Deals with the understanding of surface area and volume of solid geometrical figures-cube, cuboid, cylinder, cone and sphere

1. MENSURATION-sk
STD X
MAHARASHTRA STATE BOARD OF EDUCATION,
MUMBAI

2. MENSURATION
Mensuration is a branch of mathematics which
deals with the surface area and volume of solid,
plane and geometrical figures.

3. MENSURATION
The area of a figure is the number of unit
squares that cover the surface of a closed
figure. Area is measured in square units such as
square centimetres, square feet, square inches,
etc.

4. MENSURATION
In math, volume can be defined as the 3-
dimensional space enclosed by a boundary or
occupied by an object. ... Here, for example,
the volume of the cuboid or rectangular prism,
with unit cubes has been determined in cubic
units.

5. MENSURATION

6. MENSURATION
CURVED SURFACE
AREA (LATERAL
SURFACE AREA)
TOTAL SURFACE
AREA
AND THE AREA OF TOP AND
BOTTOM SURFACES
CURVED SURFACE AREA
(LATERAL SURFACE
AREA)

7. MENSURATION- surface area
CUBEA cube is a three-dimensional solid (has length, breadth,
height) object bounded by six square faces, with three
meeting at each vertex. The cube is the only regular
hexahedron and is one of the five Platonic solids. (solids
having regular faces- refer adjacent figure)
Area of each surface = (side)2
Lateral surface area = 4 x (side)2
Total surface area = 6 x (side)2

8. MENSURATION- surface area
CUBOID
A cuboid is a convex polyhedron bounded by six
quadrilateral faces,
Having dimensions l, b and h
Area of a blue surface = l x h
Area of a pink surface = b x h
area of a green surface = l x b
Lateral surface area = 2 (lh +bh)
=2 h (l +b)
Total surface area =2 (lh +hb + lb)
= 2(lb +bh +hl)

9. MENSURATION- surface area
h
CYLINDER A cylinder has traditionally been a three-dimensional solid. It
is the idealized version of a solid physical tin can having lids
on top and bottom. Dimensions h, r
L= Circumference of a circle= 2𝜋𝑟
𝐴 = 𝜋𝑟2
Area of rectangle= l x h= 2 𝜋𝑟 h
Lateral surface area = 2 𝜋𝑟 h
Total surface area =2 𝜋𝑟 h + 2 𝜋 r2
=2 𝜋𝑟 (h + r)
L

10. MENSURATION- surface area
CONE
A cone is a three-dimensional geometric shape that tapers smoothly
from a flat base to a point called the apex or vertex.
Dimensions: h, r, l
Curved surface area = 𝜋 𝑟 l
Total surface area = 𝜋 𝑟 l +
= 𝜋𝑟 (r +l)
Remember : l2 = r2 + h2
𝜋𝑟2

11. MENSURATION- surface area
SPHERE AND HEMISPHERE
A sphere is a geometrical object in
three-dimensional space that is the
surface of a ball.
Dimension: r
Curved surface area = 4 𝜋 𝑥 𝑟 2
Curved surface area = 2 𝜋 𝑥 𝑟 2
Total surface area = 3 𝜋 𝑥 𝑟 2

12. MENSURATION-volume
In math, volume can be defined as the 3-
dimensional space enclosed by a boundary or
occupied by an object. ...
GENERAL FORMULA: volume = A(BASE) x HEIGHT
VOLUME OF A CUBE = A(SQUARE) x H
V(CUBE) = side x side x H
V (cube ) = (side)3 ( H = side)

13. MENSURATION-volume
GENERAL FORMULA: volume = A(BASE) x HEIGHT
VOLUME OF A CUBOID =A(rectangle) x H
V(CUBOID) = L x B x H
VOLUME OF A CYLINDER = A(CIRCLE) x H
V(CYLINDER) = 𝜋 𝑥 𝑟 2 x h

14. MENSURATION-volume
cone
The volume of a cone means the third part
of the volume of a cylinder having the same
base and the same height.
It takes three cones to fill up a cylinder.

15. MENSURATION-volume
sphere
The sphere volume is 2/3 of
the volume of a cylinder with the
same radius and height equal to the
diameter.
V(CYLINDER) = 𝜋 𝑥 𝑟 2 x h
V(SPHERE) =
2
3
x 𝜋 𝑥 𝑟 2 x 2 r
V(SPHERE) =
4
3
x 𝜋 𝑥 𝑟 3

16. MENSURATION-volume
hemisphere
V(CYLINDER) = 𝜋 𝑥 𝑟 2 x h
V(SPHERE) =
2
3
x 𝜋 𝑥 𝑟 2 x 2 r
V(SPHERE) =
4
3
x 𝜋 𝑥 𝑟 3
V(HEMISPHERE) =1
2
x(
4
3
x 𝜋 𝑥 𝑟 3
= 2
3
x 𝜋 𝑥 𝑟 3

17. Formulae to find SURFACE AREA AND VOLUME
3D –SOLID
FIGURE
DIMENSIONS SURFACE AREA VOLUME
CURVED TOTAL
CUBE Side 4 x side2 6 x side2 Side3
CUBOID
l,b,h
2 h (l +b) 2(lb +bh +hl)
L x b x h
CYLINDER
r,h 2 𝜋𝑟 h 2 𝜋𝑟 (r + h) 𝜋 x 𝑟 2 x h
CONE
r,h,l 𝜋𝑟 l 𝜋𝑟 (r +l) 1
3
x 𝜋 x 𝑟 2 h
SPHERE
r 4𝜋 x 𝑟 2 4𝜋 x 𝑟 2 4
3
x 𝜋 x 𝑟 3
HEMISPHERE
r 2𝜋 x 𝑟 2 3𝜋 x 𝑟 2 2
3
x 𝜋 x 𝑟 3

18. APPLICATION
SURFACE AREA AND VOLUME
For a cone: r =1.5cm, h = 5cm
To find: volume of a cone
Formula: v(cone) =
1
3
x 𝜋 𝑥 𝑟 2H
=
1
3
x 3.14 x 1.5 x 1.5 x 5
=
1
3
𝑥
314
100
x
15
10
x
15
10
x 5
=11.775 cubic cm (157 X 75)
Find the volume of a cone if the radius of its base is
1.5cm and its perpendicular height is 5 cm.

19. APPLICATION
SURFACE AREA AND VOLUME
Find the surface area of a ball.
42

20. APPLICATION
SURFACE AREA AND VOLUME
Find the total surface area of a cylinder if the radius of its
base is 5 cm and its height is 40 cm
Cylinder: r =5 cm h = 40 cm
To find: total surface area
of the cylinder.
Total surface area (cylinder)= 2 𝜋𝑟 (h + r)
= 2 x
314
100
x 5 (5+40)
=2 x
314
100
x 5 x 45 cubic cm
= 1413 cubic cm

21. APPLICATION
SURFACE AREA AND VOLUME
Atoymadefromahemisphere,acylinderandaconeis
shown.Findthetotalareaofthetoy.
TOTAL S AREA = CS (HEMISPHERE) +CS (CYLINDER) + CS (CONE)
=2 𝜋 𝑥 𝑟 2 +2 𝜋𝑟 hc + 𝜋𝑟 l
= 𝜋 r (2r + 2hc + l)
Calculation for slant height (l) : l2 = r2 + hcone
2
=32 + 42
= 9+16
= 25
L = 5
total area= 𝜋 r(2X3 + 2 X 40 + 5)
=
22
7
X 3 (91)
= 22 X 3 X 13 sq. cm
=858 sq. cm
Hemisphere: R = 3 cm,
Cylinder: r =3 cm Hc = 40 cm
Cone: r =3 cm, hcone = 4 cm, l=?
To find: surface area of the toy

22. APPLICATION
SURFACE AREA AND VOLUME
The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. it is
melted and a cone of height 24 cm is made. Find the
radius of its base.
Cuboid: l= 44 cm, b = 21cm, h = 12cm
Cone r = ?, h =24 cm
Volume of cuboid = volume of cone
lXbXh =
1
3
x 𝜋 𝑋 𝑟 2h
44X21X12=
1
3
X
22
7
X 𝑟 2X248
44𝑋21𝑋12𝑋7
22 𝑋 8
= r 2
21 X 21 = 𝑟 2
21 = radius of the base of a cone.

23. APPLICATION
SURFACE AREA AND VOLUME
A cylinder and a cone have equal bases. The height of the
cylinder is 3 cm and the area of its base is 100 cm2
. The
cone is placed upon the cylinder. Volume of the solid
figure so formed is 500 cm3
. Find the total height of the
figure.
Cylinder: 𝜋 𝑥 𝑟 2 = 100cm2,
h = 3 cm
Cone: H=?
V(cylinder +cone) = 500 cm3
Total height of the figure= 6+3 = 9cm

24. APPLICATION
SURFACE AREA AND VOLUME
In a cylindrical glass, diameter = 14cm and h=30 cm, containing water, a metal
sphere of diameter 2 cm is immersed. Find the volume of the water.
Cylinder: d1 = 14cm, h = 30 cm, sphere: d2 = 2 cm.
To find: volume of water in the cylinder
Volume of water = volume of cylinder - volume of sphere
= 𝜋 𝑥 𝑟 2 x h -
4
3
x 𝜋 𝑥 𝑟 3
= 𝜋(49X30 -
4
3
X 1)
= 𝜋(1470 - 1.33)
= 𝜋(1468.67) cubic cm
In the cylinder

25. APPLICATION -SURFACE AREA AND VOLUME
Find the volume and the surface area of the toy shown.
Given: for a cone: r = 3 cm, h1 = 4 cm
for a hemisphere: r = 3 cm
To find total volume and total surface area
Total volume= v(cone) +
v(hemisphere)
=
1
3
x 𝜋 𝑥 𝑟 2h1 +
2
3
x 𝜋 𝑥 𝑟 3
= 𝜋 𝑥 𝑟 2(
1
3
h1 +
2
3
𝑟)
=3.14 X9(1.33+2)
………(note: h1=4cm)
=28.26(3.33)= 94.1 cm3
Length of lateral surface:
L2 = r2 + h1
2 , l2 = 42 + 32 , l = 5
Total surface are = surface area of
cone + surface area of hemisphere
= 𝜋𝑟 l +2 𝜋 𝑥 𝑟 2
= 𝜋𝑟 (I + 2r)
= 3.14 X3(5+ 6)
= 9.42(11)
= 103.62 cm2

26. APPLICATION
SURFACE AREA AND VOLUME
Using information given in the figure, find how
many jugs of water can the cylindrical pot hold?
(measurements are in cm).
10
3.5
10
R=7

27. APPLICATION
SURFACE AREA AND VOLUME
The radius of a tablet in a cylindrical wrapper is 7 mm and its
thickness is 5 mm. if the height of the wrapper is 10cm and diameter
is 14 mm then find the number of tablets in the wrapper
Wrapper (cylindrical): diameter:
14mm, height 10 cm = 100 mm
Tablets(cylindrical) : r = 7 mm,
height=5 mm

28. APPLICATION -SURFACE AREA AND VOLUME
Find the ratio of the volumes of a
cylinder and a cone having equal
radius and equal height
The radii of two cylinders are in the
ratio 2:3 and their heights are 5:3.
Find the ratio of their volumes
(𝜋 𝑥 𝑟 2 x h)1 : (𝜋 𝑥 𝑟 2 x h)2
𝜋 X 2X 2X5 : 𝜋 X3 X 3 X3
20: 27
Find the ratio of the volumes of a
cylinder, a cone and hemisphere
having equal radius and equal
height
𝜋 𝑥 𝑟 2 x h :
1
3
x 𝜋 𝑥 𝑟 2 h :
2
3
x 𝜋 𝑥 𝑟 3
3 : 1 : 2

29. REVISION
1. How much oil a cuboid can, having dimensions
20cm, 20 cm and 30 cm contain?
( 1 litre = 1000 cm3 )
2. How much cloth is needed to stitch a conical cap
having radius of its base as 10 cm and slant
height 21 cm.
3. How many solid cylinders of radius 10cm and
height 6 cm can be made by melting a solid
sphere of radius 30 cm?
4. Find the curved surface area of a cone of radius
7 cm and height 24cm.

30. REVISION
1. Find the volume of a cube having length of
side 6 cm.
2. In a cylinder if radius is halved and height is
doubled then volume will be – same /
double/ halved/four times?
3. The curved surface area of a cylinder is 440
cm2 and its radius is 5 cm then find its height.

31. THANK
YOU