Question

The width of Ram's ground is 4 /5 of its length. if its perimeter is 80 m, find its dimensions

Answer 1

$\huge{\mathfrak {\red{Q}{\underline{\underline{uestion}}}}}$

The width of Ram's ground is 4/5 of is length. If its perimeter is 80 m, find its dimensions.

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${ \red{\mathfrak{\huge{A}}}}{\underline {\underline{\mathfrak{\huge{nswer}}}}}$




→$\underline{ \underline {\mathfrak {Length = 22.23 \: m }}}$

→$\underline{ \underline {\mathfrak {Breadth = 978.12 \: m }}}$




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$\huge\mathfrak \pink{ \overline{ \underline{ \: Brainliest \: Answer \: }}}$




${ \sf Let, Length \: be \:} x \sf\: m \\ {\sf Then,} \: \frac{4}{5} x \: \sf m \\ \\ \sf Perimeter = 2(length × breadth) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \sf 2} \bigg(x + { \sf \frac{4}{5} }x \bigg) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \sf2}x + { \sf\frac{8}{5} }x \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: By \: the \: given \: condition, \\ \\ \\ \implies 2x + \frac{8x}{5} = 80 \\ \implies 10x \: + 8x = 400 \\ \implies 18x = 400 \\ \implies x = \frac{400}{18} = 22.23 \\ \\ \\ \therefore \sf Length \: is \: { \bf {\red {22.23 \:m }}}\: and \: {\sf width \: is }\: \frac{4}{5} \: × \: 22.23 \: = { \bf \red{ \: 978.12 \: m }}$

$\huge \orange{ \boxed{ \boxed{ \sf{ \therefore \: Length = 22.23 \: m}}}}$




$\huge \orange{ \boxed{ \boxed{ \sf{ \therefore \: Breadth = 978.12 \: m}}}}$




✔✔ Hence, it is solved ✅✅




$\huge \blue{ \boxed{ \boxed{ \mathscr{THANKS}}}}$

Answer 2

\begin{lgathered}{ \sf Let, Length \: be \:} x \sf\: m \\ {\sf Then,} \: \frac{4}{5} x \: \sf m \\ \\ \sf Perimeter = 2(length × breadth) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \sf 2} \bigg(x + { \sf \frac{4}{5} }x \bigg) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \sf2}x + { \sf\frac{8}{5} }x \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: By \: the \: given \: condition, \\ \\ \\ \implies 2x + \frac{8x}{5} = 80 \\ \implies 10x \: + 8x = 400 \\ \implies 18x = 400 \\ \implies x = \frac{400}{18} = 22.23 \\ \\ \\ \therefore \sf Length \: is \: { \bf {\red {22.23 \:m }}}\: and \: {\sf width \: is }\: \frac{4}{5} \: × \: 22.23 \: = { \bf \red{ \: 978.12 \: m }}\end{lgathered}Let,LengthbexmThen,54​xmPerimeter=2(length×breadth)=2(x+54​x)=2x+58​xBythegivencondition,⟹2x+58x​=80⟹10x+8x=400⟹18x=400⟹x=18400​=22.23∴Lengthis22.23mandwidthis54​×22.23=978.12m​