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Mensuration class 8 notes

In this page we will explain the topics for the chapter 11 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	P= 2(L+B) L and B are Length and Breadth of the rectangle	A=L×B
Square	P=4a a is the side of the square	A=a²
Triangle	P=Sum of sides	A=(1/2)×(Base)×(Height/Altitude)
Parallelogram	P=2(Sum of Adjacent sides)	A=(Base) ×( Height)
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

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

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	10⁶cm³	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 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

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

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