Question

The length of a rectangle exceeds its width by 4m. If the width is decreased by 2m and the length is decreased by 6m, the area is decreased by 68 sq.m. Find the dimensions of the rectangle?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The length of a rectangle exceeds its width by 4m.

Let assume that

Width of rectangle be 'x' m

So,

Length of rectangle be 'x + 4' m

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{Width = x \: \: meter} \\ &\sf{Length = x + 4 \: \: meter} \end{cases}\end{gathered}\end{gathered}$

We know that,

Area of rectangle = Length × width

So,

$\boxed{ \bf{ \: Area = x(x + 4)}} - - (1)$

According to statement

If the width is decreased by 2m and the length is decreased by 6m, the area is decreased by 68 sq.m.

Now,

Length of rectangle = x + 4 - 6 = x - 2 m

Width of rectangle = x - 2 m

So,

$\boxed{ \bf{ \: Area =( x - 2)(x - 2)}} - - (2)$

Thus,

$\rm :\longmapsto\:x(x + 4) - (x - 2)(x - 2) = 68$

$\rm :\longmapsto\: {x}^{2} + 4x - ( {x}^{2} - 2x - 2x + 4) = 68$

$\rm :\longmapsto\: {x}^{2} + 4x - ( {x}^{2} - 4x + 4) = 68$

$\rm :\longmapsto\: {x}^{2} + 4x - {x}^{2} + 4x - 4 = 68$

$\rm :\longmapsto\: 8x - 4 = 68$

$\rm :\longmapsto\: 8x = 68 + 4$

$\rm :\longmapsto\: 8x = 72$

$\rm :\longmapsto\: x = 9 \: m$

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{Width = x = 9 \: \: meter} \\ &\sf{Length = x + 4 = 13\: \: meter} \end{cases}\end{gathered}\end{gathered}$