Question

$\dag \sf{Question \: in \: the \: attachment}$​

Answer 1

Answer:

❏ Solution:-

$\begin{gathered}\bf Given \begin{cases}\text{Length (l) =40 cm} \\ \text{Breadth (b) = 22 cm} \\ \text{Side of Square = ?}\end{cases}\end{gathered}$

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

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✰1'st part of the Problem:-

Logic Behind the solution:-

if a rectangular wire is reshaped is in the shape of a square then the perimeter of the rectangle or the length of the wire will remain same for the square,

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ 22\:cm }\put(8.1,5.3){ 40 \:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Now the wire is in the shape of a rectangle of length 40 cm and breadth 22 cm,

$\therefore$Perimeter of the rectangular wire is ,

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2(length+breadth)$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2(40+22)\:cm$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2\times62\:cm$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=124\:cm$

Now, the wire is reshaped in the shape of square,

$\setlength{\unitlength}{1.05 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.6,5.8){ B }\put(9.1,5.8){ C }\put(9.05,9.1){ D }\put(4.5,7.5){ a\:cm }\put(6,6){\line(1,0){3}}\put(6,9){\line(1,0){3}}\put(9,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Let, the side of the square is = a cm.

∴ perimeter of the square is = 4a cm.

Now, $\sf\longrightarrow Perimeter_{\red{rectangle}}=Perimeter_{\red{square}}$

$\sf\longrightarrow 120=4a$

$\sf\longrightarrow a=\frac{\cancel{124}}{\cancel4}$

$\sf\longrightarrow \boxed{\red{\large{a=31\:cm}}}$

∴ Side of the square is = 31 cm.

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

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✰2'nd part of the Problem:-

➝Area of the rectangular wire ,

$\sf\implies Area_{\red{rectangle}}=(length\times breadth)$

$\sf\implies Area_{\red{rectangle}}=(40\times 22)$

$\sf\implies Area_{\red{rectangle}}=880\:cm^2$

➝Area of the square shaped wire ,

$\sf\implies Area_{\red{square}}=(side)^2$

$\sf\implies Area_{\red{square}}=(31)^2$

$\sf\implies Area_{\red{square}}=961\:cm^2$

$\therefore$Square shaped wire would have more area than rectangular shaped wire by

= (961-880) cm²

= 81 cm²

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\large \bold{\underline{\underline{Request}}}$

Please see this answer on web to see all the diagrams ;)

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HOPES IT HELPS ;)

Answer 2

Answer:

❏ Solution:-

$\begin{gathered}\bf Given \begin{cases}\text{Length (l) =40 cm} \\ \text{Breadth (b) = 22 cm} \\ \text{Side of Square = ?}\end{cases}\end{gathered}$

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

✰1'st part of the Problem:-

Logic Behind the solution:-

if a rectangular wire is reshaped is in the shape of a square then the perimeter of the rectangle or the length of the wire will remain same for the square,

$\setlength{\unitlength}{0.8cm}\begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ 22\:cm }\put(8.1,5.3){ 40 \:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Now the wire is in the shape of a rectangle of length 40 cm and breadth 22 cm,

$\therefore$Perimeter of the rectangular wire is ,

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2(length+breadth)$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2(40+22)\:cm$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=2\times62\:cm$

$\sf\longrightarrow Perimeter_{\red{rectangle}}=124\:cm$

Now, the wire is reshaped in the shape of square,

$\setlength{\unitlength}{1.05 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.6,5.8){ B }\put(9.1,5.8){ C }\put(9.05,9.1){ D }\put(4.5,7.5){ a\:cm }\put(6,6){\line(1,0){3}}\put(6,9){\line(1,0){3}}\put(9,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Let, the side of the square is = a cm.

∴ perimeter of the square is = 4a cm.

Now, $\sf\longrightarrow Perimeter_{\red{rectangle}}=Perimeter_{\red{square}}$

$\sf\longrightarrow 120=4a$

$\sf\longrightarrow a=\frac{\cancel{124}}{\cancel4}$

$\sf\longrightarrow \boxed{\red{\large{a=31\:cm}}}$

∴ Side of the square is = 31 cm.

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

✰2'nd part of the Problem:-

➝Area of the rectangular wire ,

$\sf\implies Area_{\red{rectangle}}=(length\times breadth)$

$\sf\implies Area_{\red{rectangle}}=(40\times 22)$

$\sf\implies Area_{\red{rectangle}}=880\:cm^2$

➝Area of the square shaped wire ,

$\sf\implies Area_{\red{square}}=(side)^2$

$\sf\implies Area_{\red{square}}=(31)^2$

$\sf\implies Area_{\red{square}}=961\:cm^2$

$\therefore$Square shaped wire would have more area than rectangular shaped wire by

= (961-880) cm²

= 81 cm²

$\setlength{\unitlength}{0.8cm} \begin{picture}(10,5)\thicklines\qbezier(1,1)(1,1)(9,1)\end{picture}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\large \bold{\underline{\underline{Request}}}$

Please see this answer on web to see all the diagrams ;)

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

HOPES IT HELPS ;)