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Mensuration class 8 notes



In this page we will explain the topics for the chapter 9 of Mensuration Class 8 Maths.We have given quality Mensuration class 8 notes along with video to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share and comment at the end of the page.
Table of Content

Mensuration

It is branch of mathematics which is concerned about the measurement of length, area and Volume of plane and Solid figure
Mensuration class 8 notes

Perimeter & Area of Common & Known Shapes

Perimeter
  • The perimeter of plane figure is defined as the length of the boundary
  • It units is same as that of length i.e. m ,cm,km  
1 Meter
10 Decimeter
100 centimeter
1 Decimeter
 10 centimeter
100 millimeter
1 Km
10 Hectometer
100 Decameter
1 Decameter
10 meter
1000 centimeter
 
Area
  • The area of the plane figure is the surface enclosed by its boundary
  • It unit is square of length unit. i.e. m2 ,  km2  
1 square Meter
100 square Decimeter
10000 square centimeter
1 square Decimeter
 100 square  centimeter
10000 square  millimeter
1 Hectare
100 square  Decameter
10000 square meter
1 square myraimeter
100 square kilometer
108 square meter
 

Shapes where Area and Perimeter are known

 
Shapes
Perimeter
Area

Rectangle

Area of Rectangle

P= 2(L+B)

L and B are Length and Breadth of the   rectangle

A=L×B

Square

Area of Square
 

P=4a
a is the side of the square
A=a2

Triangle
Area of Triangle  

  P=Sum of sides
A=(1/2)×(Base)×(Height/Altitude)
Parallelogram
 
Parallelogram a type of quadrilateral
  P=2(Sum of Adjacent sides)
A=(Base) ×( Height)
Circle
Area of Circle  
 
P=2πr
r is the radius of the circle
A=πr2
 
Watch this tutorial for example around Perimeter

Area of Trapezium

Trapezium is a quadrilateral with a pair of parallel sides.
 
Area of Trapezium
Isosceles trapezium
Trapezium when non-parallel sides of it are of equal length
Area of the Trapezium can be found by dividing the trapezium into two parts of three parts depending
On the shapes of the trapezium
 
Area of trapezium when it can be divied into two triangle and one rectangle
(So two triangles and One rectangle)
Here h is the distance between the parallel sides
a   and b are the parallel sides 
A= (1/2)hx + (1/2)hy + ah
=(1/2)h( x+y+ 2a)
Now b=x+y+a
So
=(1/2)h( a+b)
Similarly
Area of trapezium when it can be divied into One triangle and One rectangle
(One triangle and One rectangle)
It can be proved easily here also
A=(1/2)h( a+b)
So to find the area of a trapezium we need to know the length of the parallel sides and the
perpendicular distance between these two parallel sides. Half the product of the sum of
the lengths of parallel sides and the perpendicular distance between them gives the area of
trapezium
Example
A trapezoid's two bases are 10 m and 6m, and it is 4m high. What is its Area?
Solution:
A=(1/2)h( a+b)
So
A= (1/2)4(10+6)
=32 m2

Area of General Quadrilaterals

 
We can always divide a general quadrilateral into two triangles by drawing the diagonal. This split the quadrilateral into two triangles and Sum of areas of the triangles is equal to the area of the quadrilaterals
Area of Quadilaterals  
Here is the above figure, Diagonal AC divides the quadrilateral ABCD into two triangles.  We have drawn perpendicular to this diagonal from opposite vertices. They are of height h1 and h2 respectively
So Area of Quadrilateral ABCD will be given as
=Area of the Triangle ACD + Area of the Triangle ABC
=(1/2)×AC×h1 + (1/2)×AC×h2
=(1/2)AC(h1+ h2)
=(1/2)d(h1+ h2)
d is the length of the diagonal
 
We already know the formula for the area of special quadrilaterals (Rectangle, square and parallelogram). Those can also be derived from the above general quadrilaterals formula
We can similarly derive the formula for the Rhombus
Area of Rhombus is given by
A= (1/2)×d1×d2
Where d1 and d2 are the diagonals of the Rhombus.
Examples
1) The area of a rhombus is 240 cm2  and one of the diagonals is 10 cm. Find the other diagonal.
Solution:
Area of Rhombus is given by
A= (1/2)×d1×d2
240=(1/2)10×d
d=45 cm

Area of Polygons

We will encounter figures which are having more than 4 sides.  We can derive the area of the polygons by using the same approach i.e. dividing into triangles or quadrilaterals
 

Solid Shapes

Solid shapes are three dimensional figures.  We can identify common two dimensional figures on there faces
Example
Shape of Cube
It is a cube which has square shapes on the all the faces
Shape of Cylinder
It is cylinder and has circular figure at two bases
 

Volume,Surface area in case of Solid Figures

 
Surface Area
Surface area of a solid is the sum of the areas of its faces
Lateral Surface Are
The faces excluding the top and bottom) make the lateral surface area of the solid
Volume
Amount of space occupied by a three dimensional object (Solid figure) is called its volume.
we use square units to find the area of a two dimensional region. In case of volume we will use cubic units to find the volume of a solid, as cube is the most convenient solid shape (just as square is the most convenient shape to measure area of a region)
Volume is sometimes refer as capacity also
Volume Units
1 cm3
1mL
1000 mm3
1 Litre
 1000ml
1000 cm3
1 m3
106 cm3
1000 L
1 dm3
1000 cm3
1 L
 

How to find the surface Area and Volume of the solid Figures

Surface Area is the area of the faces. So we need to find the area of the each of the faces and sum them to find the total surface area
In-case of Cuboid, all the faces are rectangles
Surface Area of Solid Figure in Menusration
Surface Area
=2×(Area of the rectangle having length and Breath  as h and b  )+2×(Area of the rectangle having length and Breath  as l and b  ) +2×(Area of the rectangle having length and Breath  as l and h  )
=2( LB + BH + LH ).
Volume is the space occupied by the solid figure. We can measure it by counting cubes in the space.
 if the base and top of the solid are congruent and parallel to each other and its edges are perpendicular to the base then volume can be deduced as
Volume of solid=area of base × height
 

Surface Area and Volume of Cube and Cubiod

 
Surface Area and Volume of Cube and Cubiod
Type
Measurement
Surface Area of Cuboid of Length L, Breadth B and Height H
2( LB + BH + LH ).
Lateral surface area of the cuboids
2( L + B ) H
Diagonal of the cuboids
$\sqrt {L^2 + B^2 + H^2}$
Volume of a cuboids
LBH
Length of all 12 edges of the cuboids
4 (L+B+H).
Surface Area of Cube of side L
6L2
Lateral surface area of the cube
4L2
Diagonal of the cube
$L \sqrt {3}$
Volume of a cube
L3
 

Surface Area and Volume of Right circular cylinder

Surface Area and Volume of Right circular cylinder
 
Radius
The radius (r) of the circular base is called the radius of the cylinder
Height
The length of the axis of the cylinder is called the height (h) of the cylinder
Lateral Surface
The curved surface joining the two base of a right circular cylinder is called Lateral Surface.
How to calculate the Surface Area  of Right circular cylinder
 
 
Type
Measurement
Curved or lateral Surface Area of cylinder
2πrh
Total surface area of cylinder
2πr (h+r)
Volume of Cylinder
 π r2h
 
Example:
Find the height of a cylinder whose radius is 14 cm and the total surface area is 1936 cm2
Assume π= 22/7
Solution
Let height of the cylinder = h, radius = r = 14cm
Total surface area = 2πr (h + r)
2× (22/7)× 14 × (h + 14) = 1936
h = 8 cm
 
Example
Given a cuboid tank, in which situation will you find surface area and in
which situation volume.
(a) To find how much water it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
 
Solution
(a) Volume as we are talking about capacity
(b) Surface area
(c) Again Volume

Summary

Here is Mensuration Class 8 Notes Summary mensuration class 8 formula

 

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