Question

Lenght of rectangle is 1cm more than its width /and its perimeter is 14

Answer 1

$\bf{\huge{\underline{\boxed{\tt{\red{Answer}}}}}}$

Given :-

• Lenght of rectangle is 1cm more than its width
• Perimeter of the rectangle is 14 cm

To find :-

• Dimensions of the rectangle i.e length and width of the rectangle.
• Area of the rectangle.

Sóĺúťíóń :-

Let the width of the rectangle be x cm

•°•As per the first condition,

• Lenght of rectangle is 1cm more than its width

•°•Length of the rectangle = x + 1 cm

We know that perimeter of a rectangle is calculated via using the formula,

$\bf{\large{\underline{\boxed{\tt{\blue{Perimeter\:of\:a\:rectangle\:=\:2(L +B) }}}}}}$

Plug in the values,

14 = 2 ( x + 1 + x)

14 = 2x + 2 + 2x

14 - 2 = 2x + 2x

12 = 4x

$\bf\large\frac{12}{4}$ = x

3 = x

Width = 3 cm

Substitute, the value of width i.e x in the value of length.

Length = x + 1 = 3 + 1 = 4 cm

•°• Dimensions of the rectangle :

• Length = 4 cm
• Breadth = 3 cm

Now, we will move on to calculating the area of the rectangle as we have the dimensions we can easily calculate it using the formula.

We know that area of a rectangle is calculated via the formula,

$\bf{\large{\underline{\boxed{\tt{\green{Area\:of\:rectangle\:=\:(Length) (Breadth)}}}}}}$

Plug in the values,

Area of rectangle = (4 × 3)cm

Area of rectangle = 12 sq.cm

$\bf{\huge{\underline{\boxed{\tt{\purple{Verification:}}}}}}$

Length of the rectangle = 4 cm

Breadth of the rectangle = 3 cm

Perimeter of rectangle = 14 cm

Plug the values in the formula for perimeter of a rectangle,

Perimeter = 2 ( l + b)

14 = 2 ( 4 + 3)

14 = 2 (7)

14 = 14

LHS = RHS.

Hence verified.

For area of rectangle :-

Area = 12 sq.cm

Plug in the values of length,breadth and area of the rectangle in the formula of area of the rectangle,

Area = length × breadth

12 = 4 × 3

12 = 12

LHS = RHS.

Hence verified.

I found the question to be incomplete, so as just reading the question, perimeter of the rectangle is already given, either we were supposed to find the area of the rectangle or it's dimensions. Here, I have calculated both i.e the dimensions and the area of the rectangle for avoiding any kind of obscurence.