Math, asked by AryabhattaBrainly, 1 year ago

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

Answers

Answered by Anonymous
3
\huge{\mathfrak {\red{Q}{\underline{\underline{uestion}}}}}<br />



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


___________________________________



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




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

\underline{ \underline {\mathfrak {Breadth = 17.784 \: m }}}




___________________________________



 \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, <br />\\ \\ \\ \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{ \: 17.784 \: m }}



___________________________________



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




\huge \orange{ \boxed{ \boxed{ \sf{ \therefore \: Breadth = 17.784 \: m}}}}




✔✔ Hence, it is solved ✅✅




\huge \blue{ \boxed{ \boxed{ \mathscr{THANKS}}}}
Answered by Anonymous
6

\underline{\Huge\mathfrak{Question-}}

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

\underline{\Huge\mathfrak{Answer-}}

Given ;-

\underline{\small\mathfrak{Length}} = 22.23 m.

\underline{\small\mathfrak{Breadth}} = 17.784 m.

Now ,

\underline{\small\mathfrak{Assuming\:the\:Lenght\:as\:'y'\:m\:}}

So , the value becomes ;

 = &gt; \frac{4}{5} x \: m

We know that ;-

\underline{\small\mathfrak{Perimeter\:=\:2(Lenght + Breadth)\:}}

 = &gt; 2(x + \frac{4}{5}x) = 2x + \frac{8}{5}x

Now , given that ;-

 = &gt; 2x + \frac{8x}{5} = 80

 = &gt; 10x + 8x = 400

 = &gt; 18x = 400 = &gt; x = \frac{400}{18} = 22.23

Now ,

\underline{\small\mathfrak{The\:length\:}} = 22.23 m.

\underline{\small\mathfrak{The\:width\:}} = 4/5 × 22.33 = 17.748 m.

Similar questions