Let O be center and r be radius of the circle.
OP ⊥ AB and OQ ⊥ AB
∵ AB || CD so P, O, Q will lie on same line
PQ = 17 cm
AB = 10 cm
CD = 24 cm
∵ OP ⊥ AB and
OQ ⊥ CD
∴ AP = BP = \(\frac { 1 }{ 2 }\)AB
= \(\frac { 1 }{ 2 }\) × 10 = 5 cm
and CQ = QD = \(\frac { 1 }{ 2 }\) × CD
= \(\frac { 1 }{ 2 }\) × 24 = 12 cm
∵ PQ = 17 cm
Let OP = x cm then OQ = 17 – x cm
In right angled ∆OPA
by Pythagoras theorem
OA2 = OP2 + AP2
r2 = x2 + (5)2 = x2 + 25 …(i)
Similarly on right angled ∆OQC
OC2 = OQ2 + CQ2
r2 = (17 – x)2 + (12)2
= (17 – x)2 + 144 …(ii)
Equating value of r’ from equations (i) and (ii)
x2 + 25 = (17 – x)2 + 144
= 289 – 34x + x2 + 144
25 = 289 – 34x + 144
34x = 289 + 144 – 25
= 433 – 25
34x = 408
x = 12 cm
Putting value of x in equation (i)
r2 = x2 + 25 = (12)2 + 25
= 144 +25
r2 = 169
r = \(\sqrt { 169 }\) = 13 cm
Thus, radius of circle = 13 cm.