Question

the length of the rectangle exceeds its breadth by 4m. perimeter of rectangle is 40 meter. what are the length and breadth ​

Answer 1

Answer :-

• Dimensions of the rectangle are 8m and 12m respectively.

Given :-

• The length of the rectangle exceeds its breadth by 4m and the perimeter of the rectangle is 40m.

To Find :-

• Length and breadth of the rectangle.

Solution :-

Let

• Breadth of the rectangle be x
• Length of the rectangle be x + 4

→ Length exceeds the breadth by 4m.

Here

• Perimeter of the rectangle is 40m

As we know that

Perimeter of the rectangle is

2 (l + b)

Where

• l = length
• b = breadth

According to question :-

⇒ 2 (x + x + 4) = 40

⇒ 2 (2x + 4) = 40

⇒ 2x + 4 = 40/2

⇒ 2x + 4 = 20

⇒ 2x = 20 - 4

⇒ 2x = 16

⇒ x = 16/2

⇒ x = 8

Now

• Breadth = x = 8m
• Length = x + 4 = 8 + 4 = 12m

Hence, the dimensions of the rectangle are 8m and 12m respectively.

Answer 2

ANSWER :

• Length = 12 m

• Breadth = 8 m

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GIVEN :

• The length of the rectangle exceeds its breadth by 4 m

• Perimeter = 40 m

To Find :

• Length and Breadth

Solution :

→ Perimeter of rectangle = 2( l + b )

• Let the breadth be x then length will be x + 4

• According to the question :

$\implies \sf 2(\:x\:+\:4\:+\:x) \:=\: 40$

$\implies \sf 2(2x\:+\:4) \:=\: 40$

$\implies \sf 4x\:+\:8 \:=\: 40$

$\implies \sf 4x \:=\: 40\:-\:8$

$\implies \sf 4x \:=\: 32$

$\implies \sf x \:=\: \dfrac{32}{4}$

$\implies \sf x \:=\: 8$

• Breadth = x = 8 m

• Length = x + 4 = 8 + 4 = 12 m

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