Question

The width of rectangular stadium is 1/2 of its length of the area of stadium is 1800 sw.m what is difference between its length of breadth

Answer 1

Given : Width of the Stadium is 1/2 of its Length . It's Area is 1800 m² .

To Find : Find the Length and Breadth of Stadium

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SolutioN :

$\maltese$ According to the Question :

$\longmapsto$ Let the Length be y .So :

$\qquad \; {\pmb{\sf{ Length = y \; m }}}$

$\longmapsto$ Width is 1/2 of the length .So :

$\qquad \; {\pmb{\sf{ Width = \dfrac{y}{2} \; m }}}$

$\maltese$ Formula Used :

$\qquad \; {\orange{\bigstar \; {\purple{\underbrace{\underline{\red{\sf{ Area = Length \times Breadth }}}}}}}}$

$\maltese$ Calculating the Value of y :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = Length \times Breadth } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 1800 = y \times \dfrac{y}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 1800 = y \times \dfrac{y}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 1800 = \dfrac{ {y}^{2} }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 1800 \times 2 = {y}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 3600 = {y}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sqrt{3600} = y } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\frak{ y = 60 }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered}$

$\maltese$ Calculating the Dimensions :

• Length = y = 60 m
• Breadth = y/2 = 60/2 = 30 m

$\therefore \;$ Length of the Stadium is 60 m and width is 30 m .

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Answer 2

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

We are given that the width of the Rectangle is half of its length.

so let we have,

the length of the rectangle = A

and the width of the Rectangle = A/2

We know that the perimeter of the Rectangle = 2 (length + breadth)

Perimeter = 2 (A + A/2)

Perimeter = 2 *3A/2 = 3A

But we are given that the perimeter of the rectangle is 63 inches.

so 63 = 3A

A = 63/3 = 21 inches.

So the length of the rectangle = 21 inch

and the width = 21/2 = 10.5 inch