The length of a rectangle is 15 cm more than its width eand it’s perimeter is 150 cm . find the deminsions of the rectangle ?

2 thoughts on “The length of a rectangle is 15 cm more than its width eand it’s perimeter is 150 cm . find the deminsions of the rectangle ?”

1. [tex]\huge{\underbrace{\text{Question}}}[/tex]

   The length of a rectangle is 15 cm more than its width and it’s perimeter is 150 cm . find the deminsions of the rectangle.

   [tex]\huge{\underbrace{\text{Answer}}}[/tex]

   Given:–

   [tex]\sf{\rightarrow{The\:length\:is\:15\:more \:than \: the \: width \: of \: the \: rectangle}}.[/tex]

   [tex]\sf{\rightarrow{Perimeter \: of \: the \: rectangle \: is \: 150cm}}[/tex]

   To Find:–

   [tex]\sf{\rightarrow{The \: deminsions \: of \: the \: rectangle}}[/tex]

   Solution:–

   Let the length of the rectangle be x

   Therefore,

   The width of the rectangle = (x+15)

   As we know that,

   [tex]\boxed{\sf{\blue{\bold{Perimeter \: of \: a \: rectangle} = 2(l + b)}}}[/tex]

   Where,

   • l stands for length
   • b stands for breadth/width

   According to the question,

   [tex]\sf{2\{(x) + (x + 15)\} = 150}[/tex]

   [tex] \\ \sf{\implies{2(x + x + 15) = 150}}[/tex]

   [tex] \\ \sf{\implies{2(2x + 15) = 150}}[/tex]

   [tex] \\ \sf{\implies{4x + 30= 150 }} [/tex]

   [tex] \\ \sf{\implies{4x = 150 – 30}}[/tex]

   [tex] \\ \sf{\implies{x = \frac{120}{4}}} [/tex]

   [tex] \\ \sf{\therefore{\orange{x = 30}}}[/tex]

   Hence, length = 30cm

   Therefore,

   Width of the rectangle = (30+15)cm=45cm

   [tex] \underline{\boxed{\rm{Hence, length \: of \: the \: rectangle = 30cm}}}[/tex]

   [tex] \underline{\boxed{\rm{Width \: of \: the \: rectangle = 45cm}}}[/tex]

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2. Given :

   • Length of a rectangle = 15 cm
   • Width and the perimeter = 150 cm

   To find

   • Deminsions Of The Rectangle

   Solution :

   Let ,

   The width of the rectangle be x cm

   Then ,

   Length = x ( x + 15 ) cm

   Perimeter = 2 ( Length + width )

   = 2 ( x + 15 + x ) cm

   = 2 ( 2x + 15 ) cm

   = 4x + 30 cm

   But ,

   The perimeter is given as 150 cm

   Therefore ,

   4x + 30 = 150

   ==] 4x = 150 – 30

   ==] 4x = 120

   [tex]x = \frac{120}{4} = 30 \: cm[/tex]

   Hence , width of the rectangle = 30 cm and length of the rectangle = ( 30 + 15 ) cm = 45 cm

   Check

   We have ,

   Length = 45 cm

   Width = 30 cm

   Therefore ,

   Perimeter = 2 ( Length + width )

   = 2 ( 45 + 30 ) cm

   = 2 × 75 = 150 cm

   Which is the same given in the problem

   Also , Length is 15 cm more than the width , which is also the same as given in the problem .

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