Perpendicular chords in circle

Let $M$ and $N$ be the projections of $O$ onto $AD$ and $BC$ , respectively. Then $M$ is the midpoint of $AD$ and $N$ is the midpoint of $BC$ . Then $BN = CN = 3$ , so $NF = 1$ . Since $OMFN$ is a rectangle, $OM = 1$ . Then by Pythagoras on right triangle $OMA$ , $AM = 2 \sqrt{6}$ . Also, $AM$ is half of $AD = r + s$ , so $r + s = 4 \sqrt{6}$ .

Construct the right angled triangle $AF'C'$ , such that $AF'=DF=r$ and $C'F'=CF=2$ , as shown.

Note that $\angle BAC'=90^{\circ}$ (as $\angle BAD=\angle BCD$ and $\angle DAC' =\angle ADC$ ). This means that $BC'$ is a diameter. Now $BC'^2=AC'^2+AB^2=(r^2+2^2)+(4^2+s^2)$ , which means that $r^2+s^2=80$ (as $BC'=2OA=10$ ). On the other hand, it is clear that $\triangle FDC$ is similar to $\triangle FBA$ , which means that $rs=4(2)=8$ . Thus $r+s= \sqrt{(r^2+s^2)+2rs}=\sqrt{80+16}=\sqrt{96}=9.79796$ and hence $\left \lceil{100(r+s)}\right \rceil =980$ .