Question

the lenght of a rectangle exceeds the width by 14 cm perimetre of rectangle is 160 cm find lenght and width of the rectangle. ​

Answer 1

Answer:

Hello Santa19!

Your answer is 47 cm

Step-by-step explanation:

Given that

perimeter of rectangle =160cm

Width of rectangle =x

Length =x+14

Then perimeter of rectangle

$= 2(l + w) \\ \\ 2(x + x + 14) = 160 \\ \\ 2x (2x+ 14) = 160 \\ \\ \frac{ { \cancel{160}} \: \: ^{80} }{ \cancel2} \\ \\ = 80$

$2x + 14 = 80 \\ \\ 2x = 80 - 14 \\ \\ 2x = 66 \\ \\ \therefore \: x = \frac{ { \cancel{66}} \: \: ^{33} }{ \cancel2} \\ \\ = 33 \: cm$

Widtg=33cm

Length = 33+14=47cm

Hope it helps you from my side

Answer 2

Given : Length of the Rectangle exceeds the Breadth by 14 cm . Perimeter ov the rectangle is 160 cm .

To Find : Find the Length and Breadth of Rectangle

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SolutioN : For Calculating this let's form the Equation first and than we can solve it easily .Let's Solve :

$\maltese$ Formula Used :

• ${\underline{\boxed{\sf{ Perimeter = 2 \bigg( Length + Breadth \bigg) }}}}$

$\maltese$ According to the Question :

$\longmapsto$ Let the Breadth be y .So,

$\qquad \; {\pmb{\sf{ Breadth = y \; cm }}}$

$\longmapsto$ Breadth exceeds the length by 14 cm .So,

$\qquad \; {\pmb{\sf{ Length = y + 14 \; cm }}}$

$\maltese$ Calculating the Value of y :

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Perimeter = 2 \bigg( Length + Breadth \bigg) } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { 160 = 2 \bigg\{ \bigg( y + 14 \bigg) + y \bigg\} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { 160 = 2 \bigg( 2y + 14 \bigg) } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { 160 = 4y + 28 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { 160 - 28 = 4y } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { 132 = 4y } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { \dfrac{132}{4} = y } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { \cancel\dfrac{132}{4} = y } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\frak{ y = 33 }}}}} \; {\pink{\bigstar}} \\ \\ \\ \end{gathered}$

$\maltese$ Calculating the Dimensions :

• Breadth = y = 33 cm
• Length = y + 14 = 33 + 14 = 47 cm

$\therefore \;$ Length of the Rectangle is 47 cm and Breadth is 33 cm .

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