8 :D

9. diamondboy

@pooja195 @ganeshie8

10. diamondboy

help

11. pooja195

@Directrix @ganeshie8 @jim_thompson5910

12. pooja195

@zepdrix

13. Nnesha

well do you know how to find area of a circle ? i think you have to find area of a circle

14. diamondboy

yes I do but how do I go about that

15. diamondboy

Do I add the areas of the smaller circles and subtract it from that of the big circle

16. diamondboy

If thats d case what would be the radius of my big circle? 2?

17. Nnesha

radius of big woouldn't be 2 i guess :hmmm

18. Nnesha

*dies* l8r good luck cya

19. diamondboy

Ok

20. diamondboy

Thanks

21. zepdrix

|dw:1444014668003:dw|Is this what the problem looks like? :o the two circles are lined up vertically like this?

22. diamondboy

yes

23. diamondboy

byt the diameters, if extended, would be perpendicula to each other

24. zepdrix

|dw:1444014787277:dw|Ok cool :) Then I think you have the right idea. The larger radius is 2, and the smaller radii are each 1, ya?

25. diamondboy

ok

26. diamondboy

27. zepdrix

Hmm that sounds right! :)$\large\rm A=\pi (2)^2,\qquad\qquad a=\pi (1)^2$$\large\rm A-a-a=4\pi - 2\pi=2\pi\approx 6.28$

28. diamondboy

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29. diamondboy

thats how it actually looks like anyways

30. Jhannybean

$A_{big}-(A_{c_1}+A_{c_2})= A_{shaded}$

31. diamondboy

Ok thanks guys....I would have given you both a medal if I could