Question

The length and width of the rectangle are in the ratio 4:3. If the perimeter of the
rectangle is 56 m find the length and width of the rectangle.
.
Please answer with proper steps tomorrow is my exam​

Answer 1

Given: Ratio of length and breadth of a rectangular is 4:3. & Perimeter of rectangle is 56m.

Need to find: Dimensions of rectangle?

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❍ Let's consider length and width of rectangle be 4x and 3x.

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As we know that,

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$\begin{gathered}\star\:{\underline{\boxed{\frak{Perimeter_{\:(rectangle)} = 2(Length + width)}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 56 = 2(4x + 3x)\\\\\\ :\implies\sf 56 = 2(7x) \\\\\\ :\implies\sf 56 = 14x \\\\\\ :\implies\sf x = \cancel{\dfrac{56}{14}} \\\\\\ :\implies{\underline{\boxed{\frak{\purple{x = 4}}}}}\:\bigstar\\\\\end{gathered}$

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Therefore,

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Length of rectangular park, 4x = 16m

Widthth of rectangular park, 3x = 12m

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$\therefore\:{\underline{\sf{Hence,\:Dimensions\:of\:rectangle\:are\:\bf{16\:m}\: \sf{and}\: \bf{12\:m}\: \sf{respectively}.}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 16m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 12m}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

Answer 2

$\large\mathfrak{\pmb{{\underline{Given}}:}}$

• The length and breadth of the rectangle are in ratio　4 : 3
• Perimeter of the rectangle is 56m

$\large\mathfrak{\pmb{{\underline{To~find}}:}}$

• The length and breadth of the rectangle

$\large\mathfrak{\pmb{{\underline{Solution}}:}}$

• Let the length and breadth of the rectangle be 4x and 3x respectively

$\sf Perimeter \: of \: a \: rectangle = 2(l + b) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 56m= 2(4x + 3x) \\ \sf : \implies 56 = 2(7x) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 56 = 14x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 14x = 56 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies x = \frac{\cancel{56}~~^{4}}{\cancel{14}~~^{1}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies \blue{\underline{ \boxed{ \sf{ \pmb {x = 4}}}}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Now we have x = 4

$\large\therefore\mathfrak{\pmb{{\underline{Required~answer}}:}}$

• $\sf Length = 4x = 4 \times 4 = \blue{\underline{ \underline{ \sf{ \pmb{ \: 16m \: }}}}}$
• $\sf Breadth = 3x = 3 \times 4 = \blue{\underline{ \underline{ \sf{ \pmb{ \: 12m \: }}}}}$

$\large\mathfrak{\pmb{{\underline{Verification}}:}}$

• Length = 16m
• Breadth = 12m

$\sf Perimeter \: of \: a \: rectangle = 2(l + b) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf = 2(16 + 12) \\ \sf = 2(28) \: \: \: \: \: \: \: \: \: \: \\ \sf = \blue{\underline{ \boxed{ \sf{ \pmb {56m}}}}} \: \: \: \: \: \: \: \: \:$

• Hence Verified !