Question

The length of a rectangle is 2cm longer than its width. If the area of the rectangle is 48cm^2. Find its perimeter

Answer 1

Given that:

• The length of a rectangle is 2 cm longer than its width.
• The area of the rectangle is 48 cm².

To Find:

• The perimeter of the rectangle.

Let us assume:

• Breadth of the rectangle be x.
• Length = x + 2

Formula:

• Area of a rectangle = Length × Breadth
• Perimeter of a rectangle = 2(Length + Breadth)

Finding the value of x:

According to the question.

↠ 48 = (x + 2)x

↠ 48 = x² + 2x

↠ x² + 2x - 48 = 0

↠ x² + 8x - 6x - 48 = 0

↠ x(x + 8) - 6(x + 8) = 0

↠ (x - 6) (x + 8) = 0

↠ x = 6 or x = - 8

Length is always positive.

So, x = 6

Finding the perimeter of the rectangle:

↣ Perimeter = 2(x + 2 + x)

Putting the value of x.

↣ Perimeter = 2(6 + 2 + 6)

↣ Perimeter = 2(14)

↣ Perimeter = 28

Hence,

• The perimeter of the rectangle is 28 cm.

Answer 2

⚘ Question :-

• The length of a rectangle is 2 cm longer than it's width. If the area of the rectangle is 48 cm². Find it's perimeter.

⚘ Answer :-

• Perimeter of rectangle is 28 cm.

Explanation:

⚘ Given :-

• Length = Width + 2 cm

• Area of rectangle = 48 cm²

⚘ To Find :-

• Perimeter of rectangle = ?

⚘ Solution :-

• Let width of rectangle be m cm.

• As it is stated in question that the length of a rectangle is 2 cm longer than it's width. So, length of rectangle is (m + 2) cm.

★ F I N D I N GㅤV A L U EㅤO Fㅤ'm'ㅤ::

We know that,

$\large{\boxed{\sf{\blue{Area_{(rectangle)} = L\:\times\:W}}}}$

Where,

• L denotes length of rectangle

• W denotes width of rectangle

We have,

• L = (m + 2) cm

• W = m cm

• $\sf Area_{(rectangle)}$ = 48 cm²

According to the question by using the formula we get,

➨ $\sf 48 = (m + 2)\:\times\:m$

➨ $\sf 48 = m(m + 2)$

➨ $\sf 48 = m^2 + 2m$

➨ $\sf m^2 + 2m - 48 = 0$

By splitting the middle term we get,

➨ $\sf m^2 + (8 - 6)m - 48 = 0$

➨ $\sf m^2 + 8m - 6m - 48 = 0$

➨ $\sf m(m + 8) - 6(m + 8) = 0$

➨ $\sf (m - 6)\:(m + 8) = 0$

➨ $\sf m - 6 = 0 \quad | \quad m + 8 = 0$

➨ $\sf m = 0 + 6 \quad | \quad m = 0 - 8$

➨ $\large\bf\red{m = 6}\quad | \quad\red{m = -8}$

$\Big[$ Width can't be negative $\Big]$

∴ Width of rectangle is 6 cm.

Now,

➠ Length of rectangle = (m + 2) cm

Put m = 6 in above equation we get,

➠ Length of rectangle = (6 + 2) cm

➠ Length of rectangle = 8 cm

∴ Length of rectangle is 8 cm.

★ F I N D I N GㅤP E R I M E T E Rㅤ::

We know that,

$\large{\boxed{\sf{\pink{Perimeter_{(rectangle)} = 2(L + W)}}}}$

Where,

• L denotes length of rectangle

• W denotes width of rectangle

We have,

• L = 8 cm

• W = 6 cm

According to the question by using the formula we get,

➨ $\sf Perimeter_{(rectangle)} = 2(8 + 6)$

➨ $\sf Perimeter_{(rectangle)} = 2(14)$

➨ $\sf Perimeter_{(rectangle)} = 2\:\times\:14$

➨ $\large\bf\purple{Perimeter_{(rectangle)} = 28\:cm}$

∴ Perimeter of rectangle is 28 cm.

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