A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 M. Is it possible to do so? If yes, at what distances from the two gates should the poles be erected?
Answers
Let P be the pole to be erected and A, B be the opposite fixed gates.
Given, PA – PB = 7
Let PA = a, PB = b
Hence a – b = 7
⇒ a = 7 + b ...(1)
In right ΔPAB,
AB2 = AP2 + BP2 (By Pythagoras theorem)
132 = a2 + b2
169 = (7 + b)2 + b2
169 = 49 + 14b + 2b2
2b2 + 14b – 120 = 0
b2 + 7b – 60 = 0
b2 + 12b - 5b – 60 = 0
b(b + 12) - 5(b + 12) = 0
(b + 12)(b – 5) = 0
Hence b = 5 or – 12
But ‘b’ cannot be negative
Therefore, b = 5
⇒ a = 7 + 5 = 12
Hence, PA = 12 m and PB = 5 m
SOLUTION :
Given : AP - BP = 7 m, Diameter (AB) = 13 m
Let P be the location of the pole. Let x be the distance of the pole from the gate B i.e BP = x cm
Difference of the distance of the pole from the two gates = AP - BP = 7 cm
AP - x = 7
AP = (x + 7) m
In Δ PAB, ∠APB = 90°(angle in a semicircle is a right angle)
By Pythagoras theorem
AB² = AP² + BP²
13² = (x + 7)² + x²
169 = x² + 7² + 2×7 × x + x²
169 = 2x² + 49 + 14x
2x² + 14x + 49 - 169 = 0
2x² + 14x - 120 = 0
2(x² + 7x - 60) = 0
x² + 7x - 60 = 0
x² + 12x - 5x - 60 = 0
[By middle term splitting]
x(x + 12) - 5(x + 12) = 0
(x + 12) (x - 5) = 0
(x + 12) or (x - 5) = 0
x = - 12 or x = 5
Since, x cannot be negative because the distance between the pole cannot be negative, so x ≠ - 12
Therefore , x = 5
Distance of the pole from the gate B i.e BP = x cm = 5 cm
Distance of the pole from the gate A i.e AP = (x + 7) cm = (5 + 7) = 12 cm
Therefore , AP = 12 cm & BP = 5 cm.
Hence, the distance from the gate A to pole is 12 m and from gate B to the pole is 5 m
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