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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 distance from two diametrically opposite fixed gates $A$ and $B$ on the boundary is $7$ meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

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\frac{96}{7} m, 6.71 m

\frac{90}{7} m, 0.71 m

\frac{96}{5} m, 1.51 m

\frac{86}{7} m, 5.51 m

answer is A.

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Detailed Solution

Given that,
Diameter of the park = 13m
Distance of the pole from the two gates = 7m
Question Image

We have to find the distance of the pole from the two gates.
Taking

P

as point at which pole erected
Diameter

= AB = 13 m

Radius

= r = \frac{13}{2} m

AP = l_{1}

BP = l_{2}

l_{1} = rθ

l_{2} = r (π - θ)

l_{1} - l_{2} = 7

\Rightarrow r (2 θ - π) = 7

\Rightarrow 2 θ = \frac{14}{13} + \frac{22}{7}

\Rightarrow θ = \frac{192}{91}

π - θ = \frac{22}{7} - \frac{192}{91}

\Rightarrow

π - θ = \frac{94}{91}

The given condition will be possible.
The distance of pole from A will be,

d = rθ = \frac{96}{7} m

The distance of pole from B,

D = r (π - θ)

\Rightarrow D = \frac{47}{7} m

\Rightarrow D = 6.71 m

Thus, the distance from and B will be

\frac{96}{7} m

and 6.71m.
Therefore the correct option is 1.

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