Example 8 - Chapter 4 Class 10 Quadratic Equations

Last updated at May 26, 2023 by Teachoo

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Example 8
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Let P be the pole
Gates A & B are diametrically opposite
So, AB = Diameter of circle = 13 m
Also given that,
Difference of the distance of the pole from the two gates is 7 metres
BP AP or AP BP is 7 m
Let us take AP BP = 7 i.e. AP = BP + 7
Let BP = x
So, AP = BP + 7 = x + 7
Now,
Since AB is a diameter
APB = 90
Now,
APB is a right angle triangle
Using Pythagoras theorem
Hypotenuse2 = Height2 + Base2
BP2 + AP2 = AB2
x2 + (x + 7)2 = 132
x2 + x2 + 49 + 2 7=169
x2 + x2 + 49 + 14x = 169
2x2 + 14x + 49 169 = 0
2x2 + 14x 120 = 0
Divide the equation by 2
2 2/2+14 /2 120/2 = 0
x2 + 7x 60 = 0
Comparing equation with ax2 + bx + c = 0
Here, a = 1, b = 7, c = 60
We know that
D = b2 4ac
D = (7)2 4 (1) ( 60)
= 49 4 ( 60)
= 49 + 240
= 289
Hence , roots to equation are
x = ( )/2
Putting values
x = ( 7 289)/(2 1)
x = ( 7 17 )/2
Solving
So, x = 5, & x = 12
Since x is distance, it must be positive
Hence, x = 5.
Hence, BP = x = 5 m
& AP = x + 7 = 5 + 7 = 12 m

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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