The width of a field is 5m less than its length. If the area is 204 m^

1.	The width of a field is 5m less than its length. If the area is 204 m^2, find the dimmensions of the field
Answer» Answer:- $\red{\bigstar}$ Dimensions of the field are $\large\leadsto\boxed{\tt\green{17 \times <klux>12</klux> \: m^2}}$ • Given:- Width of field is 5m less than its length. Area of the field is 204 m². ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ • To Find:- Dimensions of the field ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ • Solution:- Let the length of the field be 'x'. Now, given that the width is 5m less than the length. ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Therefore, The width of the field will be 'x-5' ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ★ Figure:- $\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(2,3.5){\sf\large x m}\put(-1.4,1.4){\sf\large x-5 m}\put(2,1.4){\large\bf <klux>204</klux> $m^2$}\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}$ We know, $\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{Area = Length \times Width}}}$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf 204 = (x) \times (x-5)$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf 204 = x^2 - 5x$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf x^2 - 5x - 204 = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ • Splitting the middle term:- ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf x^2 - 17x + 12x - 204 = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf x(x - 17) + 12(x - 17) = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf (x+12) (x-17) = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Now, ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ✴ $\sf x + 12 = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\bf x = -12$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ✴ $\sf x - 17 = 0$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\bf x = 17$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ✯ Length cannot be negative. Hence, ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ★ $\large{\bf\pink{x = 17}}$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Now, Width will be x - 5 ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ➪ $\sf 17 - 5$ ⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ★ $\large{\bf\pink{x = 12}}$ Hence, Length = 17m Width = 12m⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Therefore, the dimensions of the the field are 17 × 12 m².

The width of a field is 5m less than its length. If the area is 204 m^2, find the dimmensions of the field

Answer»

Answer:-

$\red{\bigstar}$ Dimensions of the field are $\large\leadsto\boxed{\tt\green{17 \times <klux>12</klux> \: m^2}}$

• Given:-

Width of field is 5m less than its length.

Area of the field is 204 m².

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• To Find:-

Dimensions of the field

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Solution:-

Let the length of the field be 'x'.

Now, given that the width is 5m less than the length.

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore,

The width of the field will be 'x-5'

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ Figure:-

$\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(2,3.5){\sf\large x m}\put(-1.4,1.4){\sf\large x-5 m}\put(2,1.4){\large\bf <klux>204</klux> $m^2$}\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}$