Question

12. The length of a rectangle is greater than twice is breadth by 1 cm. fits
breadth is doubled and length is decreased by 4 cm, its area remains same
Find the dimension of the original rectangle​

Answer 1

$\large\bf{✯ \: understanding \: the \: concept✯}$

The length of the rectangle is greater than twice its breadth by 1 cm.

and if the breadth is doubled and the length decreased by 4 cm , its area remains the same

$\large\fbox\red{To\:find:}$

Here we need to find the area of the original rectangle .

$\sf\pink{Lets\:do\:it\:mate!}$

let the breadth of the rectangle be b

and the length be l

Therefore length (l) = 2b+1

Now,

if the breadth(b) is doubled then,

breadth ( b) = 2x

and length (l) = l - 4

then the area remains same .

$\sf{(l \times b) = (l \times b)}$

$\sf{⇒(2b - 1) \times b = (2b + 1 - 4)(2b)}$

$\sf{⇒2b {}^{2} + b = 4b {}^{2} - 6b}$

$\sf{⇒2b {}^{2} = 7b} \\ \sf{⇒b {}^{2} = \frac{7b}{2} } \\ \sf{⇒b {}^{2} = 3.5b}$

$\sf{⇒b = \frac{3.5b}{b} } \\ \sf{⇒b \: = 3.5 \: }$

Therefore,

breadth = 3.5 cm

Now lets find the length

$\sf{⇒l = 2b + 1} \\ \sf{⇒l =2 \times 3.5 + 1} \\ \sf{⇒l = 8}$

Answer:

The breadth of the rectangle is 3.5 cm and the length of the rectangle is 8 cm .

Hope it helps uh dear mate