Question

The length of a rectangle is four times its width. If the area is 100 m2 what is the length of the rectangle?


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Answer 1

$\huge\mathbb{\pink{QUESTION ♡- }}\star{}$

The length of a rectangle is four times its width. If the area is 100 m2 what is the length of the rectangle?

$\huge\mathcal{\purple{A}}{\mathcal{\green{N}}}{\mathtt{\pink{S}}}{\mathcal{\blue{W}}}{\mathcal{\purple{E}}}{\mathcal{\green{R}}}{\pink{!}}{\blue{!}}$

Let, the width of the rectangle be (b) = x

The, Length of the rectangle (l) = 4x

$Area \: of \: the \: rectangle = length × breadth \\ \\ ⟹100 = 4x \times x \\ ⟹100 = 4 {x}^{2} \\ ⟹ \frac{100}{4} = {x}^{2} \\ ⟹25 = {x}^{2} \\ ⟹ \sqrt{25} = x \\ ∴x = 5$

$Now \: length \: of \: rectangle = 4x \\ = 4 × 5 \\ </p><p>= 20 \: m$

Answer 2

AnswEr-:

• $\sf{\underline{\red{\dag{ \:Length \:of\:Rectangle\:20\:m\:.}}}}$

Explanation-:

$\mathrm {\bf{ Given-:}}$

• The length of a rectangle is four times its width.

• The Area of Rectangle is 100 m².

$\mathrm {\bf{ To\:Find -:}}$

• The Length of Rectangle.

$\sf{\bf{\dag{ Solution \:of\:Question -:}}}$

$\mathrm {\bf{Let's \: Assume-:}}$

• The width of Rectangle be x m

Then Given that ,

• The length of a rectangle is four times its width.

Then ,

• Length of Rectangle is 4x m

Therefore,

• $\mathrm{\bf{\purple{Dimensions \:\: -:}}} \begin{cases} \sf{\blue{The\:Length \:of\:the\:Rectangle \:is\:= \frak{4x\:m}}} & \\\\ \sf{\red{Width \:of\:Rectangle \:is \:=\:\frak{x \: m}}}\end{cases}$

$\underbrace {\mathrm { \bf{ Understanding \:of\:Concept \:-:}}}$

• We have to find the Length of Rectangle when the Area and some word is given for Length and Breadth of Rectangle .

• Firstly put the assumed value in the Formula for Area of Rectangle then by doing this ,

• We can get the Length and Breadth of Rectangle.

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$\sf{\bf{\dag{Finding \:Length \:of\:Rectangle -:}}}$

As , We Know that ,

• $\underline{\boxed{\star{\sf{\red{ Area_{(Rectangle)}  \: = \: Length \times Width \:sq.units}}}}}$

• $\mathrm{\bf{\purple{Here \:\: -:}}} \begin{cases} \sf{\blue{The\:Length \:of\:the\:Rectangle \:is\:= \frak{4x\:m}}} & \\\\ \sf{\red{Width \:of\:Rectangle \:is \:=\:\frak{x \: m}}}\end{cases}$

Now by Putting known Values in Formula for Area of Rectangle-:

• $\longmapsto {\mathrm { 4x \times x = 100 m^{2} }}$

• $\longmapsto {\mathrm { x \times x = \dfrac{100}{4} }}$

• $\longmapsto {\mathrm { x \times x = \dfrac{\cancel {100}}{\cancel {4}} }}$

• $\longmapsto {\mathrm { x \times x = 25 }}$

• $\longmapsto {\mathrm { x^{2} = 25 }}$

• $\longmapsto {\mathrm { x = \sqrt {25} }}$

• $\underline {\boxed{\pink{\frak { x = 5 m }}}}$

Now , By Substituting X = 5 -:

• $\mathrm{\bf{\purple{Dimensions \:\: -:}}} \begin{cases} \sf{\blue{The\:Length \:of\:the\:Rectangle \:is\:= \frak{4x= 4 \times 5 =20\:m}}} & \\\\ \sf{\red{Width \:of\:Rectangle \:is \:=\:\frak{x \:=\:5 m}}}\end{cases}$

Hence ,

• $\sf{\underline{\red{\dag{ \:Length \:of\:Rectangle\:20\:m\:.}}}}$

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$\boxed {\bf{\large {\sf |\:\:{\underline { More \:To\; Know -:}}\:\:| }}}$

• $\underline{\mathrm {Area\:of\:Rectangle \:-: Length \: \times Breadth \:sq.units}}$

• $\underline{\mathrm {Perimeter \:of\:Rectangle \:-: 2 ( Length \: + Breadth )\:units}}$

• $\underline{\mathrm {Area\:of\: Square \:-: Side \: \times Side \:sq.units}}$

• $\underline{\mathrm {Perimeter \:of\: Square \:-: 4 \: \times Side \:units}}$

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