Question

the side of squre field is 50cm lang .find 1)area 2)perimeter​

Answer 1

$\color{skyblue}{\bigstar \large\underline{\boxed{\bf\pink{Question:-}}}}$

The side of square field is 50cm long . find 1) area 2) perimete

$\color{skyblue}{\bigstar \large\underline{\boxed{\bf\pink{Answer:-}}}}$

1) Given :-

• Side of square = 50 cm

$\sf \green{Area \tiny(Square)} \\ \sf \red{ \mapsto \: side \times side }$

$\sf \orange{ \mapsto50 \times 50 \: cm}$

$\sf \blue{ \mapsto \: 2500\: cm }$

$\pink{ \underline{ \boxed{ \sf { \therefore \: The \: area\: of \: square = 2500 \: cm^2}}}}$

✧══════════•❁❀❁•══════════✧

2) Given :-

• Side of square = 50 cm

$\sf \purple{Perimeter _ {(Rectangle)}}$

$\sf\green{\mapsto 4 × side}$

$\sf \color{violet} \mapsto \: 4 \times 50 \: cm \:$

$\sf \color{yellow} \mapsto \: 200 \: cm$

$\pink{ \underline{ \boxed{ \sf { \therefore \: The \: Perimeter \: of \: square = 200cm}}}}$

Answer 2

Answer:

${\large{\sf{\pmb{\underline{\purple{Given:-}}}}}}$

• ● The side of square is 50 cm.

${\large{\sf{\pmb{\underline{\purple{To \: Find:-}}}}}}$

• ● 1) Area of Square
• ● 2) Perimeter of Square

${\large{\sf{\pmb{\underline{\purple{Using \: Formula:-}}}}}}$

${\dag{\underline{\boxed{\sf{\pink{Area \: of \: Square= {(Side)}^{2} }}}}}}$

${\dag{\underline{\boxed{\sf{\pink{Perimeter \: of \: Square =4 \times Side }}}}}}$

${\large{\sf{\pmb{\underline{\purple{Solution:-}}}}}}$

${\underline{\pmb{\frak{\bigstar \: Finding \: the \: area \: of \: Square...}}}}$

${: \implies{\sf{Area \: of \: Square= \bf{(Side)}^{2} }}}$

• ● Substituting the values

${: \implies{\sf{Area \: of \: Square= \bf{(50 \: cm)}^{2} }}}$

${: \implies{\sf{Area \: of \: Square= \bf{50 \: cm \times 50 \: cm}}}}$

${: \implies{\sf{Area \: of \: Square= \bf{2500 \: {cm}^{2}}}}}$

${\dag{\underline{\boxed{\sf{\pink{Area \: of \: Square= {2500 \: {cm}^{2}}}}}}}}$

● Henceforth,The Area of Square is 2500 cm².

$\begin{gathered} \\ \end{gathered}$

${\underline{\pmb{\frak{\bigstar \: Finding \: the \: perimeter \: of \: Square...}}}}$

${: \implies{\sf{Perimeter \: of \: Square = \bf{4 \times Side }}}}$

• ● Substituting the values

${: \implies{\sf{Perimeter \: of \: Square = \bf{4 \times 50 \: cm }}}}$

${: \implies{\sf{Perimeter \: of \: Square = \bf{200 \: cm }}}}$

${\dag{\underline{\boxed{\sf{\pink{Perimeter \: of \: Square =200 \: cm }}}}}}$

● Henceforth,The Perimeter of Square is 200 cm.

${\large{\sf{\pmb{\underline{\purple{Diagram:-}}}}}}$

${\underline{\pmb{\frak{\bigstar \: Here's \: the \: diagram \: of \: Square...}}}}$

${\star \: {\sf{\big(For \: App \: Users \big)}}}$

$\begin{gathered}\begin{gathered}\begin{gathered} {\begin{gathered} \sf{50 \:cm}\huge\boxed{ \begin{array}{cc} \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \end{array}} \\ \: \: \: \: \: \sf{50 \: cm} \end{gathered}}\end{gathered}\end{gathered}\end{gathered}$

• ~Side of Rectangle = 50 cm

$\begin{gathered} \\ \end{gathered}$

${\star \: {\sf{\big(For \: Web \: Users \big)}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 50\ cm}\put(4.4,2){\bf\large 50\ cm}\end{picture}$

• ~Request :- Please see the answer from website Brainly.in..

${\large{\sf{\pmb{\underline{\purple{Know \: More:-}}}}}}$

$\begin{gathered} \small\begin{gathered}\bigstar \: \bf\underline{More \: Useful \: Formulae } \: \bigstar  \\ \begin{gathered}{\boxed{\begin{array} {cccc}{\sf{{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}} \\  \\{\sf{{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}  \\  \\{\sf{{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}} \\  \\ {\sf{{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}} \\   \\ {\sf{{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\  \\ {\sf{{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\  \\{\sf{{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\  \\ {\sf{{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\  \\ {\sf{{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\  \\ {\sf{{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\  \\ {\sf{{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\  \\{\sf{{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}\end{array}}}\end{gathered}\end{gathered}\end{gathered}$