Question

the length of a rectangle is 5 metres less than twise the breadth. If the perimeter is50 metres, find the length and breadth of the rectangle

Answer 1

Step-by-step explanation:

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Answer 2

Proper Question :

⠀⠀⠀⠀⠀▪︎ The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters , find the length and breadth of the rectangle .

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Given : The length of a rectangle is 5 metres less than twice the breadth & the perimeter is 50 meters .

Exigency To Find : The Length & Breadth of Rectangle .

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❍ Let's Consider the Breadth be x m

⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀⠀ The length of a rectangle is 5 metres less than twice the breadth .

$\qquad:\implies \sf Length _{(Rectangle)} \:= \: 2 \times Breadth \: - 5 \:m \:$

$\qquad:\implies \sf Length _{(Rectangle)} \:= \: 2 \times x \: - 5 \:$

$\qquad:\implies \sf Length _{(Rectangle)} \:= \: 2x \: - 5 \:$

$\qquad \implies \pmb{\underline{\purple{\:Length _{(Rectangle)} \:= \: (2x \: - 5)\:m \:\: }} }\:\:\bigstar$

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad\bigstar\:\:\bf Perimeter\:of\:Rectangle\::$

$\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Rectangle)} \:: 2( l +  b)\:\:unit }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , l is the Length of Rectangle, b is the Breadth of Rectangle & it is already Given that Perimeter of Rectangle is 50 m .

$\qquad:\implies \sf Perimeter _{(Rectangle)} \:=\: 2( l +  b)$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf Perimeter _{(Rectangle)} \:=\: 2( l +  b)$

$\qquad:\implies \sf 50 \:=\: 2\bigg( (2x- 5) +  x \bigg)$

$\qquad:\implies \sf 50 \:=\: 2(2x- 5) +  2(x)$

$\qquad:\implies \sf 50 \:=\: 4x- 10 +  2x$

$\qquad:\implies \sf 50 \:=\: 4x +  2x - 10$

$\qquad:\implies \sf 50 \:=\: 6x - 10$

$\qquad:\implies \sf 50 + 10\:=\: 6x$

$\qquad:\implies \sf 60\:=\: 6x$

$\qquad:\implies \sf 6x\:=\: 60$

$\qquad:\implies \sf 6x\:=\: 60$

$\qquad:\implies \sf x\:=\: \dfrac{60}{6}$

$\qquad:\implies \sf x\:=\: \cancel {\dfrac{60}{6}}$

$\qquad:\implies \sf x\:=\: 10$

$\qquad :\implies \pmb{\underline{\purple{\:x = 10\:m }} }\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀▪︎⠀⠀The Length of Rectangle is x = 10 m

⠀⠀⠀⠀⠀▪︎⠀⠀The Breadth of Rectangle is (2x - 5) = [2 (10) - 5 ] = 20 - 5 = 15 m

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf \:Hence,\:\:Length \:and\:Breadth \:of\:Rectangle \:are\:\bf 10 \:m\:\: \sf \& \: \bf 15 \: m \:\:\sf ,respectively. \:}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \leadsto \sf Area_{(Rectangle)} = Length \times Breadth$

$\qquad \leadsto \sf Perimeter _{(Rectangle)} = 2 (Length + Breadth)$

$\qquad \leadsto \sf Area_{(Square)} = Side \times Side$

$\qquad \leadsto \sf Perimeter _{(Square)} = 4 \times Side$

$\qquad \leadsto \sf Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times (a + b )$

$\qquad \leadsto \sf Area_{(Parallelogram)} = Base \times Height$

$\qquad \leadsto \sf Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height$

$\qquad \leadsto \sf Area_{(Rhombus)} = \dfrac{1}{2} \times Diagonal _{1}\times Diagonal_{2}$

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