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 :
Options:
A . 5 m
B . $$\frac{{54}}{{11}}$$ m
C . 6 m
D . 11 m
Answer: Option C 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, $$\eqalign{ & \Rightarrow 9x - 4y = 1 \cr & \Rightarrow 4y = 9x - 1 \cr & \Rightarrow y = \frac{{9x - 1}}{4}.....(ii) \cr} $$ From (i) and (ii), we get : $$\eqalign{ & \Rightarrow 29{x^2} = 6{\left( {\frac{{9x - 1}}{4}} \right)^2} - 1 \cr & \Rightarrow 29{x^2} = 6\left( {\frac{{81{x^2} + 1 - 18x}}{{16}}} \right) - 1 \cr & \Rightarrow 243{x^2} + 3 - 54x - 8 = 232{x^2} \cr & \Rightarrow 11{x^2} - 54x - 5 = 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {11x + 1} \right) = 0 \cr & \Rightarrow x = 5\,\,m \cr & \therefore y = \frac{{9x - 1}}{4} = \frac{{9 \times 5\,\,m - 1}}{4} = 11\,\,m \cr & \text{Required difference} \cr & = \left(11-5\right) m \cr & = 6\,m } $$
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