Question

A rectangular playground is 420 sq.m. If it's length is increases by 7 m and breadth is decreased by 5 m , the area remains the same . Find the length and breadth of the playground​

Answer 1

$\sf\large\underline\green{Let:-}$

$\sf{\implies The\: length\:_{(playground)}=x}$

$\sf{\implies The\: breadth\:_{(playground)}=y}$

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

$\sf{\implies Area\:_{(playground)}=420\:m^2}$

$\sf\large\underline\green{To\: Find:-}$

$\sf{\implies The\: dimensions\:_{(playground)}=?}$

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

To calculate the dimensions of rectangular playground , at first we have to set up equation as per the given Question by applying formula. Then solve those equation after that you get the value of x and y where x is length of playground and y is the breadth of playground.

$\sf\small\underline\blue{Calculation\: for\: first\: equation:-}$

$\tt{\implies Area\:_{(playground)}=length*breadth}$

$\tt{\implies x*y=420}$

$\tt{\implies x=\dfrac{420}{y}------(i)}$

$\sf\small\underline\blue{Calculation\: for\:second\: equation:-}$

$\tt{\implies Length\:_{(increased\:by\:7)}*breadth\:_{(decreased\:by\:5)}=Area\:_{(remain\: same)}}$

$\tt{\implies (x+7)*(y-5)=x*y}$

$\tt{\implies xy-5x+7y-35=xy}$

$\tt{\implies -5x+7y=35-----(ii)}$

• In eq (ii) putting the value of x=420/y :-]

$\tt{\implies -5*\dfrac{420}{y}+7y=35}$

$\tt{\implies \dfrac{-2100}{y}+7y=35}$

$\tt{\implies \dfrac{-2100+7y^2}{y}=35}$

$\tt{\implies 7y^2-35y-2100=0}$

• Here dividing by 7 on both sides:-]

$\tt{\implies y^2-5y-300=0}$

• Now splitting the middle term here:-]

$\tt{\implies y^2-20y+15y-300=0}$

$\tt{\implies y(y-20)+15(y-20)}$

$\tt{\implies (y+15)(y-20)=0}$

$\tt{\implies \therefore\:y=-15\:\:,20}$

• Here negative values can't be the dimensions of rectangle so, we take y=20 here:-]

$\tt{\implies breadth\:_{(playground)}=20m}$

• Now putting the value of y=20 in eq (i) :-]

$\tt{\implies x=\dfrac{420}{y}}$

$\tt{\implies x=\dfrac{420}{20}}$

$\tt{\implies x=21m}$

$\sf\large{Hence,}$

$\sf{\implies The\: length\:_{(playground)}=21\:m}$

$\sf{\implies The\: breadth\:_{(playground)}=20\:m}$

Answer 2

Area of a rectangular playground = 420m²

Let, the length be x

the breadth be y

ATP

x × y = 420 (i)

(x+7)(y-5) = 420 (ii)

$(i) x \times y = 420 \\ = ) \: xy = 420 \\ = ) \: x = \frac{420}{y}$

Putting the value of x in equ (ii)

$(ii) \: (x + 7)(y - 5) = 420 \\ = ) \: ( \frac{420}{y} + 7)(y - 5) = 420 \\ = ) \: 420 - \frac{2100}{y} + 7y - 35 = 420 \\ = ) \: - \frac{2100}{y} + 7y = 420 - 420 + 35 \\ = ) \frac{ - 2100 + 7 {y}^{2} }{y} = 35 \\ = ) \: - 2100 + 7 {y}^{2} = 35y \\ = ) \: 7 {y}^{2} - 35y - 2100 = 0 \\ now \: we \: have \: to \: divide \: the \: numbers \: by \: 7 \\ = ) \: {y}^{2} - 5y - 300 = 0 \\ now \: we \: have \: to \: use \: the \: middle \: term \: method \\ = ) \: {y}^{2} - (20 - 15)y - 300 = 0 \\ = ) \: {y}^{2} - 20y + 15y - 300 = 0 \\ = ) \: y(y - 20) + 15(y - 20) = 0 \\ = ) \: (y - 20)(y + 15) = 0 \\ y = 20 \: and \: y = - 15 \\ so \: in \: this \: case \: we \: will \: take \: the \: value \: of \: y \: as \: 20.$

Putting the value of x in equ (i)

$(i) \: x \times y = 420 \\ = ) \: 20x = 420 \\ = ) \: x = \frac{420}{20} = 21$

Ans:- Original length = 21m

Original breadth = 20m