Question

A pillar has to be fitted at the point of the boundary of a circular park of diameter 13 m. Such that two gates A and B situated at both ends of diameter having a difference in distance of 7 m from this pillar. Is this possible? If yes, find the distance of the pillar from both gates

Answer 1

Let pillar is fitted at point C and distance from gate B to point C = x m.

According to the question, the difference in the distance from gate A and B to the pillar is 7 m.

AC = (x + 7) m

AB is diameter

∠ACB = 90° (∵ angle in semi circle is right angle)

Now in right angled triangle ACB

AB² = AC² + BC² (By pythagorus Theorem)

⇒ 132 = (x + 7)² + x²

⇒ 169 = x² + 49 + 14x + x²

⇒ 169 = 2x² + 14x + 49

⇒ 2x² + 14x + 49 – 169 = 0

⇒ 2x² + 14x – 120 = 0

⇒ x² + 7x – 60 = 0

Now b² – 4ac = (7)² – 4 × 1 × (-60) = 49 + 240 = 289 > 0

Thus, given quadratic equation has two real roots and so pillar can be fitted at boundary of park.

Now x² + 7x – 60 = 0

⇒ x² + (12 – 5)x – 60 = 0

⇒ x² + 12x – 5x – 60 = 0

⇒ (x² + 12x) – (5x + 60) = 0

⇒ x (x + 12) – 5(x + 12) = 0

⇒ (x + 12) (x – 5) = 0

⇒ x + 12 = 0 and x – 5 = 0

⇒ x = – 12 and x = 5

x is distance between pillar and gate B so Ignore x = -12

x = 5 m and x + 7 = 5 + 7 = 12 m.

Hence distance from pillar to gate A = 12 m.
and from pillar to gate B = 5 m.