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 the possible to do so? If yes, at what distances from the two gates should the pole be erected?
Answers
SOLUTION :
Given : AP - BP = 7 m, Diameter (AB) = 13 m
Let P be the location of the pole. Let x be the distance of the pole from the gate B i.e BP = x cm
Difference of the distance of the pole from the two gates = AP - BP = 7 cm
AP - x = 7
AP = (x + 7) m
In Δ PAB, ∠APB = 90°(angle in a semicircle is a right angle)
By Pythagoras theorem
AB² = AP² + BP²
13² = (x + 7)² + x²
169 = x² + 7² + 2×7 × x + x²
169 = 2x² + 49 + 14x
2x² + 14x + 49 - 169 = 0
2x² + 14x - 120 = 0
2(x² + 7x - 60) = 0
x² + 7x - 60 = 0
x² + 12x - 5x - 60 = 0
[By middle term splitting]
x(x + 12) - 5(x + 12) = 0
(x + 12) (x - 5) = 0
(x + 12) or (x - 5) = 0
x = - 12 or x = 5
Since, x cannot be negative because the distance between the pole cannot be negative, so x ≠ - 12
Therefore , x = 5
Distance of the pole from the gate B i.e BP = x cm = 5 cm
Distance of the pole from the gate A i.e AP = (x + 7) cm = (5 + 7) = 12 cm
Hence, the distance from the gate A to pole is 12 m and from gate B to the pole is 5 m
HOPE THIS ANSWER WILL HELP YOU......
Answer:
The given situation is shown in the attached figure.
Given:
Diameter, AB=17 m
AC−BC=7
Solution:
From property of circles, ∠ACB=90
o
Using Pythagoras theorem,
AC
2
+BC
2
=AB
2
............(i)
AC−BC=7.............(ii)
Substituting AC from (ii) in (i), we get
(7+BC)
2
+BC
2
=17
2
BC
2
+7BC−120=0
BC
2
−8BC+15BC−120=0
(BC−8)(BC+15)=0
BC=−15,8
BC is a distance and cannot be negative
BC=8 m
AC=7+BC=15 m
Hence, it is possible to do so and the distance of the gates from the poles are 8 m and 15 m.