A big rectangular plot of area 4320 m² is divided into 3 square-shaped smaller plots by fencing parallel to the smaller side of the plot. However some area of land was still left as a square could not be formed. So, 3 more square-shaped plots were formed by fencing parallel to the longer side of the original plot such that no area of the plot was left surplus. What are the dimensions of the original plot ?

Solution(By Examveda Team)

Let the side of each square formed by fencing parallel to breadth be x metres and that of each square formed by fencing parallel to length be y metres.

Then,
$$\eqalign{ & 3{x^2} + 3{y^2} = 4320 \cr & \Rightarrow {x^2} + {y^2} = 1440.....(i) \cr} $$
And,
$$\eqalign{ & x\left( {3x + y} \right) = 4320 \cr & \Rightarrow 3{x^2} + xy = 3\left( {{x^2} + {y^2}} \right) \cr & \Rightarrow xy = 3{y^2} \cr & \Rightarrow x = 3y.....(ii) \cr} $$
From (i) and (ii), we have :
$$\eqalign{ & {\left( {3y} \right)^2} + {y^2} = 1440 \cr & \Rightarrow 10{y^2} = 1440 \cr & \Rightarrow {y^2} = 144 \cr & \Rightarrow y = 12 \cr} $$
$${\text{So, }}x = 36$$
Length of rectangular plot :
$$\eqalign{ & = 3x + y \cr & = \left( {3 \times 36 + 12} \right)m \cr & = 120\,m \cr} $$
Breadth of rectangular plot = x = 36 m