Question

perimeter of rectangle is 36 cm.Diagonal is root 164
find length and breadth​

Answer 1

l=P/2﹣w

This is the formula of the question

Answer 2

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{perimeter \: of \: rectangle \: = \: 36 \: cm} \\ &\sf{diagonal \: of \: rectangle \: = \sqrt{164} \: cm} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{length \: of \: rectangle} \\ &\sf{breadth \: of \: rectangle} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{\bf \:Perimeter \:of\:Rectangle=2(Length + Breadth)}$

$\boxed{\bf \:  {(diagonal)}^{2} \: = {(Length)}^{2} + {(Breadth)}^{2} }$

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$\large\underline{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{length \: of \: rectangle \: be \: x \: cm} \\ &\sf{breadth \: of \: rectangle \: be \: y \: cm}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bold{❥︎Step :- 1 }}$

$\sf \:  ⟼ ✬ \: Perimeter \: of \: rectangle \: = \: 36 \: cm$

$\sf \:  ⟼ 2(Length + Breadth) = 36$

$\sf \:  ⟼ \: 2(x + y) = 36$

$\sf \:  ⟼ \: x + y = 18$

$\bf\implies \:y \: = 18 - x - - - - - (1)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

$\bf \:  ⟼  ✬ \: Diagonal \: of \: rectangle \: = \sqrt{164} \: cm$

 ✬ We know that

$\bf \:  {(diagonal)}^{2} \: = {(Length)}^{2} + {(Breadth)}^{2}$

$\bf\implies \: { (\sqrt{164} )}^{2} = {x}^{2} + {y}^{2}$

$\sf \:  ⟼ \: 164 = {x}^{2} + {y}^{2}$

$\sf \:  ⟼164 = {x}^{2} + {(18 - x)}^{2} - - ( \because \: of \: using \: (1))$

$\sf \:  ⟼ {x}^{2} + 324 + {x}^{2} - 36x = 164$

$\sf \:  ⟼ {2x}^{2} - 36x + 160 = 0$

 ✬ On dividing by 2 both sides, we get

$\sf \:  ⟼ {x}^{2} - 18x + 80 = 0$

$\sf \:  ⟼ \: {x}^{2} - 10x - 8x + 80 = 0$

$\sf \:  ⟼x(x - 10) - 8(x - 10) = 0$

$\sf \:  ⟼ \: (x - 10)(x - 8) = 0$

$\bf\implies \:x \: = \: 10\: or \: x \: = \: 8$

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Case :- 1

☆ When x = 10, Put in equation (1), we get y = 8

Case :- 2

☆ When x = 8, Put in equation (1), we get y = 10.

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$\begin{gathered}\begin{gathered}\bf Hence \begin{cases} &\sf{length \: of \: rectangle \: = \: 10 \: cm} \\ &\sf{breadth \: of \: rectangle \: = \: 8 \: cm}\end{cases}\end{gathered}\end{gathered}$

Or

$\begin{gathered}\begin{gathered}\bf Hence \begin{cases} &\sf{length \: of \: rectangle \: = \: 8 \: cm} \\ &\sf{breadth \: of \: rectangle \: = \: 10 \: cm}\end{cases}\end{gathered}\end{gathered}$