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## anonymous 4 years ago Express the ratio of the area of the larger circle to the area of the smaller circle in simplest radical form. Larger circle: radius of 2 + √3 and smaller circle's radius: 2-√3.

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1. anonymous

|dw:1328379356570:dw|

2. dumbcow

is this a repost? what was unclear on explainations in prev post?

3. anonymous

Why is it 7 + 4√3?

4. anonymous

I tried pi (2)^2 + √3^2

5. dumbcow

oh you forgot the middle term (2+sqrt3)^2 = 4 + 2sqrt3 +2sqrt3 + 3

6. dumbcow

(a+b)^2 does Not equal (a^2 +b^2)

7. anonymous

middle term? what? D:

8. dumbcow

FOIL ? (2+sqrt3)(2+sqrt3) = 4+4sqrt3 +3

9. anonymous

ohhh.

10. anonymous

thank you!

11. dumbcow

:)

12. anonymous

Okay so I have now: $\pi \times 7 + 4\sqrt{3} \div \pi \times 7-4\sqrt{3} \times \pi \times 7+4\sqrt{3} \div \pi \times 7+4\sqrt{3}$

13. dumbcow

yes except the pi cancels out...pi/pi = 1 so you can get rid of it

14. anonymous

ohh..

15. anonymous

Is that the same for 7/7?

16. dumbcow

no because its not a multiplicative constant...or its part of the addition so it can't cancel it helps to put parenthesis around them, (7+4sqrt3) is like 1 term so only if (7+4sqrt3) is on bottom can it cancel sorry if that was confusing

17. anonymous

Okay so: = $(7 + 4\sqrt{3})(7 + 4\sqrt{3}) \div (7 - 4\sqrt{3})(7 + 4\sqrt{3})$ = $49 + 28\sqrt{3} + 28\sqrt{3} + 16\sqrt{9} \div 49 + 28\sqrt{3} - 28\sqrt{3} - 16\sqrt{9}$ = $121\sqrt{15} \div$

18. anonymous

$33\sqrt{15}$

19. anonymous

The answer is supposed to be: 97 + 56√3

20. dumbcow

correct until last line...where did 121 come from?

21. dumbcow

also the numbers inside the radicals doNOt change when adding

22. dumbcow

oh i see what you did ... 2sqrt3 + 3sqrt2 does NOT equal 5sqrt5 keep the terms with radicals separate you can only combine sqrt2 with other sqrt2 's

23. dumbcow

Top: 16sqrt9 = 16*3 = 48 28sqrt3 + 28sqrt3 = 56sqrt3 49 + 48 = 97 --> 97 + 56sqrt3 Bottom: 49 - 48 = 1 28sqrt3 - 28sqrt3 = 0 --> 1

24. anonymous

ohhhh. Thank you!

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