Question

The length of a rectangle exceeds it breadth by 5m. If the breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140m². Find it's dimensions.​

Answer 1

Answer:

➝ Length = 25m.

➝ Breadth = 20m.

Step-by-step explanation:

Let the Breadth of the rectangle be 'x'.

⇒ The length of the rectangle is 'x + 5'.

Now, the breadth is doubled, and the length is reduced by 9.

⇒ Breadth = 2x

⇒ Length = [(x + 5) - 9]

Now, the new area is 140m² more than the previous. Therefore:

$\rm \Longrightarrow 140 + Area \ of \ original \ rectangle = Area \ of \ new \ rectangle$

$\rm \Longrightarrow 140 + \Big(x(x + 5)\Big) = \Big((x + 5 - 9)(2x)\Big)$

$\rm \Longrightarrow 140 + \Big(x^2 + 5x\Big) = \Big((x - 4)(2x)\Big)$

$\rm \Longrightarrow 140 + x^2 + 5x = 2x^2 - 8x$

$\rm \Longrightarrow 140 = 2x^2 - 8x - x^2 - 5x$

$\rm \Longrightarrow 140 = x^2 - 13x$

$\rm \Longrightarrow x^2 - 13x - 140 = 0$

$\rm \Longrightarrow x^2 + 7x - 20x - 140 = 0$

$\rm \Longrightarrow x(x + 7) - 20(x + 7) = 0$

$\rm \Longrightarrow (x + 7) (x - 20) = 0$

Therefore, x = -7, or 20. But dimensions cannot be negative.

Therefore x = 20m.

Original length = x + 5 = 20 + 5 = 25m.

Original breadth = x = 20m,

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow the\:length\:of\:rectangle\:exceeds\:its\:breadth\:5m$

$\sf\therefore let \:breadth\:be\:'x'$

$\sf\therefore length= x+5$

$\sf\dashrightarrow \:if\:breadth\:were\:doubled\:and\:length\:gets\:reduced\:by\:9m.then,$

$\sf\therefore breadth=2x$

$\sf\therefore length= (x+5)-)$

$\sf\dashrightarrow area\:of\:rectangle\:increased\:by\:140m^2$

$\sf\dashrightarrow note:- original\:rectangle \:is\:equal\:to\:the\:new\:rectangle$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow THE\:DIMENSIONS\:OF\:THE\:RECTANGLE.$

EQUATION FORMED,

$\rm{\boxed{\bf{ \star\:\: area\:of\:rectangle(1)+140m^2= area\:of\:rectangle (2)\:\: \star}}}$

$\large\underline\bold{SOLUTION,}$

TAKING THE EQUATION FORMED, WE GET,

$\sf\therefore area\:of\:rectangle(1)+140m^2= area\:of\:rectangle (2)$

$\sf\implies 140 + (x \times (x + 5)) = ((x + 5 - 9) \times (2x)$

$\sf\implies 140+ x^2+5x= (x-4) \times 2x$

$\sf\implies 140 + x^2 + 5x = 2x^2 - 8x$

$\sf\implies 140 +5x+8x= 2x^2-x^2$

$\sf\implies 140 +13x= x^2$

$\sf\implies x^2 - 13x - 140 = 0$

$\sf\implies x^2 + 7x - 20x - 140 = 0$

$\sf\implies x(x + 7) - 20(x + 7) = 0$

$\sf\implies (x + 7) (x - 20) = 0$

$\sf\implies x+7=0,x-20=0$

$\sf\implies x=-7,x=20$

$\sf\implies x= -7,20$

$\sf\therefore NOTE:-\: DIMENSIONS\: CAN'T\:BE\:NEGATIVE.$

$\sf\therefore x=20m$

$\large{\boxed{\bf{ \star\:\: x=20m\:\: \star}}}$

$\sf\therefore length =x+5$

$\sf\implies x=20m$

$\sf\implies 20+5$

$\sf\implies 25m$

$\sf\therefore breadth= x$

$\sf\therefore x=20m$

$\large\underline\bold{THE\:LENGTH\:OF\:THE\:ORIGINAL\:RECTANGLE\:IS\: 25m.}$

$\large\underline\bold{THE\:BREADTH\:OF\:THE\:ORIGINAL\:RECTANGLE\:IS\: 20m}$

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