Question

A pole has to be erected at a point on the boundary of a circular park of diameter 17 meter in such a way that the differences of its distances from two diametrically opposite fixed gates where the pole is to be erected.​

Answer 1

Answer:

Solution;-

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

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.

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

Answer.

The answer can be proved also.

AB² = AP² + BP²

17² = 15² + 8²

289 = 225 + 64

289 = 289

L.H.S = R.H.S  

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