Question

the length of the rectangular playground is 2m larger than its breath if the area of playground is 195m square find the length and breath of field
​

Answer 1

• Length is 15 m.
• Breadth is 13 m.

Step-by-step explanation:

Given:-

• Area of rectangular playground is 195 m².

To find:-

• Length and breadth of rectangular playground

Solution:-

Let, Breadth of rectangular playground be x.

And Length of rectangular playground be x + 2. [We take length be x + 2 because it is given that length is larger than it's breadth by 2m ]

$\boxed{\sf \bold{Area \: of \: rectangle = Length \times Breadth}}$

Put, Length, Breadth and area of rectangular playground in formula :

$\longrightarrow$ (x + 2) × x = 195

$\longrightarrow$ x² + 2x = 195

$\longrightarrow$ x² + 2x - 195 = 0

$\longrightarrow$ x² + 15x - 13x - 195 = 0

$\longrightarrow$ x(x + 15) - 13(x + 15) = 0

$\longrightarrow$ (x + 15) (x - 13) = 0

____

• x + 15 = 0

x = -15

• x - 13 = 0

x = 13

Breadth cannot be negative. So, Breadth of rectangular playground is 13 m

And We have also taken, Length be x + 2 = 13 + 2 = 15. Thus, Length of rectangular playground is 15 m.

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

${ \frak { \underline\purple{\qquad Given\: :\qquad}}} \:$

• Length of playground is 2 m larger than it's width and it's area is 195m²

${ \frak { \underline\purple{\qquad To\:Find\: :\qquad}}} \:$

• Length and width of the playground.

${ \frak { \underline\purple{\qquad Diagram \::\qquad}}} \:$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x + 2 m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large x m}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

${ \frak { \underline\purple{\qquad Solution\: :\qquad}}} \:$

$\:\:\:\:\:\:\bullet\:\sf{Let \:it's \:width(w)\: = x \:m}$

$\:\:\:\:\:\:\bullet\:\sf{\therefore\:Length(l)\: =\: x \:+ 2\: m}$

${\red\bigstar\:\underline{\boxed {\bold \green {Area_{(rectangle)}\:\leadsto\:width(w)\:\times\:length(l)}}}}$

Putting all values in the formula :-

$\sf{x\:\times\:(x\:+\:2)\:=\:195}$

$\sf{x^2\:+\:2x\:=\:195}$

$\sf{x^2\:+\:2x\:=\:195}$

$\sf{x^2\:+\:2x\:-\:195\:=\:0}$

☯ Quadratic equation formed !! ☯

☯ Let's solve it !! ☯

$\sf{x^2\:+\:15x\:-\:13x\:-\:195\:=\:0}$

$\sf{x(x\:+\:15)\:-\:13(x\:+\:15)\:=\:0}$

$\sf{(x\:+\:15)\:(x\:-\:13)\:=\:0}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

• Value of x for , (x + 15) = 0

➱ x = 0 - 15

➱ x = -15

• Value of x for , (x - 13) = 0

➱ x = 0 + 13

➱ x = 13

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Width cannot be negative !! ☯

$\underline{\boxed {\bold {\therefore \purple {x\:=\:width(w)\:=\:13}}}}\:\pink\bigstar$

• Length

➪ $\sf{ x \:+ \:2}$

➪ $\sf{13\:+\:2}$

➪ $\sf\underline{15}$

$\underline{\boxed {\bold {\therefore \orange {x\:+\:2\:=\:length(l)\:=\:15}}}}\:\green\bigstar$

${ \frak { \underline\purple{\qquad Note :\qquad}}} \:$

• Dear user if you are not able to see the diagram from app. Please see it from the the site (brainly.in). It will be correctly displayed there.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━