Area and Perimeter of the Rectangle

Area and Perimeter of the Rectangle are explained below:

Length is denoted by l, width (breadth) is denoted by w(b) and diagonal is denoted by d. There are two lengths, two widths and two diagonals.

area & perimeter of the rectangle

Formulas for Area and Perimeter of the Rectangle are given below :

Perimeter of the rectangle = 2(l + w) units
Length of Rectangle = p/2 - w units
Width(w) of Rectangle = p/2 - l units
Diagonal of Rectangle = √(l 2 + w 2 ) units
Area of Rectangle = l x w sq.units
Length of Rectangle = A/w units
Width of Rectangle = A/l units

Some solved examples on Area and Perimeter of the Rectangle

1) Find the perimeter of a rectangle whose length and width are 25 m and 15 m respectively.
Solution :
Length = l = 25 m and Width = w = 15 m
∴ P = 2 ( l + w )
⇒ = 2 ( 25 + 15 )
⇒ = 2 x 40
⇒ = 80 m
Perimeter = 80 m
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2)How many rectangles can be drawn with 36 cm as the perimeter, given that the sides are positive integers in cm ?
Solution :
Perimeter = 36
⇒ 2 ( l + w ) = 36
⇒ l + w = 18
Since length and width are positive integers in centimeters. Therefore, possible dimensions are :
( 1, 17 )cm , ( 2, 16 ) cm , ( 3, 15 ) cm, (4 , 14) cm, ( 5,13) cm, ( 6, 12) cm , ( 7, 11) cm, ( 8, 10 ) cm, ( 9, 9) cm.
Hence, there are 9 rectangles.
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4) The length of a rectangular field is twice its width. A man jogged around it 5 times and covered a distance of 3 km. What is the length of the field?
Solution :
In completing one round of the field distance covered is equal to the perimeter of the field.
∴ Distance covered in 5 rounds = 5 x Perimeter
= 5 x 2 ( l + w )
= 10 x ( l + w)
= 10 x ( 2w + w ) [ length = l = 2w ]
= 10 x 3w
= 30 w
But, the total distance covered is given = 3 km = 3000 m.
∴ 30 x w = 3000
⇒ w = 3000 / 30
⇒ w = 100 m
⇒ Length = 2 w = 2 x 100 = 200 m.
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Some solved examples on Area of Rectangle

1) Find the area, in hectare, of a field whose length is 240 m and width 110 m.

Solution :
Length = l = 240 m and width = w = 110 m
∴ Area of the field = l x w
⇒ = 240 x 110
= 26,400 m
2
= 26,400 / 10,000 [ Since 10,000 m
2 = 1 hectares ]
Area of field = 2.64 hectare
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2) A door frame of dimensions 4 m x 5 m is fixed on the wall of dimension 11 m 11 m. Find the total labour charges for painting the wall if the labour charges for painting 1 m
2 of the wall is $ 2.50.
Solution :
Painting of the wall has to be done excluding the area of the door
Area of the door = l x w
= 4 x 5
= 20 m
2
Area of wall including door = side x side
= 11 x 11
= 121 m
2
Area of wall excluding door = (121 - 20) m
2
= 101 m
2
Total labour charges for painting the wall
=$ 2.50 x 101
= $ 252.50
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3) Find the breadth of a rectangular plot of land, if its area is 440 m
2 and the length is 22 m. Also find its perimeter.
Solution :
Area of the rectangular sheet = 440 m
2
Length (l) = 22 m
Area of the rectangle = l x w (where w = width of the rectangular plot
Therefore, width (w) = Area/l = 440/22 = 20 m
Perimeter of sheet = 2(l + w)
= 2(22 + 20)m
= 84 m
So, the width of the rectangular plot is 20 m and its perimeter is 84 m.
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3) A rectangular garden is 90 m long and 75 m broad. A path 5 m wide is to be built out around it. Find the area of the path.

Solution :

Then, clearly,
Area of the path = Area of rectangle EFGH – Area of rectangle ABCD
From the figure, we have
EF = 90 + 5 + 5
= 100 m
and FG = 75 + 5 + 5 = 85 m
Now, area of rectangle EFGH = 100 x 85 = 8500 m
2
And area of rectangle ABCD = 90 x 75 = 6750 m
2
Therefore,
Area of path = Area of rectangle EFGH – Area of rectangle ABCD
= 8500 – 6750
= 1750 m
2
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Mensuration : Area and Perimeter of the Rectangle

Perimeter and Area of Irregular Shape
Area and Perimeter of the Rectangle
Area of Square (perimeter of square)
Perimeter of Parallelogram(Area of Parallelogram)
Area of Rhombus(Perimeter of rhombus)
Area of Trapezoid (Trapezium)
Triangle Area (Perimeter of triangle)
Herons Formula

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