a pole has to be erected at a point on the boundary of a circular park of diameter 13m 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 two gates should the pole be erected?
Answers
Since A and B are the end points of the diameter, they subtend a right-angled triangle at the point P, where pole is erected. Since, the hypotenuse is equal to 13m, the other two sides of right triangle need to be 12m and 5m. And we also observer that the difference between the two sides is 7m. Pole is erected at a distance of 5m and 12m form the gates.
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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