PQ and RS are two parallel chords of a circle of length 14 cm and 48 cm, respectively, and lie on the same side of the centre O. If the distance between the chords is 17 cm, what is the radius (in cm) of the circle?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 25
PQ = 14 cm
RS = 48 cm
O is the center.
The perpendicular from the center of a circle to a chord bisects the chord.
Pythagoras theorem -
Hypotenuse2 = Perpendicular2 + Base2
Let OX ⊥ RS and OY ⊥ PQ.
Let OX be d.
XY = 17
OY = (d + 17)
Applying Pythagoras' theorem in ΔOXS -
OS2 = OX2 + XS2
OS2 = d2 + 242
In ΔOYQ -
OQ2 = OY2 + YQ2
OQ2 = (d + 17)2 + 72
d2 + 242 = (d + 17)2 + 72 [OS = OQ = radius of the circle]
d2 + 576 = d2 + 172 + 2 × d × 17 + 49
576 = 289 + 34d + 49
34d = 576 - 289 - 49
34d = 238
d = 238/34
d = 7
OS2 = 72 + 242
OS2 = 49 + 576
OS2 = 625
OS = 25
Radius of the circle = 25 cm