Question

the length of the rectangle is 4cm than the width. the area of the rectangle is 32cm². find the width and length of the rectangle

Answer 1

Answer :-

• Dimensions of the rectangle are 4cm and 8cm respsctively.

Given :-

• The length of the rectangle is 4cm than the width and the area of the rectangle is 32cm².

To Find :-

• Dimensions of the rectangle.

Solution :-

Let

• Width of the rectangle be x
• Length of the rectangle will be x + 4

Given

• The area of the rectangle is 32cm².

As we know that

Area of a rectangle is l × b

Where

• l = length
• b = breadth

According to question :-

⇒ x (x + 4) = 32

⇒ x² + 4x = 32

⇒ x² + 4x - 32 = 0

⇒ x² - 4x + 8x - 32 = 0

⇒ x (x - 4) + 8 (x - 4) = 0

⇒ (x + 8) (x - 4) = 0

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• x + 8 = 0

⇒ x = 0 - 8

⇒ x = -8

• x - 4 = 0

⇒ x = 0 + 4

⇒ x = 4

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x = 4 [ breadth can't be negative ]

Now

• Breadth of the rectangle = x = 4cm
• Length of the rectangle = x + 4 = 4 + 4 = 8cm

Hence, the dimensions of the rectangle are 4cm and 8cm respectively.

Answer 2

Given :

• Length of the rectangle is 4cm than the width
• Area of the rectangle is 32cm²

To Find :

• Width and length of rectangle

Solution :

• Let the width be x , then length will be x + 4
• Also we , know area of rectangle is : length × breadth

• According to the question :

$\implies \sf x(x\:+\:4)\:=\:32$

$\implies \sf x²\:+\:4x\:=\:32$

$\implies \sf x²\:+\:4x\:-32\:=\:0$

$\implies \sf x²\:+\:8x\:-\:4x\:-\:32\:=\:0$

$\implies \sf x(x\:+\:8)\:-\:4(x\:+\:8)\:=\:0$

$\implies \sf (x\:-\:4)(x\:+\:8)\:=\:0$

• Now , there are two possibilities :

1. x - 4 = 0 ; x = 4 cm
2. x + 8 = 0 ; x = - 8 cm

• Since , dimensions cannot be negative , therefore +ve one is considered .

• Also , length = x + 4 = 4 + 4 = 8 cm

• Width of rectangle is 4 cm

• Length of rectangle is 8 cm

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