+400 Chords $\overline{PR}$ and $\overline{QS}$ of a circle are perpendicular. Find $RS$.
PQ = 3, QR = 5, PS = 7
+129829 Note that triangle QVR is similar to triangle PVS
And QR /PS = 5/7
Therefore QV / PV = (5/7)
QV = (5/7)PV
PQ^2 = PV^2 + QV^2
PQ^2 = PV^2 +[ (5/7)PV]^2
3^2 = PV^2 + (25/49)PV^2
9 = [74/49]PV^2
[441] / 74 = PV^2
21 / sqrt [74 ] = PV
QV = (5/7)* 21 /[sqrt [74]
QV = 15 / sqrt [74]
PS^2 = PV^2 + VS^2 QR^2 = QV^2 + VR^2
7^2 = [ 441/74] + VS^2 5^2 = [225/74] + VR^2
49 - [441/74 ] =VS^2 25 - [225/74] = VR^2
3185/74 = VS^2 1625/74 = VR^2
RS =sqrt [ VS^2 + VR^2]
RS = sqrt [ [ 3185 + 1625 ] / 74 ]
RS = sqrt [ 4810 / 74 ]
RS = sqrt [ 65 ]
