Question

The length of the rectangle is greater than the breadth by 3 cm . If the length is Increased by 9 cm and the breadth is reduced by 5 cm , the area remains the same . Find the demension of the rectangle.

Answer 1

Answer:

The dimensions of the rectangle are 18 cm and 15 cm.

Step-by-step explanation:

Given :-

• The length of the rectangle is greater than the breadth by 3 cm.
• If the length is increased by 9 cm and the breadth is reduced by 5 cm, the area remains same.

To find :-

• The dimensions of the rectangle.

Solution :-

Consider,

• Breadth of the rectangle = x cm

★ The length of the rectangle is greater than the breadth by 3 cm.

Then,

• Length of the rectangle = (x+3) cm

Formula used :-

${\boxed{\sf{Area\:of\: rectangle=length\times\: breadth}}}$

Therefore,

Area of the rectangle,

= (length×breadth)

=[ (x+3)× x ]cm²

= (x² + 3x) cm²

★ If the length is increased by 9 cm and the breadth is reduced by 5 cm, the area remains same.

Then,

• Length = (x+3+9) = (x+12) cm
• Breadth = (x-5) cm

Area of the rectangle,

= (x+12)(x-5) cm²

=( x² -5x +12x -60) cm²

=( x² +7x -60) cm²

According to the question ,

x² + 3x = x² + 7x -60

→ 3x = 7x -60

→ 3x - 7x = -60

→ -4x = -60

→ x = 15

Therefore,

★ Breadth of the rectangle = 15 cm

★ Length of the rectangle = (15+3) = 18 cm

Answer 2

$\bf{\underline{Question:-}}$

The length of the rectangle is greater than the breadth by 3 cm . If the length is Increased by 9 cm and the breadth is reduced by 5 cm , the area remains the same . Find the demension of the rectangle.

$\bf{\underline{Solution:-}}$

Let,

Breadth be x cm

Then,

Lenght will be x + 3 cm

• Area of rectangle = x ( x+3) cm²---equ(¡)
• Given that breadth is reduced by 5cm = X - 5 cm
• Area of rectangle = ( x + 12 ) ( x - 5 ) ---equ(¡¡)

Now,

• Solving equ(¡) & equ(¡¡)

$\sf → x ( x + 3) = (x +12) (x-5)$

$\sf → x^2 + 3x = x^2 + 7x -60$

$\sf → 4x = 60$

$\sf → x = 60/4$

$\sf → x = 15$

$\bf{\underline{Hence:-}}$

• Breadth = 15cm
• Length = x + 3 = 15 + 3 = 18cm