Q31 of 168 Page 7

A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Let P be the position of the pole and A and B be the opposite fixed gates.



PA = ‘a’ mts


PB = ‘b’ mts


PA – PB = 7 m


⇒ a – b = 7


⇒ a = 7 + b .........(1)


In Δ PAB,


AB2 = AP2 + BP2


⇒ (17) = (a)2 + (b)2


⇒ a2 + b2 = 289


⇒ Putting the value of a = 7 + b in the above,


(7 + b)2 + b2 = 289


⇒ 49 + 14b + 2b2 = 289


⇒ 2b2 + 14b + 49 – 289 = 0


⇒ 2b2 + 14b – 240 = 0


Dividing the above by 2, we get.


⇒ b2 + 7b – 120 = 0


⇒ b2 + 15b – 8b – 120 = 0


⇒ b(b + 15) – 8(b + 15) = 0


⇒ (b – 8) (b + 15) = 0


⇒ b = 8 or b = –15


Since this value cannot be negative, so b = 8 is the correct value.


Yes it is possible to erect a pole.


Putting b = 8 in (1), we get.


a = 7 + 8


a = 15 m


Hence PA = 15 m and PB = 8 m


So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m.


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