Math, asked by sundarkrishna11, 1 year ago

A pole has to be erected at a point on the boundary of a circular park of diameter 17m in such a way that the differences of its distances from two diametrically opposite fixed gates A & B on the boundary is 7 meters . Find the distances from the two gates where the pole has to be erected.

Answers

Answered by Golda
357
Solution;-

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 
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Answered by Adhi55555visi
26

Answer:

Solution;-

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)

Mark as brilliant

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.

Mark as brilliant

The answer can be proved also.

AB² = AP² + BP²

17² = 15² + 8²

289 = 225 + 64

289 = 289

L.H.S = R.H.S 

Mark as brilliant

, Answer is from the gate A THE POLE P IS 15 METRE LONG

AND FROM THE GATE B THE POLE 8 METRE LONG

Mark as brilliant

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