Question

The length of a rectangle is 5 cm more than its breadth. If the perimeter of the rectangle is 40cm. Find the length and breadth of the rectangle. Answer this question step by step. I will mark you as brainlist.​

Answer 1

$\large\bf\underline {To \: find:-}$

• Length and breadth of rectangle.

$\large\bf\underline{Given:-}$

• length is 5cm more than it's breadth
• perimeter of rectangle = 40cm

$\huge\bf\underline{Solution:-}$

• Let breadth of rectangle be x

So,

• Length of rectangle = x + 5

$\underline{ \boxed{ \bf \: perimeter \: of \: rectangle = 2(l + b) }} \star$

Perimeter of rectangle = 40cm

➛ 2(x + x + 5) = 40

➛ 2(2x + 5) = 40

➛ 2x + 5 = 40/2

➛ 2x + 5 = 20

➛ 2x = 20 - 5

➛ 2x = 15

➛ x = 15/2

➛ x = 7.5

So,

• ❥ Breadth of rectangle (x) = 7.5cm
• ❥ Length of rectangle( x + 5) = 12.5cm.

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✦ Rectangle Formula's

»★ Area of rectangle = l × b

»★ Perimeter of rectangle = 2(l + b)

»★ Diagonal of rectangle = √l² + b²

»★ Length of rectangle = Area/Breadth

»★ Breadth of rectangle = Area/Length

where,

• ≫ l = Length
• ≫ b = Breadth

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Answer 2

Given :

• Length of a rectangle is 5 cm more than it's Breadth.

• The perimeter of the reactangle is 40 cm.

To Find :

• The length and the Breadth of the reactangle.

Figure :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(7.7,3){\large\sf{A}}\put(7.4,2){\sf{\large{x cm}}}\put(7.7,1){\large\sf{B}}\put(9.3,0.7){\sf{\large{(x+5) cm}}}\put(11.1,1){\large\sf{C}}\put(11.1,3){\large\sf{D}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\end{picture}$

Solution :

• Let the breadth of the reactangle be x cm.

• Then it's length = x + 5 cm

Perimeter = 2 (length + breadth)

➩ 40 = 2 (x + (x + 5))

➩ 40/2 = 2x + 5

➩ 2x + 5 = 20

➩ 2x = 20 - 5

➩ 2x = 15

➩ x = 15/2cm or 7.5cm

Therefore,

• Breadth = x = 7.5 cm

• So, length = x + 5cm = 7.5 + 5 = 12.5 cm