Question

please slove for me please please ​

Answer 1

$\red{ \bold{ \huge{ \underline{ \underline{ \overline{ \blue{\mid{ \: QUESTION \: { \mid}}}}}}}}}$

$\bold{The \: \: perimeter \: \: of \: \: a \: \: rectangular \: \: field} \\ \bold{is \: \: \frac{3}{8} \: \: km. \: \: If \: \: the \: \: length \: \: is \: \: twice \: \: as \: \: its \: \: }\\ \bold{width. \: Find \: \: the \: \: area \: \: of \: \: the \: \: rectangular}\\ \bold{field \: \: in \: \: m {}^{2} \: . }$



$\red{ \bold{ \huge{ \underline{ \underline{ \overline{ \blue{\mid{ \: SOLUTION \: { \mid}}}}}}}}}$



$\bold {\underline{ \: \: Given \: : }} \\ \\ \bold{Perimeter \: \: of \: \: the\: \: rectangular \: \: field= \frac{3}{8} \: km}$


$\bold{ \underline{ \: \: To \: \: find \: : }} \: \: \: \: \: \: \: \bold{Area \: \: of \: \: the \: \: field}$


$\bold{Let,} \\ \\ \bold{Width \: \: of \: \: the \: \: field = x \: \: km} \\ \\ \bold{So,} \\ \\ \bold{The \: \: length \: \: of \: \: the \: \: field = 2x \: \: km}$


$\bold{According \: \: To \: \: Question} \\ \\ \bold{2(Length + Width) = Perimeter} \\ \\ \bold{ = > 2(2x + x) = \frac{3}{8} } \\ \\ \bold{ = > 2 \times 3x = \frac{3}{8} } \\ \\ \bold{ = > 6x = \frac{3}{8} } \\ \\ = > \bold{x = \frac{3}{8} \times \frac{1}{6} } \\ \\ \bold{ = > x = \frac{1}{16} }$



$\bold{Hence, \: \: } \\ \\ \bold{Width \: \: of \: \: the \: \: field = x = \frac{1}{16} \: \: km } \\ \\ \bold{Length \: \: of \: \: the \: \:field = 2x = (2 \times \frac{1}{16} )\: km = \frac{1}{8 } \: \: km}$



$\bold{Now,} \\ \\ \bold{Area =Length \times Width } \\ \\ \bold{ = \frac{1}{8} \times \frac{1}{16} \: \: sq.km } \\ \\ \bold{ = \frac{1}{128} \: \: sq.km } \\ \\ \bold{7812.50 \: sq.m}$



$\blue{ \bold{ \huge{ \underline{ \underline{ \overline{ \mid{ \: ANSWER \: { \mid}}}}}}}}$


$\boxed{ \boxed{ \bold{ \purple{ \: \: Area \: \: of \: \: the \: \: rectangular \: \: field= 7812.50 \: \: sq.m \: }}}}$


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✝$\mathfrak{ \purple{ \: \: THANKS... \: \: }}$✝