Question

A pole has to be erected at a point on the boundary of a circular path park of diameter 13 m in such a way that the difference of its distance from two diametrically opposite fix gate a and b on the boundary is 7m is it possible to do so? If yes at what distance from the two gates should be paul be created

Answer 1

it is an NCERT example of maths so check it from there ch 4

Answer 2

Hey there!

Here you go,

Let P be the pole to be erected and A, B be the opposite fixed gates.

Given, PA – PB = 7

Let PA = a, PB = b

Hence a – b = 7

⇒ a = 7 + b �...(1)

In right ΔPAB,

AB2 = AP2 + BP2 (By Pythagoras theorem)

132 = a2 + b2

169 = (7 + b)2 + b2

169 = 49 + 14b + 2b2

2b2 + 14b – 120 = 0

b2 + 7b – 60 = 0

b2 + 12b - 5b – 60 = 0

b(b + 12) - 5(b + 12) = 0

(b + 12)(b – 5) = 0

Hence b = 5 or – 12

But ‘b’ cannot be negative

Therefore, b = 5

⇒ a = 7 + 5 = 12

Hence, PA = 12 m and PB = 5 m