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.
Share
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.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
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:
[tex]\rm \Longrightarrow 140 + Area \ of \ original \ rectangle = Area \ of \ new \ rectangle[/tex]
[tex]\rm \Longrightarrow 140 + \Big(x(x + 5)\Big) = \Big((x + 5 - 9)(2x)\Big)[/tex]
[tex]\rm \Longrightarrow 140 + \Big(x^2 + 5x\Big) = \Big((x - 4)(2x)\Big)[/tex]
[tex]\rm \Longrightarrow 140 + x^2 + 5x = 2x^2 - 8x[/tex]
[tex]\rm \Longrightarrow 140 = 2x^2 - 8x - x^2 - 5x[/tex]
[tex]\rm \Longrightarrow 140 = x^2 - 13x[/tex]
[tex]\rm \Longrightarrow x^2 - 13x - 140 = 0[/tex]
[tex]\rm \Longrightarrow x^2 + 7x - 20x - 140 = 0[/tex]
[tex]\rm \Longrightarrow x(x + 7) - 20(x + 7) = 0[/tex]
[tex]\rm \Longrightarrow (x + 7) (x - 20) = 0[/tex]
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✔
[tex]\large\underline\bold{GIVEN,}[/tex]
[tex]\sf\dashrightarrow the\:length\:of\:rectangle\:exceeds\:its\:breadth\:5m[/tex]
[tex]\sf\therefore let \:breadth\:be\:'x'[/tex]
[tex]\sf\therefore length= x+5[/tex]
[tex]\sf\dashrightarrow \:if\:breadth\:were\:doubled\:and\:length\:gets\:reduced\:by\:9m.then,[/tex]
[tex]\sf\therefore breadth=2x[/tex]
[tex]\sf\therefore length= (x+5)-)[/tex]
[tex]\sf\dashrightarrow area\:of\:rectangle\:increased\:by\:140m^2[/tex]
[tex]\sf\dashrightarrow note:- original\:rectangle \:is\:equal\:to\:the\:new\:rectangle[/tex]
[tex]\large\underline\bold{TO\:FIND,}[/tex]
[tex]\sf\dashrightarrow THE\:DIMENSIONS\:OF\:THE\:RECTANGLE.[/tex]
EQUATION FORMED,
[tex]\rm{\boxed{\bf{ \star\:\: area\:of\:rectangle(1)+140m^2= area\:of\:rectangle (2)\:\: \star}}}[/tex]
[tex]\large\underline\bold{SOLUTION,}[/tex]
TAKING THE EQUATION FORMED, WE GET,
[tex]\sf\therefore area\:of\:rectangle(1)+140m^2= area\:of\:rectangle (2)[/tex]
[tex]\sf\implies 140 + (x \times (x + 5)) = ((x + 5 - 9) \times (2x)[/tex]
[tex]\sf\implies 140+ x^2+5x= (x-4) \times 2x[/tex]
[tex]\sf\implies 140 + x^2 + 5x = 2x^2 - 8x[/tex]
[tex]\sf\implies 140 +5x+8x= 2x^2-x^2 [/tex]
[tex]\sf\implies 140 +13x= x^2[/tex]
[tex]\sf\implies x^2 - 13x - 140 = 0[/tex]
[tex]\sf\implies x^2 + 7x - 20x - 140 = 0 [/tex]
[tex]\sf\implies x(x + 7) - 20(x + 7) = 0 [/tex]
[tex]\sf\implies (x + 7) (x - 20) = 0 [/tex]
[tex]\sf\implies x+7=0,x-20=0[/tex]
[tex]\sf\implies x=-7,x=20[/tex]
[tex]\sf\implies x= -7,20[/tex]
[tex]\sf\therefore NOTE:-\: DIMENSIONS\: CAN'T\:BE\:NEGATIVE.[/tex]
[tex]\sf\therefore x=20m[/tex]
[tex]\large{\boxed{\bf{ \star\:\: x=20m\:\: \star}}}[/tex]
[tex]\sf\therefore length =x+5 [/tex]
[tex]\sf\implies x=20m[/tex]
[tex]\sf\implies 20+5[/tex]
[tex]\sf\implies 25m[/tex]
[tex]\sf\therefore breadth= x [/tex]
[tex]\sf\therefore x=20m[/tex]
[tex]\large\underline\bold{THE\:LENGTH\:OF\:THE\:ORIGINAL\:RECTANGLE\:IS\: 25m.}[/tex]
[tex]\large\underline\bold{THE\:BREADTH\:OF\:THE\:ORIGINAL\:RECTANGLE\:IS\: 20m}[/tex]
_________________