Question

The length of a rectangle is 5 more than the 2 times of its breadth. If the perimeter of the rectangle is at least 640 cm. ​

Answer 1

Answer:

length is 211.67 cm

breadth is 428.34 cm

Step-by-step explanation:

Let the breadth of the rectangle be x

then length of the rectangle = 2x + 5

Perimeter = 640 cm

l + b = 640

2x + 5 + x = 3x + 5 = 640

3x = 640 - 5

3x = 635

x = 635/3

x = 211.66 cm

breadth = 211.67 cm

length = 211.67 * 2 + 5 = 428.34 cm

Answer 2

Given : The length of a rectangle is 5 more than the 2 times of its breadth & the perimeter of the rectangle is 640 cm.

Exigency to find : Length and Breadth of a Rectangle.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❒ Let's consider the Breadth of Rectangle be x cm .

Given that ,

• The length of a rectangle is 5 more than the 2 times of its breadth

Then ,

• Length of Rectangle is ( 2x + 5 ) cm .

$\cal{\dag{ As,\:We\:know\:that\::}}$

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

⠀⠀⠀⠀Here , l is the Length of Rectangle & b is the Breadth of Rectangle.

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

$\qquad \longmapsto \sf 640 = 2( x + 2x + 5)$

$\qquad \longmapsto \sf \cancel {\dfrac{640}{2}} = ( x + 2x + 5)$

$\qquad \longmapsto \sf 320 = ( x + 2x + 5)$

$\qquad \longmapsto \sf 320 = ( 3x + 5)$

$\qquad \longmapsto \sf 320-5 = 3x$

$\qquad \longmapsto \sf 315 = 3x$

$\qquad \longmapsto \sf \cancel {\dfrac{315}{3}}= x$

$\qquad \longmapsto \boldsymbol {\underline {\purple { x =105 \:cm}}}$

Therefore,

• The Length of Rectangle is (2x + 5 ) = ( 2(105) + 5) =(210 +5) = 215 cm .
• The Breadth of Rectangle is x = 105 cm .

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Length \:\& \:Breadth \:of\:Rectangle \:is\:\bf{215\:cm\:\:\&\:105\: cm}}}}$