Question

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


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Asker

@Abhyajha

Answer 1

hey mate
here's the solution

Answer 2

Let P be the required point at which the pole is erected.

Let BP= x m.

⇒ AP - BP = 7 m.

   AP - x = 7

  AP = (x + 7) m.

Now,

We have AB = 13 m. and ∠P = 90°[Since it is an angle in semi-circle].

By Pythagoras theorem, In right-angled ΔABP:

⇒ AB^2 = AP^2 + BP^2

⇒ (13)^2 = (x + 7)^2 + x^2

⇒ 169 = x^2 + 49 + 14x + x^2

⇒ 169 - 49 = 2x^2 + 14x

⇒ 120 = 2x^2 + 14x

⇒ 2x^2 + 14x - 120 = 0

⇒ x^2 + 7x - 60 = 0

⇒ x^2 + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5,-12{The side of the triangle should not be negative}

⇒ x = 5.

Then ,

⇒ x + 7

⇒ 5 + 7

⇒ 12.

Therefore, It is possible to erect a pole at a distance of 5 m from gate B and 12 m from gate A.

Hope this helps!