Question

The length of a rectangle is 4 units less than twice its breadth. If the length is decreased by 3 units and the
breadth is increased by 5 units the area of the rectangle is increased by 21 units. Find the length and breadth of
the rectangle​

Answer 1

$\large\bf\underline \green{Given:-}$

• length of rectangle is 4units less then twice it's breadth.

• If the length is decreased by 3 units and the breadth is increased by 5 units the area of the rectangle is increased by 21 units.

$\large\bf\underline \green{To \: find:-}$

• Length and breadth of rectangle.

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

• Let the breadth of rectangle be x units.

• So, length is 2x-4

According to question:-

If the length is decreased by 3 units and the

breadth is increased by 5 units the area of the rectangle is increased by 21 units

⠀⠀⠀⠀⠀➝ (2x-4-3)(x+5) = x(2x-4)+21

⠀⠀⠀⠀⠀➝ (2x - 7)(x+5) = x(2x-4)+21

⠀⠀⠀⠀⠀➝ 2x(x+5)-7(x+5) = 2x²-4x +21

⠀⠀⠀⠀⠀➝2x² +10x -7x -35 = 2x² -4x +21

⠀⠀⠀⠀⠀➝ 3x + 4x = 21 + 35

⠀⠀⠀⠀⠀➝ 7x = 56

⠀⠀⠀⠀⠀➝ x = 56/7

⠀⠀⠀⠀⠀➝ x = 8

So, the breadth of rectangle = 8 units

Length of rectangle = 2x- 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀➝2×8 - 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀➝16 - 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀➝ 12 units

Answer 2

$\red{\underline{\underline{Answer:}}}$

$\sf{Length \ and \ breadth \ of \ rectangle}$

$\sf{are \ 8 \ units \ and \ 12 \ units \ respectively.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ length \ of \ a \ rectangle \ is}}$

$\sf{4 \ units \ less \ than \ twice \ of \ it's \ breadth.}$

$\sf{\implies{If \ the \ length \ is \ decreased \ by}}$

$\sf{3 \ units \ and \ the \ breadth \ by \ 5 \ units}$

$\sf{the \ area \ of \ the \ rectangle \ is \ increased}$

$\sf{by \ 21 \ units.}$

$\sf\pink{To \ find:}$

$\sf{Length \ and \ breadth \ of \ rectangle.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ length \ of \ rectangle \ be }$

$\sf{x \ units \ and \ breadth \ be \ y \ units.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{x+4=2y}$

$\sf{\therefore{x-2y=-4...(1)}}$

$\boxed{\sf{Area \ of \ rectangle=length (l)\times \ breadth (b)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{(x-3)(y+5)=xy+21}$

$\sf{xy+5x-3y-15=xy+21}$

$\sf{5x-3y=21+15}$

$\sf{\therefore{5x-3y=36...(2)}}$

$\sf{Multiply \ equation (1) \ by \ 5, \ we \ get}$

$\sf{5x-10y=-20...(3)}$

$\sf{Subract \ equation (2) \ from \ equation (3)}$

$\sf{5x-10y=-20}$

$\sf{-}$

$\sf{5x-3y=36}$

_______________________

$\sf{-7y=-56}$

$\sf{\therefore{y=\frac{56}{7}}}$

$\boxed{\sf{\therefore{y=8}}}$

$\sf{Substitute \ y=8 \ in \ equation (1), \ we \ get}$

$\sf{x-2(8)=-4}$

$\sf{\therefore{x-16=-4}}$

$\sf{\therefore{x=-4+16}}$

$\boxed{\sf{\therefore{x=12}}}$

$\sf\purple{\tt{\therefore{Length \ and \ breadth \ of \ rectangle}}}$

$\sf\purple{\tt{are \ 8 \ units \ and \ 12 \ units \ respectively.}}$