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 = √(l² + w²) 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 m²
= 26,400 / 10,000 [ Since 10,000 m² = 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 m² 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 m²
Area of wall including door = side x side
= 11 x 11
= 121 m²
Area of wall excluding door = (121 - 20) m²
= 101 m²
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 m² and the length is 22 m. Also find its perimeter.
Solution :
Area of the rectangular sheet = 440 m²
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 m²
And area of rectangle ABCD = 90 x 75 = 6750 m²
Therefore,
Area of path = Area of rectangle EFGH – Area of rectangle ABCD
= 8500 – 6750
= 1750 m²
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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

Mensuration

7th grade math

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