A pole has to be erected at a point on the boundary of a circular ground of diameter
20 m in such a way that the difference of its distances from two diametrically opposite
fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at
what distance from the two gates should the pole be erected?
Answers
Answer:
Step-by-step explanation:
Let P be the position of the pole and A and B be the opposite fixed gates.
PA - PB = 7 m
⇒ a - b = 7
⇒ a = 7 + b .........(1)
In Δ PAB,
AB² = AP² + BP²
⇒ (17) = (a)² + (b)²
⇒ a² + b² = 289
⇒ Putting the value of a = 7 + b in the above,
⇒ (7 + b)² + b² = 289
⇒ 49 + 14b + 2b² = 289
⇒ 2b² + 14b + 49 - 289 = 0
⇒ 2b² + 14b - 240 = 0
Dividing the above by 2, we get.
⇒ b² + 7b - 120 = 0
⇒ b² + 15b - 8b - 120 = 0
⇒ b(b + 15) - 8(b + 15) = 0
⇒ (b - 8) (b + 15) = 0
⇒ b = 8 or b = -15
Since this value cannot be negative, so b = 8 is the correct value.
Putting b = 8 in (1), we get.
a = 7 + 8
a = 15 m
Hence PA = 15 m and PB = 8 m
So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m
Answer.
The answer can be proved also.
AB² = AP² + BP²
17² = 15² + 8²
289 = 225 + 64
289 = 289
L.H.S = R.H.S
