We at **ask-math **believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

**We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.**

**Affiliations with Schools & Educational institutions are also welcome.**

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

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

__________________________________________________________________

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.

_________________________________________________________________

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.

_________________________________________________________________

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

________________________________________________________________

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

________________________________________________________________

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.

_________________________________________________________________

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

________________________________________________________________

• 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

7th grade math

Home

GMAT

GRE

1st Grade

2nd Grade

3rd Grade

4th Grade

5th Grade

6th Grade

7th grade math

8th grade math

9th grade math

10th grade math

11th grade math

12th grade math

Precalculus

Worksheets

Chapter wise Test

MCQ's

Math Dictionary

Graph Dictionary

Multiplicative tables

Math Teasers

NTSE

Chinese Numbers

CBSE Sample Papers