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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 mThis 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»

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



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