Question

Area of square field is 5184 metre square . A rectangular field whose length is twice of it's breadth has it's perimeter of square field. Find the area of rectangular field.

Answer 1

Answer:

Breadth is 864 and length is 2×864=1728. Area of rectangular field L×B=864×1728=1492992

Answer 2

Answer:

${\large{\underline{\underline{\textsf{\textbf{Given\:: -}}}}}}$

• ↝ Area of square field

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

${\large{\underline{\underline{\textsf{\textbf{To Find\:: -}}}}}}$

• ↝ Area of Rectangle

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

${\large{\underline{\underline{\textsf{\textbf{Using Formulae\:: -}}}}}}$

$\small{\bigstar{\underline{\boxed{\bf{\pink{Side \: of \: square = \sqrt{Area \: of \: square}}}}}}}$

$\small{\bigstar{\underline{\boxed{\bf{\pink{Perimeter \: of \: square = 4a}}}}}}$

$\small{\bigstar{\underline{\boxed{\bf{\pink{Perimeter \: of \: rectangle = 2(Lenght + Breadth)}}}}}}$

$\small{\bigstar{\underline{\boxed{\bf{\pink{Area \: of \: Rectangle = Length \times Breadth}}}}}}$

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

${\large{\underline{\underline{\textsf{\textbf{Solution\:: -}}}}}}$

$\red\bigstar$ Fistly, Finding the side of square :-

${\dashrightarrow{\sf{Side_{(Square)} = \sqrt{Area \: of \: square}}}}$

• Substuting the value

${\dashrightarrow{\sf{Side_{(Square)} = \sqrt{5184 \: {m}^{2} }}}}$

${\dashrightarrow{\sf{Side_{(Square)} = \sqrt{72 \times 72}}}}$

${\dashrightarrow{\sf{Side_{(Square)} = {72 \: m}}}}$

${\bigstar{\purple{\underline{\boxed{\bf{Side \: of \: square = {72 \: m}}}}}}}$

∴ The side of square is 72 m.

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

$\red\bigstar$ Now, Finding the perimeter of square :-

${\dashrightarrow{\sf{Perimeter_{(Square)} = 4a}}}$

• Substuting the values

${\dashrightarrow{\sf{Perimeter_{(Square)} = 4(72)}}}$

${\dashrightarrow{\sf{Perimeter_{(Square)} = 4\times 72}}}$

${\dashrightarrow{\sf{Perimeter_{(Square)} = 288 \: m}}}$

${\bigstar{\underline{\boxed{\bf{\purple{Perimeter \: of \: square = 288 \: m}}}}}}$

∴ The perimeter of square is 288 m.

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

$\red\bigstar$ Here :-

• ↝ Perimeter of Square = Perimeter of Rectangle

Let the,

• ↝ Breadth of Rectangle = x
• ↝ Lenght of Rectangle = twice of breath (2×x) = 2x

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

$\red\bigstar$ Now, Finding the sides of rectangle :-

$\small{\dashrightarrow{\sf{Perimeter_{(Rectangle)} = 2(Lenght + Breadth)}}}$

• Substuting the values

${\dashrightarrow{\sf{288 \: m = 2(2x + x)}}}$

${\dashrightarrow{\sf{288 \: m = 2(3x)}}}$

${\dashrightarrow{\sf{\dfrac{288}{2} = 3x}}}$

${\dashrightarrow{\sf{\cancel{\dfrac{288}{2}} = 3x}}}$

${\dashrightarrow{\sf{144 = 3x}}}$

${\dashrightarrow{\sf{x = \dfrac{144}{3} }}}$

${\dashrightarrow{\sf{x = \cancel{\dfrac{144}{3}}}}}$

${\dashrightarrow{\sf{x = 48 \: m }}}$

${\bigstar{\underline{\boxed{\bf{\purple{x = 48 \: m }}}}}}$

∴ The value of x is 48 m.

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

$\red\bigstar$ Thus :-

• Breadth of Rectangle = 48 m
• Lenght of Rectangle = 48×2 = 96 m

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

$\red\bigstar$ Now, Finding the area of rectangle :-

${\dashrightarrow{\sf{Area_{(Rectangle)} = Lenght \times Breadth}}}$

• Substuting the values

${\dashrightarrow{\sf{Area_{(Rectangle)} = 96 \: m\times 48 \: m}}}$

${\dashrightarrow{\sf{Area_{(Rectangle)} = 4608 \: {m}^{2} }}}$

${\bigstar{\underline{\boxed{\bf{\purple{Area \: of \: rectangle = 4608 \: {m}^{2} }}}}}}$

∴ The area of rectangular field is 4608 m².

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

${\large{\underline{\underline{\textsf{\textbf{Diagram\:: -}}}}}}$

$\red\bigstar$ Diagram of square :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 72\ m}\put(4.4,2){\bf\large 72\ m}\end{picture}$

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

$\red\bigstar$ Diagram of Rectangle :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 96\ m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 48\ m}\put(-0.5,-0.4){\bf}\put(-0.5,3.2){\bf}\put(5.3,-0.4){\bf}\put(5.3,3.2){\bf}\end{picture}$

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

${\large{\underline{\underline{\textsf{\textbf{Request\:: -}}}}}}$

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