If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. The perimeter of the original rectangle (in cm) is :

Solution(By Examveda Team)

Let the length and width of the rectangle be $$l$$ cm and b cm respectively.
Then,
$$\eqalign{ & \left( {l - 4} \right)\left( {b + 3} \right) = lb \cr & \Rightarrow lb + 3l - 4b - 12 = lb \cr & \Rightarrow 3l - 4b = 12.....(i) \cr & {\text{And, }} \cr & l - 4 = b + 3 \cr & \Rightarrow l - b = 7.....(ii) \cr} $$
Multiplying (ii) by 4 and subtracting (i) from it, we get :
$$l$$ = 16
Putting $$l$$ = 16 in (ii), we get : b = 9
∴ Perimeter of the original rectangle :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = [2\left( {16 + 9} \right)]cm \cr & = 50\,cm \cr} $$