Question

A wire is in the shape of a rectangle . Its length is 40cm and breadth is 22cm . If the same wire is rebent in the shape of square, what will be the measure of each side. Also find which shape encloses more area.

Answer 1

$\huge\underline\mathrm{Question-}$

A wire is in the shape of a rectangle . Its length is 40cm and breadth is 22cm . If the same wire is rebent in the shape of square, what will be the measure of each side. Also find which shape encloses more area.

$\huge\underline\mathrm{Answer-}$

• Measure of each side is 31 cm.
• Square encloses more area.

$\huge\underline\mathrm{Explanation-}$

$\begin{lgathered}\bold{Given} \begin{cases}\sf{Length\:of\:rectangle\:=\:40\:cm} \\ \sf{Breadth\:of\:rectangle\:=\:22\:cm}\end{cases}\end{lgathered}$

To find :

• The measure of each side if wire is rebent in the shape of square.

• Among rectangle and square, which has more are.

Solution :

We know that,

$\large{\boxed{\rm{\pink{Perimeter\:of\:rectangle\:=\:2(L+B)}}}}$

$\begin{lgathered}\bold{Where} \begin{cases}\sf{\red{L\:refers\:to\:length\:of\:rectangle}} \\ \sf{\green{B\:refers\:to\:breadth\:of\:rectangle}}\end{cases}\end{lgathered}$

Now putting the values,

$\longmapsto$ $\rm{Perimeter\:of\:wire\:=\:2(40+22)}$

$\longmapsto$ $\rm{Perimeter\:of\:wire\:=\:2(62)}$

$\longmapsto$ $\rm{Perimeter\:of\:wire\:=124\:cm}$

It is given that the rectangular wire is rebent in the shape of square, so perimeter should be equal.

$\therefore$ $\rm{Perimeter\:of\:square\:=\:Perimeter\:of\:rectangle}$

$\longmapsto$ $\rm{4a\:=\:124}$

$\longmapsto$ $\rm{a\:=\:\cancel{\dfrac{124}{4}}}$

$\longmapsto$ $\rm{a\:=\:31\:cm}$

Therefore, each side of square is 31 cm.

$\rule{200}2$

★ Area of square = a²

$\longmapsto$ 31 × 31

$\longmapsto$ 961 cm²

★ Area of rectangle = L × B

$\longmapsto$ 40 × 22

$\longmapsto$ 880 cm²

$\therefore$ Square encloses more area.

Answer 2

$\bf{\Huge{\boxed{\tt{\purple{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

A wire is in the shape of a rectangle. Its length is 40cm & breadth is 22cm. If the same wire is rebent in the shape of square.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The measure of each side & also which shape enclose more area.

$\bf{\Large{\underline{\tt{\green{Explanation\::}}}}}$

According to the question;

$\bigstar\:$ Length of the wire will be the same.

$\leadsto\sf{\pink{Perimeter\:of\:rectangle\:=\:Perimeter\:of\:square}}$

$\bf{We\:have}\begin{cases}\sf{The\:length\:of\:rectangle\:(L)=\:40cm}\\ \sf{The\:breadth\:of\:rectangle\:(B)=\:22cm}\\ \sf{Perimeter\:of\:rectangle\:=\:2(Length+breadth)}\\ \sf{Perimeter\:of\:square\:=\:4*side}\end{cases}}$

$\bigstar\:$ Let the side of square be R.

Now,

$\implies\sf{2(L+B)\:=\:4R}$

$\implies\sf{2(40cm+22cm)\:=\:4R}$

$\implies\sf{2(62cm)\:=\:4R}$

$\implies\sf{124cm\:=\:4R}$

$\implies\sf{R\:=\:\cancel{\dfrac{124}{4} }cm}$

$\implies\sf\boxed{\tt{\red{R\:=\:31cm}}}$

_________________________________

$\bf{\Large{\boxed{\rm{\bigstar\:{Area\:of\:rectangle\:\:\&\:\:square\::}}}}}$

$\bf{Formula}\begin{cases}\sf{\pink{Area\:of\:rectangle\:=\:(Length*breadth)}}}\\ \sf{\orange{Area\:of\:square\:=\:(Side*Side)}}}\end{cases}}$

Rectangle :

$\mapsto\sf{Area\:=\:(40*22)cm^{2} }$

$\mapsto\sf{\purple{Area\:=\:880cm^{2} }}$

Square :

$\mapsto\sf{Area\:=\:(31*31)cm^{2} }$

$\mapsto\sf{\purple{Area\:=\:961cm^{2} }}$

Thus,

$\bf{\Large{\boxed{\sf{The\:area\:of\:square\:is\:greater\:then\:area\:of\:rectangle.}}}}}$