Question

The perimeter of a rectangular field is 88 metre and its area is 420 square
metre. Find the length and the breadth of the rectanguler field.​

Answer 1

Given :

• Perimeter of the rectangle is 88 m.
• Area of the rectangle is 420 m²

To Find :

• Length of the rectangle.
• Breadth of the rectangle.

Solution :

Let the length of the rectangle be x m.

Let the breadth of the rectangle be y m.

Case 1 :

The perimeter of the rectangle is 88 m.

Equation :

$\sf{\longrightarrow{Perimeter\:=\:2(length\:+\:breadth)}}$

$\sf{\longrightarrow{88=2(x+y)}}$

$\sf{\longrightarrow{\dfrac{88}{2}\:=\:x+y}}$

$\sf{\longrightarrow{44=x+y}}$

$\sf{\longrightarrow{44-y=x\:\:\:\:\:(1)}}$

Case 2 :

The area of the rectangle is 420 m².

Equation :

$\sf{\longrightarrow{Area\:=\:length\:\times\:breadth}}$

$\sf{\longrightarrow{420=x\:\times\:y}}$

$\sf{\longrightarrow{420=(44-y)y}}$

$\sf{\longrightarrow{420=44y-y^2}}$

$\sf{\longrightarrow{44y-y^2=420}}$

$\sf{\longrightarrow{-44y+y^2=-420}}$

$\bold{\Big[Dividing\:by\:minus\Big]}$

$\sf{\longrightarrow{y^2-44y+420=0}}$

$\sf{\longrightarrow{y^2-30y-14y+420=0}}$

$\sf{\longrightarrow{y(y-30)-14(y-30)=0 }}$

$\sf{\longrightarrow{(y-30) (y-14) =0}}$

$\sf{\longrightarrow{y-30=0\:\:\:or\:\:\:y-14=0}}$

$\sf{\longrightarrow{y=30\:\:\:or\:\:\:\:y=14}}$

When, y = 30 m,

$\sf{\longrightarrow{44-y=x}}$

$\sf{\longrightarrow{44-30=x}}$

$\sf{\longrightarrow{14=x}}$

$\large{\boxed{\bold{Length\:of\:rectangle\:=\:x\:=\:14\:m}}}$

$\large{\boxed{\bold{Breadth\:of\:rectangle\:=\:y\:=\:30\:m}}}$

When, y = 14 m,

$\sf{\longrightarrow{44-y=x}}$

$\sf{\longrightarrow{44-14=x}}$

$\sf{\longrightarrow{30=x}}$

$\large{\boxed{\bold{Length\:of\:rectangle\:=\:x\:=\:30\:m}}}$

$\large{\boxed{\bold{Breadth\:of\:rectangle\:=\:y\:=\:14\:m}}}$

Answer 2

$\huge{\bf{\underline{\purple{Solution :}}}}$

Given,

• Perimeter of Rectangular field = 88 m
• Area of Rectangular field = 420 m²

To Find,

• Length and breadth of the rectangular field = ?

Solution,

Suppose the length of Rectangular field be a

And , Suppose the breadth be b

$\boxed{\bold{\red{According \ to \ Question :}}}$

$\implies \bold{ Perimeter \ of \ Perimeter = 2(a +b)}$

$\implies \bold{2(a + b) = 88}$

$\implies \bold{a + b = \frac{88}{2}}$

$\implies \bold{a + b = 44}$

$\implies \boxed{\bold{ a = 44 - b}}$........1) Equation

$\boxed{\bold{\red{Now \ Again \ According \ to \ Question :}}}$

$\implies \bold{Area \ of \ Rectangle = a \times b}$

$\implies \bold{(44 - b) \times b}}$

From First Equation :

$\implies \bold{(44b - b^{2} ) = 420 }$

$\implies \bold{b^{2} - 44b + 420 = 0}$

$\bold{\red{Use \ Middle \ term \ splitting \ method :}}$

$\implies \bold{b^{2} - 30b -14b + 420 = 0}$

$\implies \bold{b( b - 30) - 14 (y - 30) = 0}$

$\implies \bold{(b - 30) ( b - 14) = 0}$

$\implies \bold{ b - 30 = 0 \ or\ b - 14 = 0}$

$\implies \bold{\therefore b = 30 \ or \ b = 14}$

If Breadth = 30 then,

$\implies \bold{ 44 - b = a}$

$\implies \bold{44 - 30 = a}$

$\implies \bold{a = 14}$

∴Length = 14 and Breadth = 30

If Breadth = 14 then,

$\implies \bold{44 - b = a}$

$\implies \bold{ 44 - 14 = a}$

$\implies \bold{a = 30}$

∴Length = 30 and Breadth = 14

$\rule{200}2$