Question

A rectangular field is 50 m long and 20 m wide. There is a path of equal width all around it having an area of 121 m^2. What will be the width of the path?

(a) 10.205 m

(b) 11.205 m

(c) 12.56 m

(d) 13.76 m

This question was addressed to me in exam.

Enquiry is from Solution of Quadratic Equation by Squaring Method in portion Quadratic Equation of Mathematics – Class 10

Answer 1

Correct option is (b) 11.205 m

For explanation I would say: Let the width of the path be x metres.

Length of the field including the path = 50+2x m

Breadth of the field including the path = 14+2x m

Area of the field including the path = (50+2x)(14+2x) m^2

Area of the field excluding the path = 50×14=700 m^2

∴ area of the path = [(50+2x)(14+2x)]-700

∴ [(50+2x)(14+2x)]-700=121

700+100x+28x+4x^2-700=121

4x^2+128x-121=0

x^2+42x=\(\frac {121}{4}\)

Adding \(\frac {b^2}{4}\) on both sides, where b=42

x^2 + 42x + \(\frac {42^2}{4}=\frac {42^2}{4}+\frac {121}{4}\)

x^2 + 42x + \(\frac {1764}{4}=\frac {1764}{4}+\frac {121}{4}\)

x^2 + 42x + \(\frac {1764}{4}=\frac {1885}{4}\)

\((x+\frac {21}{2})\)^2=\((\frac {\sqrt {1885}}{2})\)^2

x + \(\frac {21}{2}\) = ±\(\frac {\sqrt {1885}}{2}\) = ±\(\frac {43.41}{2}\)

x = \(\frac {43.41}{2}-\frac {21}{2}=\frac {22.41}{2}\) = 11.205 and x = \(\frac {-43.41}{2}-\frac {21}{2}=\frac {-64.41}{2}\) = -32.20

Since, the length cannot be negative; therefore, x=11.205 m