Question

A rectangle is drawn so that the width is 2 feet shorter than the length. The area of the rectangle is 48 square feet. Find the length of the rectangle.

Answer 1

Given :

• Rectangle is drawn so that the width is 2 feet shorter than the length.
• The area of the rectangle is 48 sq feet.

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To Find :

• The Length of the rectangle

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Solution :

We know that,

$\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:$

So,

Let's assume the width of the rectangle as x and the length will be (x + 2).

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Now, Substituting the given values in the formula :

$\qquad \sf \: { \dashrightarrow (x + 2) \times x = 48 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} + 2x = 48 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} + 2x - 48 = 0 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} +8x -6x- 48 = 0 }$

$\qquad \sf \: { \dashrightarrow {x} (x +8) -6(x+8) = 0 }$

$\qquad \sf \: { \dashrightarrow (x +8) (x - 6) = 0 }$

$\qquad \sf \: { \dashrightarrow x = -8, \: \: x = 6 }$

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Since, The width can't be negative, so the width will be 6 which is positive.

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$\qquad { \pmb{ \bf{ Width _{(rectangle)} = 6\:ft}}}\:$

$\qquad { \pmb{ \bf{ Length _{(rectangle)} = 6+2=8 \: ft}}}\:$