# Thread: area of a circle in a square

1. so its 707 to 0dp

2. You're overdoing things!

You need to practice with factors,

Be very accustomed to those.
Then you have Pythagoras' theorem to master.

Firstly, zero multiplied by anything is zero...

3(0)=0+0+0

5(0)=0+0+0+0+0

and so on...

$\displaystyle (R-3)(R-15)=0$

means $\displaystyle R-3=0$ or $\displaystyle R-15=0$

So, from that, what possible values can we have for the radius ?

3. well 3-3=0 and 15-15=0 so 3 and 15?

4. Originally Posted by kitobeirens
well 3-3=0 and 15-15=0 so 3 and 15?
Yes, now does a radius of 3 make any sense, given that a side of the little rectangle is 6 ?

5. no so is it 15

6. Originally Posted by kitobeirens
no so is it 15
Yes, so plugging that R into the circle area equation gives the circle area

I'd recommend you practice from the beginning with this problem.
It would be of benefit to you.

7. so the working oout i did before was right 225-270+45=0 which has r as 15

8. so the area is 707

9. Yes, in units of square centimetres, to the nearest square centimetre.

10. how would you mathematically worked out r from r-15=0 and r-3=0 because it probably isn't great to guess that 3 is too small in other cases this might not be possible

11. Originally Posted by kitobeirens
how would you mathematically worked out r from r-15=0 and r-3=0 because it probably isn't great to guess that 3 is too small in other cases this might not be possible
Given the geometry, R has to be greater than 6.
So there is a lower bound for R as the problem was posed.
Hence, this rules out the possiblity of R being 3.

12. ok thanks for helping me

13. If the 2 givens (3 and 6 in your case) are represented by u and v, and d = diameter of circle, then:

d = [-b + SQRT(b^2 - 4c)] / 2, where b = -4(u + v) and c = 4(u^2 + v^2)

Stick 3 and 6 in there and you'll get d = 30 (hence radius = 15).

Try u=88, v=99: d = 638

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