Formula of mensuration

SOME IMPORTANT FORMULAE

Area and perimeter of Rectangle and Square

• Area of rectangle (A) = length(l) × Breath(b), A = l× b
• Perimeter of a rectangle (P) = 2 × (Length(l) + Breath(b)), P = 2 × (l + b)
• Area of a square (A) = Length (l) × Length (l), A = l×l
• Perimeter of a square (P) = 4 × Length (l), P = 4 ×l

Related Topics

• Maths Formula
• Mensuration Circle

SOLVED EXAMPLES

Question. Find the perimeter and area of a rectangle with length 8 cm and breadth 10 cm.

Sol. Given l = 8 cm and b = 10 cm

∴ Perimeter = 2 (l + b)

= 2 × (8 + 10) cm = 36 cm.

Area    = (l × b)= (8 × 10)cm 2 = 80 cm 2 .

Question. The perimeter of a rectangle is 50 cm and its breadth is 9 cm. Find the length and area of the rectangle.

Sol.Let the length be l cm. Then,

2(l + b) = Perimeter ⇒ 2 × (l + 9) = 50

∴ Length = 16 cm.

Area = (l × b) = (16 × 9) cm 2 = 144 cm 2 .

Hence, the length of rectangle is 16 cm and its area is 144 cm 2 .

Questions:

Question 1. Find the perimeter and area of a square with side 11 cm

Question 2. ………… is a direction in which one can measure distance.

Question 3. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

6 cm 8 cm

Answers:

1. 44 cm , 121 cm 2 2. Dimension 3.  both will have equal

1. Find the perimeter and area of a square with side 11 cm

Sol: Side = 11 cm

So, perimeter of the square is = 4 × Side

= 4 × 11

= 44 cm

Area of the square = (side) 2

= (11) 2

= 121 cm 2

3. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

6 cm

8 cm

Sol : Perimeter of square = 4 (Side of the square) = 4 (6 m) = 24 m

Perimeter of rectangle = 2 (Length + Breadth)

= 2 (8 m + Breadth)

= 16 m + 2 × Breadth

It is given that the perimeter of the square and the rectangle are the same.

16 m + 2 × Breadth = 24 m

5. Area of parallelogram(A) = Length(l) × (Height(h), A = l× h

6. Perimeter of a parallelogram (P) = 2 ×( (length(l) + Breadth(b))

Perimeter of a parallelogram (P) = 2 ×( (length(l) + Breadth(b)), P = 2 × (l + b)

7. Area of a triangle (A) = (Base(b)  × (Height(b)) / 2,

And for a triangle with sides measuring “a” , “b” and “c” , Perimeter = a+b+c

and s = semi perimeter = perimeter / 2 = (a+b+c)/2

And also . Area of triangle =

This formulas is also knows as “Heroe’s formula”.

8. Area of triangle(A) =

Where , A , B and C are the vertex and angle A , B , C are respective angles of triangles and  a , b , c are the respective opposite sides of the angles as shown in figure below:

9. Area of isosceles triangle =

Where , a = length of two equal side , b= length of base of isosceles triangle.

10. Area of trapezium (A) =

Where , “a” and “b” are the length of parallel sides and “h” is the perpendicular distance between “a” and “b” .

11. Perimeter of a trapezium (P) = sum of all sides

12. Area f rhombus (A) =  Product of diagonals / 2

13. Perimeter of a rhombus (P) = 4 × l

where l = length of a side

14. Area of quadrilateral (A) = 1/2 × Diagonal ×( (Sum of offsets)

15. Area of a Kite (A) = 1/2 × product of it’s diagonals

16. Perimeter of a Kite (A) = 2 × Sum on non-adjacent sides

SOLVED EXAMPLES

Question. The base of a triangle is 24 cm and its corresponding height is 15 cm. Find the area of the triangle.

Solution: Area of the triangle sq. units.

Question. Find the area of a parallelogram whose base is 4 m and height 50 cm.

Solution: Given : Base = 4 m, Height = 50 cm

∴ Area of the parallelogram = (Base × Height) sq. units. = (4 × 0.5) m 2 = 2 m 2

Question. Find the area of the rhombus, the lengths of whose diagonals are 15 cm and 19 cm.

Solution: Area of the rhombus = × (product of diagonals) = 142.5 cm 2 .

∴ The area of the given rhombus is 142.5 cm 2 .

CBSE NCERT Solutions for Class 8 Maths

class 8 maths NCERT solutions Chapter 1: Rational Numbers

class 8 maths NCERT solutions Chapter 2: Linear Equations in One Variable

class 8 maths NCERT solutions Chapter 3: Understanding Quadrilaterals

class 8 maths NCERT solutions Chapter 4: Practical Geometry

class 8 maths NCERT solutions Chapter 5: Data Handling

class 8 maths NCERT solutions Chapter 6: Square and Square Roots

class 8 maths NCERT solutions Chapter 7: Cube and Cube Roots

class 8 maths NCERT solutions Chapter 8: Comparing Quantities

NCERT Class 8 Maths solution Chapter 9: Algebraic Expressions and Identities

class 8 maths NCERT solutions  Chapter 10: Visualizing Solid Shapes

class 8 maths NCERT solutions Chapter 11: Mensuration

class 8 maths NCERT solutions  Chapter 12: Exponents and Powers

class 8 maths NCERT solutions  Chapter 13: Direct and Inverse Proportions

class 8 maths NCERT solutions  Chapter 14: Factorization

class 8 maths NCERT solutions  Chapter 15: Introduction to Graphs

class 8 maths NCERT solutions  Chapter 16: Playing with Numbers

Notes,worksheet and solved question for Maths class 8

1. class 8 maths notes on chapter Liner equation in one variable
2. class 8 maths notes on chapter algebric expression
3. class 8 maths notes on chapter Mensuration
4. class 8 maths notes on chapter Square and square roots
5. class 8 maths notes on chapter statistice
6. class 8 maths notes on chapter practical Geometry
7. class 8 maths notes on chapter commericial maths
8. class 8 maths notes on chapter solid shape
9. class 8 maths notes on chapter quadrilaterals
10. class 8 maths notes on chapter exponents
11. c lass 8 maths notes on chapter factorisation
12. class 8 maths notes on chapter inverse proporation
13. class 8 maths notes on chapter cube and cube roots

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