Question

$\huge\star \: \sf{ Question}$



The length of a rectangle is greater than the breadth by 3cm . If the length is increased by 9cm and the breadth is reduced by 5cm , the area remains the same. Find the dimensions of the rectangle ?​

Answer 1

Let the breadth be x cm.

Then the length will be (x+3) cm.

Area of the rectangle = x(x+3) cm^2. (1)

Given that length is increased by 9cm = x +3+9 = x + 12 cm.

Given that breadth is reduced by 5cm = x - 5 cm.

Area of the rectangle = (x+12)(x-5). (2)

On solving (1) and (2), we get

x(x+3) = (x+12)(x-5)

x^2 + 3x = x^2 + 7x - 60

4x = 60

x = 15.

Breadth = 15cm.

Length = 15 + 3 = 18 cm.

Answer 2

Step-by-step explanation:

$\sf \: Solution :$

Let the breadth be x cm

so length will be (x + 3) cm

We know that

$\sf ☞ \: Area \: = Length \times Breadth$

$\sf \: =x \times (x+3) \\ \sf =(x^ 2 +3x)cm^ 2$

Given that length is increased by 9 cm=

(x+3+9) = (x+12) cm

Breadth is seduced by 5 cm=

(x-5)cm

$\sf \: Area = l \times b \\ \sf =(x+12) \times (x-5) \\ \sf =x(x-5)12(a-5) \\ \sf =x^ 2 -5x+12x-60 \\ \sf =x^ 2 +7x-60$

According to the question :

$= \sf \: x^ 2 +3x =x^ 2 +7x-60 \\ \sf = \cancel{x^ 2} - \cancel{ x^ 2} +3x-7x=-60 \\ \sf = 4x \: = 60 \\ \sf = x = \cancel{\frac{60}{4} }$

$\sf \red{ Breath = 15 cm} \\ \sf \red{Length = (15+3) = 18 cm }$

Hope it'll help you friend

have a nice day :-D