Math, asked by sourishchakraborty61, 9 months ago

The width of a rectange is two-thirds its length. If the perimeter is 180 metres, find the
dimensions of the rectangle . ​

Answers

Answered by SarcasticL0ve
18

GivEn:

  • The width of a Rectangle is two-thirds its length.

  • Perimeter of Rectangle is 180 m

To find:

  • Dimensions of Rectangle.

SoluTion:

According to question,

★ The width of a rectangle is two-thirds its length.

So, Let's length of Rectangle be l.

Therefore, width of a rectangle is \sf \dfrac{2}{3}l.

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We have,

Perimeter of Rectangle = 180 m.

As we know that,

\star\;{\boxed{\sf{\purple{Perimeter_{\;(Rectangle)} = 2(l + b)}}}}

Therefore,

:\implies\sf 2 \bigg( l + \dfrac{2}{3}l \bigg) = 180

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:\implies\sf 2 \bigg( \dfrac{3l + 2l}{3} \bigg) = 180

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:\implies\sf 2 \bigg( \dfrac{5l}{3} \bigg) = 180

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:\implies\sf \dfrac{10l}{3} = 180

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:\implies\sf 10l = 180 \times 3

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:\implies\sf l = \cancel{ \dfrac{540}{3}}

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:\implies{\underline{\boxed{\sf{\pink{54\;m}}}}}\;\bigstar

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\therefore Hence, Length of Rectangle is 54 m.

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Therefore, Width of Rectangle is,

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:\implies\sf \dfrac{2}{3} \times 54 = 36 m

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DIAGRAM:

\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.3,2){\sf{\large{36 m}}}\put(9.2,0.7){\sf{\large{54 m}}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\end{picture}

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Hence, Dimensions of Rectangle ( length and breadth ) are 54 m and 36 m respectively.

Answered by Anonymous
6

AnswEr:

GivEn:-

\qquad\bullet\normalsize\sf\ Width \: of \: rectangle(w) = \frac{2}{3}Length

\qquad\bullet\normalsize\sf\ Perimeter \: of \: rectangle(P) \: = 180m

To Find:-

\qquad\bullet\normalsize\sf\ Dimensions \: \left( Length \: \& \: width \right) \\ \qquad\sf\ of \: rectangle

Solution:-

\underline{\bigstar\:\sf{According \: to \: given \: in \: question :}}

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As we know that, width of rectangle is also called breadth of rectangle and Perimeter of rectangle is two times of sum of length and breadth;

\quad\dashrightarrow\sf\ P= 2 \left(L + B \right)

Put the known values in formula;

\quad\dashrightarrow\sf\ 180 = 2 \left(L + \frac{2}{3}L \right)

\qquad\scriptsize\dag\sf\ Take \: the \: LCM

\quad\dashrightarrow\sf\ 180 = 2 \left(\frac{3 + 2}{3}L \right)

\quad\dashrightarrow\sf\ 180 = 2 \left(\frac{5}{3}L \right)

\quad\dashrightarrow\sf\ P=  \left(\frac{10}{3}L \right)

\quad\dashrightarrow\sf\frac{180 \times\ 3}{10} = L

\quad\dashrightarrow\sf\frac{\cancel{180} \times\ 3}{\cancel{10}} = L

\quad\dashrightarrow\normalsize{\underline{\boxed{\mathsf \red{L = 54m}}}}

_________________

Now; Dimensions (L & W)

⋆Length of rectangle :

\normalsize\ : \implies\sf\ Length = 54m

⋆Width of rectangle :

\normalsize\ : \implies\sf\ Width = \frac{2}{3}L

\normalsize\ : \implies\sf\ Width = \frac{2}{\cancel{3}} \times\ \: \cancel{54}

\normalsize\ : \implies\sf\ Width = 36m

\therefore\:\underline{\textsf{Hence, \: the \: dimensions \: are}{\textbf{\: 54m \: and \:  36m}}}

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