Thread: [SOLVED] Equation of an arc/pythagorus

1. [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)?

2. 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.

3. 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.

4. 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.

$A = \left(\frac{\theta}{2\pi}\right)\; \pi r^2$ The fraction is the ratio of the angle of the sector to that of the entire circle.

5. 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

6. ok I pretty much get it now ie 90%
Ill look at it latter with fresher eyes to make sure.

Thank you.