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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 - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

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?

Solution

Let be the required location on the boundary of a circular park such that its distance from gate 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, 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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Solution 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 Concept: Solutions of Quadratic Equations by Factorization.
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