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1. Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is :

A. 5 m

B. $\frac{54}{11}$ m

C. 6 m

D. 11 m

Solution:

Let the sides of the two squares be x metres and y metres respectively
Then,

\Rightarrow 29 x^{2} = 6 y^{2} - 1..... (i)

And,

\begin{aligned} \Rightarrow 9 x - 4 y = 1 \\ \Rightarrow 4 y = 9 x - 1 \\ \Rightarrow y = \frac{9 x - 1}{4} . . . . . (i i) \end{aligned}

From (i) and (ii), we get :

\begin{aligned} \Rightarrow 29 x^{2} = 6 {(\frac{9 x - 1}{4})}^{2} - 1 \\ \Rightarrow 29 x^{2} = 6 (\frac{81 x^{2} + 1 - 18 x}{16}) - 1 \\ \Rightarrow 243 x^{2} + 3 - 54 x - 8 = 232 x^{2} \\ \Rightarrow 11 x^{2} - 54 x - 5 = 0 \\ \Rightarrow (x - 5) (11 x + 1) = 0 \\ \Rightarrow x = 5 m \\ ∴ y = \frac{9 x - 1}{4} = \frac{9 \times 5 m - 1}{4} = 11 m \\ Required difference \\ = (11 - 5) m \\ = 6 m \end{aligned}

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