Question

A The length of a rectangle is 4 m more
than the width. The area of the rectangle
is 45 m2. Find the length and width.

Answer 1

Given :

• The length of a rectangle is 4m more than the width.
• The area of the rectangle is 45m²

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To Find :

• The length and width of the rectangle.

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Solution :

We know that,

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

So,

Let's assume the length of the rectangle as x and the width will be (x – 4).

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Now, Substituting the given values in the formula :

$\qquad \sf \: { \dashrightarrow x \times (x - 4) = 45 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} - 4x = 45 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} - 4x - 45 = 0 }$

$\qquad \sf \: { \dashrightarrow {x}^{2} - 9x+ 5 x - 45 = 0 }$

$\qquad \sf \: { \dashrightarrow x(x - 9) + 5(x - 9) = 0 }$

$\qquad \sf \: { \dashrightarrow (x - 9) (x + 5) = 0 }$

$\qquad \sf \: { \dashrightarrow x = 9, \: \: x = - 5}$

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Since, The length can't be negative, so the length will be 9 which is positive.

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$\qquad { \pmb{ \bf{ Length _{(rectangle)} = 9\:m}}}\:$

$\qquad { \pmb{ \bf{ Width _{(rectangle)} = 9 - 4=5\: m}}}\:$

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Answer 2

The length is 9m and the width is 5m.

What is a Rectangle?

This is defined as is a quadrilateral with four right angles and the area is calculated as Length × width.

Let's assume length is x and width is (x-4).

We have x × (x-4) = 45m²

x² - 4x = 45

x² - 4x - 45 = 0

x² - 9x + 5x -45 = 0

x(x-9) +5 (x-9)

(x+5)(x-9) = 0

x = -5 or x = 9

x is therefore 9m and the width is (9-4) = 5m

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