Question

The length of a rectangle is greater than 4 times its breadth by 5cm. if its length is reduced
by 2cm and breadth is increased by 2cm. then the addition of the areas of the two rectangle
is 1150sq.cm. Find the length and the breadth of the original rectangle (Ans: 49cm, 11cm)​

Answer 1

Answer:  l=25cm , b=5cm

Answer 2

SOLUTION :-

Original rectangle :-

Let,

Breadth = x

Length = 4x + 5

Area = $\sf x (4x+5)$

        = $\sf 4x^2+5x$

New rectangle :-

Breadth = x + 2

Length = 4x + 5 - 2

            = 4x + 3

Area = $\sf (x+2)(4x+3)$

        = $\sf 4x^2+3x+8x+6$

        = $\sf 4x^2+11x+6$

According to the question,

$\longrightarrow \sf 4x^2+5x+4x^2+11x+6 = 1150\\\\\longrightarrow 8x^2+16x +6- 1150 = 0 \\\\\longrightarrow 8x^2+16x-1144 = 0 \\\\\longrightarrow x^2+2x- 143 = 0$

$\boxed{\bf x= \dfrac{-b\pm\sqrt{b^{2}-4ac} }{2a}}$

$\longrightarrow\sf x = \dfrac{-2\pm\sqrt{2^2-(4\times 1\times -143)} }{2\times 1} \\\\\longrightarrow x = \dfrac{-2\pm \sqrt{4+572} }{2}\\\\\longrightarrow x= \dfrac{-2\pm \sqrt{576} }{2} \\\\\longrightarrow x= \dfrac{-2\pm 24}{2}$

$\bullet\sf \ If \ x = \dfrac{-2+24}{2}, then \ x = \dfrac{22}{2}=\bf { 11} \ .\\\\\bullet\sf \ if \ x = \dfrac{-2-24}{2}, then \ x = \dfrac{-26}{2}= \bf - 13 \ .$

Breadth cannot be negative.

Breadth of the rectangle = 11 cm

Length of the rectangle = 4 × 11 + 5

                                         = 44 + 5

                                         = 49 cm