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)?
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.