surface area and volume class 10

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CLASS
X-A

CUBE :-
A cube is a three
dimensional figure,
with six sides- all
Sides in shape of
Square.
Length of side is denoted by the letter ‘l’.
l

Lateral Surface Area :-
Lateral surface area refers to
the area of only the walls ( it
does not include the area of
the floor and roof).
Formula :- 4 l²
Derivation :- Since all the sides
of cube are in the shape of
square.
area of the square= l²

EXAMPLES :-
1. Find the lateral surface area of the
cube with side of 15cm.
Sol.- We are given-
l = 15cm
lateral surface area = 4l²
= 4(15
cm)²
= 4*

EXAMPLES :-
2.Find the lateral surface area of the cube
with area of one face 81cm². Also find the
length of the side.
Sol. – Area of one face = 81cm²
l² = 81cm²
l = √81cm²
l = 9cm
Lateral surface area of cube = 4l²
= 4(9cm)²
=4*81cm²
= 324cm²

Total Surface Area Of Cube :-
Formula :- 6l²
Derivation :- Since all the faces of a cube
are squares ,
Area of square = l²
No. of square = 6
Area of 6 square = Total surface
area of cube
= 6l²
Therefore , total surface area of the cube is
6l² .

EXAMPLE :-
1. Find the total surface area of the cube
with side of 7.2cm.
Sol. - We are given,
l = 7.2cm
Total surface area = 6l²
= 6(7.2cm)²
= 6*51.84 cm²
= 311.04 cm²

EXAMPLE :-
2.A gift in a shape of cube is to be wrapped in
a gift paper. Find the total cost of the
wrapper need to cover the gift whose side
is 6.8cm, at the cost of Rs.5 per m².
Sol. – We are given ,
side of the cube (l) = 6.8 cm
Total surface area = 6 (l)²
= 6 (6.8 cm)²
= 6* 46.24
cm²
= 277.44 cm²
cost of the wrapper = Rs. 5/m²
= 5*2.7744m²
= Rs. 13.87

Volume Of Cube : -
Volume of the cube refers to the
space inside the six walls.
Formula :- l * l * l
= l³
Unit :- unit³

CUBOID :-
Cuboid is a three dimensional figure,
with six sides and all sides of equal length.
In Cuboid opposite rectangles are
equal.
It’s three dimensions are :- 1.Length(l)
2. Breadth (b)
3. Height (h)
l
b
h

LATERAL SURFACE AREA:-
Lateral surface area of the cuboid refer to the area of
the four walls of it.
Formula :- 2(l+b) h
Derivation :- Area of rectangle1 = l*h
Area of rectangle2 = b*h
Area of rectangle3 = l*h
Area of rectangle 4 = b*h
Total area =2lh+2bh
= 2(l+b) h
l
b
h

TOTAL SURFACE AREA:-
Formula :- 2(lb + bh + hl )
Derivation :- Area of rectangle 1 (= lh) +
Area of rectangle 2 (=lb )+
Area of rectangle 3 (=lh ) +
Area of rectangle 4 (=lb ) +
Area of rectangle 5 (=bh ) +
Area of rectangle 6 (= bh )
= 2(l*b ) + 2 ( b*h ) + 2 (l*h )
= 2 ( lb + bh + hl )
h

EXAMPLE :-
1. Marry wants to decorate her Christmas tree. She
wants to place her tree on a wooden box covered with
coloured paper with picture of Santa clause on it . She
must know the exact quantity of paper to buy it. If
the dimensions of the box are : 80cm* 40cm* 20cm,
how many square sheets of paper of side 40cm would
she require?
Sol. –The surface area of the box = 2(lb + bh + hl )
= 2[ ( 80*40) +(40*20)
+(20*80)]
= 2 (3200 + 800 + 1600
)
= 2 * 5600 cm³ = 11200
cm³
The area of each sheet of paper= 40* 40 cm² = 1600cm²
Therefore no. of sheets require = Surface area of the
box/ Area of one sheet of paper
= 11200/ 1600 = 7
Therefore , she would require 7 sheets.

CYLINDER :-
A right circular cylinder
is a solid generated
by the revolution of a
rectangle about one
of its side.
It is a folded rectangle
with both circular
ends.
h
r

CURVED SURFACE AREA OF CYLINDER:-
Curved surface area of the cylinder :-
= Area of the rectangular sheet
= length * breadth
= perimeter of the base of the
cylinder* h
= 2πr * h
= 2πrh

1. Shubhi had to make a model of a cylindrical
kaleidoscope for her project. She wanted to use
chart paper to use chart paper to make the
curved surface of it. What would be the area of
chart paper required by her, if she wanted to
make a kaleidoscope of length-25cm with a 3.5cm
radius ?
Sol. – Radius of the base of the cylindrical
kaleidoscope (r) = 3.5cm
Height (length) of kaleidoscope (h)
= 25cm
Area of paper required = curved surface area of
kaleidoscope
= 2πrh
= 2*22/7*3.5*25 cm²

TOTAL SURFACE AREA OF CYLINDER :-
Total surface area of a cylinder :
= area of the rectangular sheet + 2 (area of the
circular regions )
= perimeter of the base of cylinder* h + 2 (area of
circular base )
= 2πrh + 2πr²
= 2 πr ( r + h )
h
r

EXAMPLE :-
1. A barrel is to be painted from inside and outside. It has no lid
.The radius of its base and height is 1.5m and 2m respective.
Find the expenditure of painting at the rate of Rs. 8 per
square meter.
Sol. – Given, r= 1.5m , h = 2m
Base area of barrel = πr²
Base area to be painted (inside and outside ) = 2 πr²
=2 * 3.14 * (1.5 )²
cm²
= 2* 3.14 * 2.25 =
14.13cm²
Curved surface area of barrel = 2 πrh
Area to be painted = 2 * 2 πrh
= 4 * 3.14 *1.5 *2 cm²
= 12 * 3.14cm² = 37.68
cm²
Total area to be painted = ( 37.68 + 14.13 ) cm² =
51.81 cm²
Expenditure on painting = Rs. 8 * 51.81
= Rs. 414.48

VOLUME OF CYLINDER :-
Volume of a cylinder can be built up
using circles
Of same size.
So, the volume of cylinder can be
obtained as :-
base area * height
= area of circular base * height
= πr²h
r

EXAMPLE :-
1. A measuring jar of one liter for measuring milk is of
right circular cylinder shape. If the radius of the base is
5cm , find the height of the jar.
Sol. – Radius of the cylindrical jar = 5cm
Let ‘h’ be its height
Volume = πr²h
Volume = 1 liter = 1000cm³
Πr²h = 1000
H = 1000/πr²
H = 1000 *7 / 22*5*5 cm
= 1000*7 / 22*25 cm
= 140 / 11 cm = 12.73 cm
Height of the jar is 12.73 cm .

1. Find the weight of a hollow cylindrical lead pipe 26cm long
and 1/2cm thick. Its external diameter is 5cm.(Weight of
1cm³ of lead is 11.4 gm )
Sol. – Thickness = 1/2cm
External radius of cylinder = R= (2+1/2)cm = 5/2cm
Internal radius of cylinder = r = (5/2 – 1/2 ) = 2 cm
Volume of lead = π(R² - r² )*h = π[ (5/2)² - 2²]
*26
= 22/7 *[25/4 – 4] *26
= 22/7*(25-16/4) *26
=11*9*13/7 = 1287/7 cm³
Weight of 1cm³ of lead = 11.4 gm
Weight of cylinder = 11.4 *1287/7 gm =
14671.8/7 gm
= 2095.9714 gm
= 2095.9714/1000
kg = 2.0959714 kg
= 2.096kg
Therefore, weight of the cylindrical

RIGHT CIRCULAR CONE :-
If a right angled triangle is revolved about one
of its sides containing a right angle, the solid
Thus formed is called a right circular cone.
The point V is the vertex of cone.
The length OV=h, height of the cone
The base of a cone is a circle with O as center
and OA as radius.
The length VA = l , is the slant height of the
cone.
V
l
h
O
r
A

CURVED SURFACE AREA OF CONE :-
It is the area of the curved part of
the cone. (Excluding the circular base )
Formula :-
1/2* perimeter of the base* slant height
= ½ * 2πr * l
= πr l
l
r

EXAMPLE :-
2.How many meters of cloth 5m wide will be required to
make a conical tent , the radius of whose base is 7m and
whose height is 24m ?
Sol. – Radius of base = 7m
Vertical height , ‘h’ = 24m
Slant height ‘l’ = √ h² + r² = √(24)² + (7)²
=√576 + 49 = √625 = 25 m
Curved surface area = πrl
= 22/7 *7*25 m² = 550 m²
Width of cloth = 5m
Length required to make conical tent = 550/5 m
= 110m

TOTAL SURFACE AREA OF CONE :-
Total surface area of the cone :-
=Curved surface area of cone + circular base
( Red coloured area + green coloured area )
=πrl + πr²
=πr ( l + r ) hh
h
l
r

EXAMPLE :-
1. Total surface area of a cone is 770cm². If the slant height of cone is 4 times
the radius of its base , then find the diameters of the base.
Sol. – Total surface area of cone = 770 cm²
= πr (r + l ) = 770
= l = 4 * radius
= = 4r
= πr (r + 4r ) = 770
= 5πr ² = 770
= r² = 770 *7 / 5 *22 = 7 * 7
= r = 7cm
Therefore, diameter of the base of the cone is 14cm.

VOLUME OF THE CONE :-
Formula :- 1/3 πr²h
Derivation :- If a cylinder and cone of sane base
Radius and height are taken , and if cone is put
Under the cylinder then it will occupy only
One –third part of it .
Therefore, volume of cone is 1/3 of the volume of
Cylinder.
= 1/3πr²h
hh
1
2
3
h
l
rr

EXAMPLE :-
1. The radius and perpendicular height of a cone are in the ratio 5 :12. if the volume of the cone is
314cm³, find its perpendicular height and slant height.
Sol. – Let the radius of the cone = 5x
Perpendicular height of the cone = 12x
Volume of the cone = 314 m³
Hence,
1/3πr²h = 314
= Πr²h = 942
= 3.14 (5x)² (12) = 942
= 3 * 314 x³ = 942
= x³ = 1
= x = 1
Therefore, perpendicular height of the cone
= 12m
And radius of the cone =
5m
Slant height of cone = √ r² + h ² = √5²
+ 12²

EXAMPLE :-
2.A wooden right circular cone has a base of radius 3cm and height 4cm.
The upper part of the cone cut is in such a way that the conical
piece will have height 1cm and base radius 0.75cm. Find the volume
of the remaining portion.
Sol. – For complete cone, r = 3cm
height ‘h’ = 4cm
Volume of the complete cone = 1/3πr²h
= 1/3 * π * 3 *3 *4 = 12 π cm³
For the upper part of cone, radius = 0.75cm , height = 1cm
= 1/3πR²H
= 1/3*π*0.75*0.75*1
= 0.1875π cm³
Volume of the remaining portion of the cone –
= Volume of the complete cone –
volume of the cut cone
= 12 π – 0.1875π = 11.8125π =
11.8125 *3.14
= 37.09 cm³

SPHERE : -
The set of all points in space equidistant from a fixed
point, is called a sphere .
The fixed point is called the center of the sphere.
A line segment passing through the center of the sphere
with its end points on the sphere is called a diameter
of the sphere.
r

SURFACE AREA OF SPHERE : -
Surface area of the
sphere :-
=4πr²
r

EXAMPLE :-
1.If the diameter of a sphere is ‘d’ and curved
surface area ‘S’, then show that S = πd². Hence,
find the surface area of a sphere whose
diameter is 4.2 cm.
Sol. – d = 2r
Curved surface area of sphere = S
= 4πr² = π * 4r²
= π(2r)²
= πd²
Here, d = 4.2cm
Surface area of the sphere = πd² = 22/7
* (4.2)²
= 55.44cm²

VOLUME OF THE SPHERE :-
Volume of the sphere :-
=4/3πR³

EXAMPLE :-
1. How many spherical bullets can be made out of lead whose edge
measures 44cm, each bullet being 4cm in diameter.
Sol. – Let the total no. of bullets be x
Radius of spherical bullet = 4/2 cm = 2cm
Volume of a spherical bullet = 4/3 π * (2)³ cm³
=(4/3 *22/7 *8 ) cm³
Volume of solid cube = (44)³ cm³
Number of spherical bullets recast = volume of cube
= 44*44*44*3*7
volume of one
bullet 4 *22*8
= 33*77
= 2541

EXAMPLE :-
2. If the radius of a sphere is doubled , what is the ratio of the volume of the
first sphere to that of the second ?
Sol. – For the first sphere ,
Radius = r
Volume = V1
For the second sphere,
Radius = 2r
Volume = V2
Then , V1 = 4/3πr³
V2 = 4/3π(2r)³ = 4/3π(8r³)
Therefore, V1 = 4/3πr³ = 1
V2 = 4/3π*8³ = 8
Ratio = 1:8

HEMISPHERE :-
A plane passing through the centre of a
sphere divides the sphere into two equal
parts .
Each part is known as hemisphere.
r

CURVED SURFACE AREA OF HEMISPHERE
Formula : - 2πr²
Derivation :-
Since,
hemisphere is half of sphere-
Therefore,
Surface area of sphere = 4πr²
Half of it = 2πr²
r

TOTAL SURFACE AREA OF HEMISPHERE :
Total surface area of
hemisphere:
= Curved surface area +
circular base
= 2πr² + πr²
= 3πr²

EXAMPLE :-
1. The internal and external diameters of a hollow hemispherical vessel are 25cm and
24 cm respectively. The cost to paint 1cm³ of the surface is Rs 0.05. Find the total
cost to paint the vessel all over .
Sol. – External area which is to be painted
= 2πR² = 2*22/7*25/2*25/2 cm²
= 6875/7 cm²
Internal area which is to be painted
= 2πr² = 2*22/7 * 24/2*24/2 cm²
= 6336/7 cm²
Area of the ring at top = 22/7 {(25)² + ( 24/2 )² }
= 22/7 [ (12.5)² + (12) ² ] = 22/7 “(12.5 +12) (12.5-
12) = 22/7 *24.5 *0.5
= 269.5/ 7 cm²
Total are to be painted= 6875 + 6336 + 269.5 = 13480.5 cm²
7 7 7 7
= 1925.78 cm²
Cost of painting @ Re. 0.05/cm² Rs.
= Rs. 1925.78 *0.05 = Rs. 96.289

FRUSTUM : -
If a cone is cut by a plane parallel to the base then the part
between the base and the plane is called frustum
of the cone
Here, EBSF is frustum from the cone ABC.
PF = R = radius
QC = r = radius
PQ = h = height
FC = l = slant height

SURFACE AREA OF THE FRUSTUM :-
#CURVED SURFACE AREA OF FRUSTUM :-
= πl (R +r ) + πR² + rπ²
l=√ h² + (r – r )²
#TOTAL SURFACE AREA OF FRUSTUM :-
= π (R + r) l
l = √ h² + ( R – r )²
R
r
h
r
R
h
l

EXAMPLE :-
1.A friction clutch is in the form of frustum of a cone the
diameter being 16cm and 10 cm and length 8cm. Find its
bearing surface .
Sol. – Let ABB’A’ be the friction clutch
Let ‘l’ be its slant height
l = √ 8² + (8-5)²
= √ 64 + 9 = √73 cm
Bearing surface = Lateral surface area of ABB’A’
= πl (R +r ) = 22/7 *√73(8 + 5 ) cm²
= 349 cm sq.

VOLUME OF THE FRUSTUM :-
Volume of frustum of
cone :
= 1/3 πh (R² + r² +
Rr)

EXAMPLE :-
1. The radii of the ends of a bucket of height 24 cm are
15 cm and 5 cm. Find its capacity.
Sol. – Capacity of the bucket = Volume of the frustum
= πh [ r² + R² + Rr ]
3
= 22 * 24 [(15)² + 5² + 15 * 5 ]
cm³
7 3
= 22 *8 [ 225 +25 + 75 ] cm³
7
= 176 * 325 cm³
7
= 8171.43 cm³

SURFACE AREAS AND VOLUMES OF
COMBINATION OF SOLIDS :-
SOME EXAMPLES :-
1.The decorative block is made up of two solids – a cube
with edge 5cm and a hemisphere fixed on the top
has a diameter of 4.2cm. Find the total surface area of
the solid.
Sol. – Total surface area of cube = 6*5*5
= 150 cm²
The surface area of block = Total surface area of block +
curved surface area of the hemisphere – area of the base
of the hemisphere
= 150 - πr² + 2πr² = (150 +πr²)
cm²
= 150 cm² + [ 22/7 * 4.2/2 *4.2/2]
cm²
=( 150 + 13.86) cm² = 163.86 cm²

EXAMPLE :-2. Mayank made a bird –bath for his garden in the
shape of a cylinder with a hemispherical
depression at one end . The height of the
cylinder is 1.45m and its radius is 30cm . Find
its total surface area.
Sol.- Let ‘h’ be the height of the cylinder
‘ r’ be the common radius of the cylinder , and
hemisphere
Total surface area of the bird bath = CSA of
cylinder +CSA of hemisphere
= 2πrh +
2πr² = 2πr (r +h)
= 2*22/7
*30(145 +30)cm²
= 33000cm²

EXAMPLE :-
3.A solid consisting of a right circular
cone of height 120cm and radius 60cm
standing on a hemisphere of radius
60cm is placed in a right circular
cylinder full of water such that it
touches the bottom. Find the volume of water left in the
cylinder
if the radius of the cylinder is 60cm and its height is
180 cm.
Sol. – We are given :-
Height of the cone ‘h’ = 120cm
Radius of the cone ‘r’ = 60cm
Radius of the hemisphere ‘R’ = 60cm
Radius of the cylinder ‘R2’ = 60cm

EXAMPLE :-
Volume of water left = Volume of cylinder-
(volume of cone +
volume of hemisphere)
= πR2² h – (1/3 r²h + 2/3πR³ )
= πR² h - [1/3π( r²h + 2R2³) ]
= πR²h – [ 1/3π (60 * 60 *120 + 2 *
60 * 60 * 60 ) ]
= πR²h – [1/3π(432000 + 432000)]
= πR² h – [ 1/3π 864000} = πR² -
(π*288000)
= π ( R²h – 288000) = π(3600 * 180 –
288000)
= π ( 648000 – 288000 ) = π *
360000
= 3.14 * 360000 = 1130400cm³

EXAMPLE :-
4. A solid iron pole consists of a
cylinder of height 220cm and base
diameter 24cm, which is surmounted by
another cylinder of height 60cm and
radius 8cm. Find the mass of the pole,
given that 1cm³ of iron has approximately
8gms mass.
Sol.- Dimension of smaller cylinder-
Radius ‘ r’ = 8cm
Height ‘h’ = 60 cm
Dimension of large cylinder-
Radius ‘ R’ = 12cm
Height ;H’ = 220cm

Volume of the statue = Volume of large cylinder +
Volume of
small
cylinder
= πR²H + πr²h
= 3.14*8*8*60 +
3.14*12*12*220
= 3.14 * 64 * 60 + 3.14 *
144 * 220
= 2009.6*60 + 690.8*144
= 12057.6 + 99476.2 =
111533.8cm³
Mass of the statue-
Mass of 1cm³ = 8gms
Mass of 111533.8 cm³ = 111533.8 *8

CONVERSION OF SOLIDS FROM ONE
FIGURE INTO OTHER :-
1.A metallic right circular cone 20cm high whose
Vertical angle is 60° is cut into two parts at the
middle of its height by a plane parallel to its base.
If the frustum so obtained be drawn into a wire
of diameter 1/16 cm, find the length of wire.
Sol.- ∆ABC H=20cm
β=60°
r= tan 30 =r/20
= 1/√3 = r/20
= r = 20/√ 3 cm
∆ADE α = 30°
R = tan 30 = R/10
= 1/√3 = R/10
= R = 10/√3 cm
Frustum BCED h = 10cm

Volume of the frustum BCED –
= 1/3π h (R² + r² + Rr )
= 1/3π * 10 [ (20/√3)² + (10/√3 )² + (20/√3)
(10/√3)]
=10/3 π ( 200/3 + 100/3 + 200/3) = 10π/3
(700/3 )
=10 * 22 * 700
3 * 7 3
= 22000/7 cm³
Area of the wire :-
d =1/16cm
r = 1/32 cm
l = ?
πr² = 22/7 * 1/32 * 1/32
= 22/7168 cm²

Length of the wire = Volume of frustum
Area of the wire
= 22000/9 cm³
22/7168 cm²
= 22000 * 7168
9 22
= 7168000/9 cm
= 71680 /9 m
= 7964.4 m

EXAMPLE :-
2.How many silver coins, 1.75cm in diameter and
thickness2mm,
must be melted to form a cuboid of dimension
5.5cm*10cm*3.5cm?
Sol.-Radius of the coin ‘r’ = 1.75/2 cm
Height of the cone ‘h’ = 2mm = 0.2cm
Volume of coin = πr²h = 22/7 *1.75/2*
1.75/2* 0.2
= 0.481cm
Length of cuboid ‘l’ = 5.5cm
Breadth of cuboid ‘b’ = 10 cm
Height of cuboid ‘h ‘ = 3.5cm

Volume of the cuboid = l * b * h
= 5.5 * 10 * 3.5
= 192.5 cm³
Number of coins needed = volume of cuboid
volume of coin
= 192.5 *100
0.481 *100
= 400 coins

surface area and volume class 10

surface area and volume class 10

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surface area and volume class 10