# Class 9 – Mensuration

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Topic Sub Topic Online Practice Test
Mensuration
• Review of mensuration
• Surface area and volume of a cone
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Mensuration
• Surface area and volume of a sphere
• Surface area and volume of combined solids
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## Study Material

 Solid: An object having length, breadth and height (thickness) is called a solid. The side portions of a solid are called lateral surfaces. The solid bodies occur in various shapes such as prism, cuboid, cube, pyramid, cylinder, cone, sphere and etc., Prism: A solid having two congruent parallel faces (bases) and whose other faces (lateral faces) are parallelograms formed by joining corresponding vertices of the bases is called a prism. Right prism: A prism whose bases are perpendicular to the lateral edges and all lateral faces are rectangles is called a right prism. Example: Triangular prism, square prism, rectangular prism, hexagonal prism. The name of the prism depends upon its base. We observe that: i) Number of lateral surfaces of a right prism = Number of sides of the base of the prism. ii) Total number of surfaces of a prism = Number of lateral surfaces + 2 (base and top) = Number of sides of prism + 2. iii) Number of edges of a prism = Number of sides of prism x 3.

iv) Sum of the lengths of edges of a prism= Number of sides x height + twice the perimeter of the base.

Let S = sum of the lengths of edges.
h = height of the solid

n= number of sides of the base.

p= perimeter of the base.

units.

Example: The sides of a triangular prism are 3cm, 4 cm and 5 cm. Its height is 10 cm. Find the total length of its edges.

Solution: Number of sides of base (triangle) = n= 3 cm

Length of sides of triangular base are 3 cm, 4 cm, 5 cm

Height of prism = h = 10 cm

Perimeter of base = (a + b + c)units

= 3 + 4 + 5

Perimeter of base (p) = 12 cm

Sum of lengths of edges (S)

Sum of lengths of edges (S) = 54 cm.

Surface area and volumeof a prism:

i) Lateral surface area of a prism =Perimeter of the base x height of the prism.
L.S.A. of prism sq. units.

ii) Total surface area of a prism = Lateral surface area + 2 x area of the base.
T.S.Aof prism = L.S.A. + 2 A sq. units.

iii) Volume of a prism = Area of the base xheight of prism.
V = A x h cubic units.

Example: The sides of a triangular prism are 12cm, 16 cm and 20 cm. Its height is
15 cm. Find i) Lateral surface area ii) Totalsurface area iii) Volume.

Solution: Sides of a triangular base of prism are12 cm, 16 cm, 20 cm

a = 12 cm, b = 16 cm, c =20 cm

Height of prism (h) = 15 cm

Perimeter of base (p) = a + b+ c = 12 + 16 + 20 = 48 cm.

Area of base(A)

Area of base (A) = 96 sq. cm

i) Lateral surface area = ph sq.units
= 48 x 15
Lateral surface area = 720 sq. cm

ii) Total surface area = L.S.A. + (2 xArea of base) sq. units.
= 720 + 2(96) = 720 + 192
Total surface area = 912 sq. cm

iii) Volume of prism = Area of base xheight cu. units
=96 x 15
Volume of prism = 1440 cubic. cm.

Cuboid: If the base of the right prism is rectangle then it is called cuboid.

Baseof the cuboid is rectangle.

Surface area and volume: Let l, b and h are the measurement of a cuboid then,

i) Lateral surface area

ii) Total surface area

iii) Volume of cuboid

Diagonal of a cuboid .

Example: Find the volume, total surface area and lateral surface area of a cuboid which is 15 m long, 12 m wide and 4.5 m high.

Solution: Here,

Volume of thecuboid cubic units

Volume of the cuboid

Total surface area of the cuboid

Total surface area of the cuboid

Lateral surface area of the cuboid

Lateral surface area of the cuboid

Example: A wall of dimensions 20 m x0.5 m x 2 m is to be constructed by using bricks with dimensions Find the number of bricks required.

Solution: Volume of the wall

Volume of each brick

Number of bricks

Number of bricks

Cube: A cuboid whose length, breadth and height are equal is called a cube.

i) Each side is represented by ‘a’or ‘s’ units.

ii) Diagonal of a cube

Surface area and volume: Let a is a side of a cube then,

i) Lateral surface area = 4a2sq. units

ii) Total surface area

iii) Volume of cube

Example: Find the volume, total surface area,lateral surface area and the length of the diagonal of a cube, whose edgemeasures 20 cm.

Solution: Here, a = 20 cm.

Volume of the cube = a3cubic units

Volume of the cube

Total surface area of the cube

Total surface area of the cube

Lateral surface area of the cube

Lateral surface area of the cube

Diagonal of the cube

Diagonal of the cube

Example: The lateral surface area of a cube is324 cm3. Find its volume and the total surface area.

Solution: Let each side of the cube be a cm.

Then, the lateral surface area of thecube

Volume of the cube = a3cubic units.

Volume of the cube

Total surface area of the cube

Total surface area of the cube

Pyramid: You might have heard of pyramids in Egypt. A pyramid is a solid object with its bases, one end as a polygon, the other end as a vertex and the lateral surfaces as triangles.
The triangles meet at a common ‘point’ called the vertex and the length of the perpendicular segment from the vertex to its base is called the height of thepyramid.

Right pyramid: If the base is a regular polygon andthe foot of the perpendicular segment from the vertex coincides with the centre of the polygon, the solid is called a right pyramid.

The length of the perpendicular segment from the vertex to any of the sides of the regular polygon (which forms the base of the right pyramid) is called the slant height of the right pyramid.

i) The area of the triangular face A (side of the polygon) (slant height)
The lateral surface area LSA (side of the polygon) (slant height)
(perimeter of the base) (slant height)

Volume of the pyramid is one-third the volume of the right prism with the same base and the same height.

Example: The base of a right pyramid is a square of side 10 cm. If the height of the pyramid is 12 cm, find the i) slant height, ii) lateral surface area, iii) volume

Solution: Square ABCD is the base and S is the vertex of the pyramid and SO is the altitude.

OS = 12 cm, AB= 10 cm, PO = 5 cm

OSP is a right angled triangle,

Lateral surface area

Lateral surface area

Volume of the pyramid

Volume of the pyramid

i) Sum of the edges of a cuboid .

ii) Sum of the edges of a cube = 12aunits.

iii) Box with lid:

a)Inner length = Outer length – twice the thickness of wood.

c)Inner height = Outer height – twice the thickness of wood.

iv) Box without lid:

a)Inner length = Outer length – twice the thickness of wood.

c)Inner height = Outer height – thickness of wood.

v) 1 cubic metre = 1 kilolitre = 1000litres.

vi) 1000 cu.cm = 1 Litre.

vii) Rise in level

Cylinder: We may recall that the lateral faces of a right prism are rectangles and its base is in the form of a polygon. In daily life, we come across several objects having no rectangular faces, for example measuring jars, powder tins, road rollers, oil tins etc. In most of the cases their lateral surfaces are curved and their ends are congruent circles. These objects are examples of solid figures called circular cylinders but generally we call them simply cylinders.

Right cylinder: If the line joining the centers of the circular bases is perpendicular to the base, the solid figure is called a rightcircular cylinder.

If the bases of a right prism arecircular it becomes a cylinder. Bases of cylinder are plane surfaces and theyare circular. The lateral surface of the cylinder is a curved surface.

Surface area and volume: Let r is the radius and h is the height of cylinder then,

i) Curved surface area (CSA)

ii) Total surface Area (TSA)

iii) Volume (V)

Example: Find the lateral surface area of aright circular cylinder, if its base has a radius of 7 cm and its height is 10cm.

Solution:You know that the lateral surface is the curved surface

Here r = 7 cm and h = 10 cm
Curved surface area

Curved surface area

Example: It is required to make from a metal sheet a closed cylindrical tank of
height 2 m and base of radius 77 cm. How many square meter of metal sheet isrequired?

Solution: Radius of the base = 77 cm

Height of the cylindrical tank = 2 m

Area of the metal sheet required =Total surface area of the tank.

Total surface area of a cylinder sq.unit

Total surface area of a cylinder

Hollow cylinder: A hollow cylinder is the solid formed by just the lateral surfaces (curved surface) of the cylinder.

Example: Water pipes, electric pipes.

In hollow cylinder both ends are open.

The uniform thickness of the hollow cylinder = (R – r) cm

Surface area and volume:

i) Volume of the hollow cylinder

ii) Curved surface area

iii) Area of ring

iv) Total surface area

Example: The ratio of the radius and height of a cylinder is 3 : 2. The radius is 21 cm. Find its curved surface area, totalsurface area and volume.

Solution:

Example: The total surface area of a cylinder is 220 cm2 and its heights is 6.5cm. Find its volume.

Total surface area = 220 cm2

i) Horizontal cross section of a cylinder is circle.

ii) Vertical cross section of a cylinder is rectangle.

1. Heights of two cylinders are equal. Their radii Rand r. Then, Ratio of radii
i) Ratio of curved surface areas
ii) Ratio of volumes
Conclusion: If the heights of two cylinders are equal, their (lateral) curved surfaces are in the ratio oftheir radii, and their volumes are in the ratio of the squares of their radii.

2. Base radii of two cylinders are equal.Their heights are H, h, then
Ratio of their heights = H : h
i) Ratio of the curved surface areas
ii) Ratio of their volumes
Conclusion:
If the base radii of two cylinders are equal, their curved surface areas and volumes are in the ratio of their heights.