Math, asked by praveen8199, 1 year ago

The perimeter of a rectangle is 40cm and its length is 2cm more than its breadth find length and breadth ?

Answers

Answered by isharani
14
we know,
perimeter of rectangle =2(l+b)
let the breadth of rectangle be x cm and length be (x+2)cm.
40cm=2(x+2+x)
40cm=2x+4+2x
40cm=4x+4
(40cm-4=4x
36cm=4x
36/4=x
9=x.
so, length =(x+2)=(9+2)=11cm.
breadth=x cm =9cm.
here is ur answer.
i hope it helps u.

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Answered by Anonymous
2

\bf{\underline{\underline \blue{Solution:-}}}

\sf\underline{\red{\:\:\: AnswEr:-\:\:\:}}

The breadth of the rectangle = 6 Cm

The length of the rectangle = 14 Cm

\sf\underline{\red{\:\:\: Given:-\:\:\:}}

The perimeter of rectangle is 40cm.

The length of the rectangle is more than double its breadth by 2

\sf\underline{\red{\:\:\: Need\:To\: Find:-\:\:\:}}

The breadth of the rectangle = ?

The length of the rectangle = ?

\bf{\underline{\underline \blue{Explanation:-}}}

\sf\underline{\pink{\:\:\: Diagram:-\:\:\:}}

\setlength{\unitlength}{2cm}\begin{picture}(16,4)\thicklines\put(8,3){\circle*{0.1}}\put(7.8,3){\large{D}}\put(7.2,2){\mathsf{\large{?cm}}}\put(8,1){\circle*{0.1}}\put(7.8,1){\large{A}}\put(9.3,0.8){\mathsf{\large{?cm}}}\put(11.1,1){\large{B}}\put(8,1){\line(1,0){3}}\put(11,1,){\circle*{0.1}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11,3){\circle*{0.1}}\put(11.1,3){\large{C}}\end{picture}

Let the breadth of the rectangle = y

\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}

Length = 2y + 2

\sf\underline{\red{\:\:\: Formula\:Used\: Here:-\:\:\:}}

\bigstar \:  \boxed{ \sf \: Perimeter\:of\:a\: rectangle = 2 \times (Length + Breadth) }\\\\

\sf\underline{\green{\:\:\: Now,Putting\:the\: values:-\:\:\:}}

\dashrightarrow \sf {Perimeter \:of\:a\: rectangle = 2[(2y + 2) + (y)]} \\\\

\dashrightarrow \sf {40 = 2 \times (3y + 2) }\\\\

\dashrightarrow \sf {40 = 6y + 4} \\\\

\dashrightarrow \sf {40 - 4 = 60}\\\\

\dashrightarrow \sf {y = \dfrac{\cancel{36}}{\cancel{6}}\:} \\\\

\dashrightarrow \sf {y = 6} \\\\

\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}

The breadth of the rectangle = y

The breadth of the rectangle = 6 Cm

\sf\underline{\green{\:\:\: And:-\:\:\:}}

The length of the rectangle = 2y + 2

The length of the rectangle = 2 × 6 + 2

The length of the rectangle = 12 + 2

The length of the rectangle = 14 Cm

\rule{200}{2}

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