Question

The width of a rectangle is four-fifth its length. If the perimeter of the rectangle is 90 m, find the length and breadth of the rectangle.​

Answer 1

Let the length of the rectangle be x . Then, the breadth of the rectangle

=

$\frac{4}{5}x$

According to the question, perimeter of the rectangle

=

$90m$

$= 2(x + \frac{4}{5}x) = 90$

$= x + \frac{4}{5}x = \frac{90}{2} (transposing \: 2 \: to \: rhs)$

$= \frac{5x + 4x}{5} = 45$

$= \frac{9x}{5} = 45$

$x = \frac{45 \times 5}{9} \: (transposing \: 9 \: and \: 5 \: to \: rhs)$

$x = 25$

Therefore length of the rectangle, x= 25m and breadth of the rectangle,

$\frac{4}{5}x = \frac{4}{5} \times 25 = 20m$

$\red{answer} \: \blue{ = 20m}$

Answer 2

Answer:

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Diagram : - }}}}}}\end{gathered}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 25 m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 20 m}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Given : - }}}}}}\end{gathered}$

• → The width of a rectangle is four-fifth its length.
• → The perimeter of rectangle is 90 m.

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:To \: Find : - }}}}}}\end{gathered}$

• → Lenght of Rectangle
• → Breadth of Rectangle

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Using \: Formula : - }}}}}}\end{gathered}$

$\small\bigstar{\underline{\boxed{\sf{Perimeter \: of \: rectangle =2\bigg(Length + Breadth \bigg)}}}}$

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Solution : - }}}}}}\end{gathered}$

$\small\bigstar$ Here, Let the :-

• Breadth = x m
• Length = ⅘x

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

$\small\bigstar$ Now, According to the question :-

$\small{\longrightarrow{\sf{Perimeter_{(Rectangle)} =2\bigg(Length + Breadth \bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg(x + \dfrac{4}{5}x \bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg(\dfrac{(x \times 5) + (4x \times 1)}{5}\bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg(\dfrac{5x + 4x}{5}\bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg( \: \dfrac{9x}{5} \: \bigg)}}}$

$\small{\longrightarrow{\sf{\dfrac{90}{2} =\dfrac{9x}{5} }}}$

$\small{\longrightarrow{\sf{\cancel{\dfrac{90}{2}} =\dfrac{9x}{5} }}}$

$\small{\longrightarrow{\sf{45=\dfrac{9x}{5} }}}$

$\small{\longrightarrow{\sf{x = 45 \times \dfrac{5}{9} }}}$

$\small{\longrightarrow{\sf{x = \cancel{45} \times \dfrac{5}{\cancel{9}}}}}$

$\small{\longrightarrow{\sf{x = 5 \times 5}}}$

$\small{\longrightarrow{\underline{\underline{\sf{x = 25 \: m}}}}}$

$\normalsize{\longrightarrow{\underline{\underline{\sf{\pink{x = 25 \: m}}}}}}$

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

$\small\bigstar$ Therefore :-

$\quad{: \implies{\sf{Lenght = \bf{\purple{25 m}}}}}$

$\quad{:\implies{\sf{Breadth = \dfrac{4}{5} \times 25 = \dfrac{100}{5} = \bf \purple{20m}}}}$

Hence, the lenght of rectangle is 25m and the breadth of rectangle 20m.

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Verification : - }}}}}}\end{gathered}$

$\small\bigstar$ Let's check our answer :-

$\small{\longrightarrow{\sf{Perimeter_{(Rectangle)} =2\bigg(Length + Breadth \bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg(20 + 25\bigg)}}}$

$\small{\longrightarrow{\sf{90m =2\bigg( \: 45\: \bigg)}}}$

$\small{\longrightarrow{\sf{90m =2 \times 45}}}$

$\small{\longrightarrow{\sf{90m = 90m}}}$

$\normalsize{\longrightarrow{\underline{\underline{\sf{\pink{LHS =RHS}}}}}}$

Hence Verified!

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

$\begin{gathered}{\underline{\underline{\maltese{\bf{\red{\:Learn \: More : - }}}}}}\end{gathered}$

$\small\boxed{\begin{gathered} \dag\quad\underline{\underline{\bf Formulas\:of\:Areas}}\quad\dag\\ \\ \dashrightarrow\sf Square=(side)^2\\ \\ \dashrightarrow\sf Rectangle=Length\times Breadth \\\\ \dashrightarrow\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \dashrightarrow\sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \dashrightarrow\sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \dashrightarrow\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \dashrightarrow\sf Parallelogram =Breadth\times Height\\\\ \dashrightarrow\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \dashrightarrow\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end{gathered}}$

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