on number 13 Im confused. I thought a-b would be pythagorus $\displaystyle a^2+b^2=c^2$ so sqrt (2r^2) .... It gives sqrt(2r)?

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- Mar 29th 2009, 11:24 AMbrentwoodbc[SOLVED] Equation of an arc/pythagorus
on number 13 Im confused. I thought a-b would be pythagorus $\displaystyle a^2+b^2=c^2$ so sqrt (2r^2) .... It gives sqrt(2r)?

http://i166.photobucket.com/albums/u...g?t=1238354412 - Mar 29th 2009, 11:28 AMstapel
I'm assuming you're referring to exercise 13(a)i, the straight-line distance between A and B.

Since the triangle (inside the quarter-circle) is isosceles right, with equal-length legs "r", the hypotenuse is, as you suggest, given by:

. . . . .$\displaystyle \sqrt{r^2\, +\, r^2}\, =\, \sqrt{2r^2}\, =\, \sqrt{2}\,r$

I would guess that $\displaystyle \sqrt{2r}$ is a typo. (Wink) - Mar 29th 2009, 11:42 AMbrentwoodbc
thanks, i just have one more question

15 a/b

for a

it says the answer is a = r^2xθ all over 2

I know a area of a triangle is BxH all over 2

and a circle is $\displaystyle A = Pi r^2$ but I dont understand how to get the answer.

http://i166.photobucket.com/albums/u...g?t=1238355434 - Mar 29th 2009, 11:46 AMJester
- Mar 29th 2009, 11:50 AMIluvMath
the ratio of the angle theta with respect to one full revolution is the same as the sector area with respect to the area of a circle. Put mathematically:

theta/2pi = SectorArea/ pi*r^2

:) - Mar 29th 2009, 12:09 PMbrentwoodbc
ok I pretty much get it now ie 90%

Ill look at it latter with fresher eyes to make sure.

Thank you.