# [SOLVED] Equation of an arc/pythagorus

• Mar 29th 2009, 11:24 AM
brentwoodbc
[SOLVED] Equation of an arc/pythagorus
on number 13 Im confused. I thought a-b would be pythagorus $a^2+b^2=c^2$ so sqrt (2r^2) .... It gives sqrt(2r)?
http://i166.photobucket.com/albums/u...g?t=1238354412
• Mar 29th 2009, 11:28 AM
stapel
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:

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

I would guess that $\sqrt{2r}$ is a typo. (Wink)
• Mar 29th 2009, 11:42 AM
brentwoodbc
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 $A = Pi r^2$ but I dont understand how to get the answer.

http://i166.photobucket.com/albums/u...g?t=1238355434
• Mar 29th 2009, 11:46 AM
Jester
Quote:

Originally Posted by brentwoodbc
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 $A = Pi times r squared$ but I dont understand how to get the answer.

http://i166.photobucket.com/albums/u...g?t=1238355434

$A = \left(\frac{\theta}{2\pi}\right)\; \pi r^2$ The fraction is the ratio of the angle of the sector to that of the entire circle.
• Mar 29th 2009, 11:50 AM
IluvMath
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 PM
brentwoodbc
ok I pretty much get it now ie 90%
Ill look at it latter with fresher eyes to make sure.

Thank you.