Let ABCD is a brick in which AD = 5 cm and also AE = 15 cm If radius of the wheel = OD = OE = r cm From the figure, OF = r – 5 cm, FD = 5 cm Now, from the right Δ OFD, OD2–OF2 = FD2 ⇒ r2–(r–5)2 = 152 ⇒ 10r – 25 = 225 ⇒ 10r = 250 ⇒ r =