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Question
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 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?
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Solution
Let P be the required location on the boundary of a circular park such that its distance from gate B is x metre that is BP x metres.
Then, AP = x + 7
In the right triangle ABP we have by using Pythagoras theorem
AP2 + BP2 = AB2
(x + 7)2 + x2 = (13)2
x2 + 14x + 49 + x2 = 169
2x2 + 14x + 49 - 169 = 0
2x2 + 14x - 120 = 0
2(x2 + 7x - 60) = 0
x2 + 7x - 60 = 0
x2 + 12x - 5x - 60 = 0
x(x + 12) - 5(x - 12) = 0
(x + 12)(x - 5) = 0
x + 12 = 0
x = -12
Or
x - 5 = 0
x = 5
But the side of right triangle can never be negative
Therefore, x = 5
Hence, P is at a distance of 5 metres from the gate B.
⇒ BP = 5m
Now, AP = (BP + 7)m = (5 + 7)m = 12 m
∴ The pole has to be erected at a distance 5 mtrs from the gate B and 12 m from the gate A.
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