Question

The sum of the length and breadth of a rectangle is 20 cm. When the length is reduced by 1 cm and the
breadth is increased by 1 cm, the area is increased by 3 cm2. Find the original dimensions of the
rectangle0

Answer 1

Given data:

                  let 'l' be the length

                   let 'b' be the breadth of the rectangle

l+b=20 cm

area of a rectangle = l×b sq units

here,

(l-1)(b+1)=3 cm sq

l=?

b=?

solution:

    $l+b=20\\b=20-l$------------(i)

$(l-1)(b+1)=3\\\\b+1=\frac{3}{l-1} \\\\b=\frac{3}{l-1}-1\\ \\b=\frac{3-(l-1)}{l-1}$-----(ii)

from (i) and (ii)

$20-l=\frac{3-(l-1)}{l-1} \\\\(20-l)(l-1)=3-(l-1)\\\\20l-20-l^2+l=3+1-l\\\\-l^2+22l-20=4\\\\l^2-22l+24=0$

$\frac{l^2-22l+24}{2}=0$