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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

• Area of Rectangle = l x w sq.units

• Length of Rectangle = A/w units

• Width of Rectangle = A/l units

1) Find the perimeter of a rectangle whose length and width are 25 m and 15 m respectively.

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 ?

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?

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

Length = l = 240 m and width = w = 110 m

∴ Area of the field = l x w

⇒ = 240 x 110

= 26,400 m

= 26,400 / 10,000 [ Since 10,000 m

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

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

Area of wall including door = side x side

= 11 x 11

= 121 m

Area of wall excluding door = (121 - 20) m

= 101 m

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

Area of the rectangular sheet = 440 m

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.

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

And area of rectangle ABCD = 90 x 75 = 6750 m

Therefore,

Area of path = Area of rectangle EFGH – Area of rectangle ABCD

= 8500 – 6750

= 1750 m

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• 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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