Question

The length of rectangle is 10 more than its breadth if the perimeter of rectangle is 400 m find its Length breadth and area

Answer 1

Here is the answer of ur question

Answer 2

$\huge \underline{\textbf{ANSWER}}$

$\large \textsf{Given\:That :}$

$\textsf{ Length of rectangle is 10m more than its breadth }$

$\textsf{Perimeter of rectangle = 400m }$

$\textsf{ So now let the Breadth 0f Rectangle = X }$

$\sf{ \therefore \: According \: to\: Question\: Length \:of \:rectangle\: = \:Breadth\: + \:10m }$

$\sf{That\: is \; \implies}$

$\sf{Length \: = \:x\: + \:10m}$

$\textsf{Now according to question \bold{Perimeter\; of \:Rectangle\: = \:400m}}$

$\textsf{So now we know that}$

$\boxed{\sf{Perimeter\: of\: Rectangle\: =\: 2(Length\: + \: Breadth)}}$    

$\textsf{ So now putting values of estimated Length and Breadth we have}$

$\sf{Perimeter\: of \:rectangle\: = \:2(\:( \:x\:+\:10m\:)\: +\: ( \: x\: )}$

$\sf{That\: is \implies}$

$\sf{400m\: = \:2(\:( \:x\:+\:10m\:)\: +\: ( \:x\: )}$

$\sf{\implies 400m\: = \:2(\:( \: 2x\:+ 10m \: )}$

$\sf{O p e ning\: the\: Bracket \:and \:finding\: the \:Product We \:have}$

$\sf{\implies 400m\: = \:4x\: +\:20m}$

$\textsf{Now taking All variable terms at LHS and all Constant terms at RHS we have }$

$\sf{\implies \:4x\: =\: 400m \: - 20m}$

$\textf{That is \implies}$

$\sf{\implies \:4x\: =\: 380m}$

$\textsf{ Now taking 4 to RHS we have}$

$\sf{\implies \:x\: =\:\dfrac{ 380m}{4}}$

$\sf{\therefore \:we\:have\:x\: =\: 95m}$

$\underline{\textsf{Putting value of X in estimated form of length and breadth we have}}$

$\bold{\sf{\underline{Length =\: x \:+\: 10m \implies \:95m \:+\: 10m = 105m}}}$

$\bold{\sf{\underline{Breadth =\: x\: = \:95m}}}$

$\textsf{ Now we know that --}$

$\boxed{\sf{Area\: of\: rectangle \:= \:Length \times Breadth}}$

$\sf{ \therefore \; area \;of \:this \:rectangle\: = \:105m \times 95m}$

$\boxed{\sf{ \therefore \; Area \;of \:rectangle\: = \:9975 {m}^{2}}}$