1.

The sum of areas of two squares is 625m^2. If the difference in their perimeter is 20m then, what will be the sides of squares?(a) 10, 15(b) 15.28, 20.28(c) 13, 67.34(d) 35.67, 46.78This question was posed to me during an interview.My question comes from Solution of Quadratic Equation by Squaring Method topic in section Quadratic Equation of Mathematics – Class 10

Answer»

Correct answer is (B) 15.28, 20.28

Explanation: Let the sides of two squares be x and y.

Their AREAS will be x^2 and y^2 respectively.

Sum of their areas is 625 m^2

x^2+y^2=625(1)

Their perimeters will be 4x and 4y

Difference in their perimeters is 20 m

4x-4y=20

x-y=5

x=5+y(2)

Substituting (2) in (1) we get,

(y+5)^2+y^2=625

y^2+10y+25+y^2=625

2y^2+10y+25=625

2y^2+10y-620=0

y^2+5y=310

Adding \(\FRAC {b^2}{4}\) on both sides, where b=5

y^2 + 5y + \(\frac {5^2}{4}=\frac {5^2}{4}\) + 310

y^2 + 5y + \(\frac {25}{4}=\frac {25}{4}\) + 310

y^2 + 5y + \(\frac {25}{4}=\frac {1265}{4}\)

\((y+\frac {5}{2})\)^2 = \((\frac {\sqrt {1265}}{2})\)^2

y+\(\frac {5}{2}\) = ±\(\frac {\sqrt {1265}}{2}\)=±\(\frac {35.56}{2}\)

y = \(\frac {35.56}{2}-\frac {5}{2}=\frac {30.56}{2}\) = 15.28 and y = \(\frac {-35.56}{2}-\frac {5}{2}=\frac {-40.56}{2}\) = -20.28

Since, length cannot be negative HENCE, y=15.28

Now, x=5+y=5+15.28=20.28


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