The area of a square is twice that of a rectangle. The perimeter of the rectangl

Question

The area of a square is twice that of a rectangle. The perimeter of the rectangle is 10 cm. If its length and breadth each is increased by 1 cm, the area of the rectangle become equal to the area of the square. The length of side of the square is :

Options:

A . $$2\sqrt 3 $$ cm

B . $$3\sqrt 2 $$ cm

C . $$4\sqrt 3 $$ cm

D . $$12 $$ cm

Answer: Option A
Let the length and breadth of the rectangle be $$l$$ cm and n cm respectively
Then,
$$\eqalign{
& 2\left( {l + b} \right) = 10 \cr
& \Rightarrow l + b = 5 \cr
& \Rightarrow b = \left( {5 - l} \right)cm \cr} $$
Area of the rectangle :
$$\eqalign{
& = l\left( {5 - l} \right)c{m^2} \cr
& = \left( {5l - {l^2}} \right)c{m^2} \cr} $$
Area of the square :
$$\eqalign{
& = 2\left( {5l - {l^2}} \right)c{m^2} \cr
& = \left( {10l - 2{l^2}} \right)c{m^2} \cr} $$
$$\eqalign{
& \therefore \left( {l + 1} \right)\left( {6 - l} \right) = \left( {10l - 2{l^2}} \right) \cr
& \Rightarrow {l^2} - 5l + 6 = 0 \cr
& \Rightarrow \left( {l - 3} \right)\left( {l - 2} \right) = 0 \cr
& \Rightarrow l = 3 \cr} $$
Area of the square :
$$\eqalign{
& = \left( {10 \times 3 - 2 \times 9} \right)c{m^2} \cr
& = 12\,c{m^2} \cr} $$
∴ Side of the square $$ = \sqrt {12} \,cm = 2\sqrt 3 $$

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