Surface area and volume
1.
2.
Curved surface area of a right circular cylinder is 4.4 sq m. if the radius of the base of the cylinder is 0.7 m find its height.
The circumference of the trunk of a tree (cylindrical), is 44dm. Find the volume of the timber obtained from the trunk
22
if the length of the trunk is 5 m. ( π = ).
7
3.
4.
If the areas of three adjacent faces of a cuboids are X, Y and Z. If its volume is V, prove that V2 =XYZ
Find the volume of an ion has in the shape of cuboids whose length, breadth and height measure 25 cm. 18 cm and 6 cm
respectively. Find also its weight in kilograms if 1 cu cm of iron weight 100 grams.
5.
6.
The surface area of cuboids is 3328 m2; its dimensions are in the ratio 4:3:2. Find the volume of the cuboids.
The volume of a rectangular slower of stone is 10368 dm3 and is dimensions are in the ratio of 3:2:1. (i) Find the dimensions
(ii) Find the cost of polishing its entire surface @ Rs. 2 per dm
2.
7.
In a cylindrical drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space
required for each bag is 2.1 cu m.
8.
The inner diameter of a cylindrical wooden tripe is 24 cm. and its outer diameter is 28 cm. the length
of wooden tripe is 35 cm. find the mass of the tripe, if cu cm of wood has a man of 0.6 g.
9.
The difference between outside and inside surface of a cylindrical metallic tripe 14 cm. long is 44 sq cm.
if the tripe is made of 99 cu cm. of metal, find the outer and inner radius of the tripe.
Answers
Ans 1
Curved surface area of the cylinder = 4.4sq.m
Radius of the cylinder = 0.7m
Let h be the height of the cylinder
∴ Curved surface area of the cylinder = 2π rh
2π rh =4.4
or,
22
2 × × 0.7 = 4.4
7
4.4 × 7
h=
m
2 × 22 × 0.7
44 × 7
m
=
44 × 7
= 1m
Here the height of the cylinder = 1m
Ans 2. Let r be the height of the cylindrical Trunk
Circumference of the trunk = 44dm
2π r = 44
∴
or,
2×
∴
22
× r = 44
7
r=
44 × 7
m
2 × 22
r = 7 dm ⇒
∴
7
m
10
Volume of the timber = π r 2 hcm unit
 22 7 7

=  × × × 5  cu m
 7 10 10 
770
=
cu m
100
= 7.7cu m
Ans 3. Let length, breadth and height of the cuboid l, b and h respectively
∴
v = lbh
(i)
Again,
x = lb
y = bh
and
z = hl
∴
xyz = (lb) (bh) (hl)
= l 2b 2 h 2
= ( lbh )
2
= v2
Hence,
Ans 4.
[u sin g (i )]
v = xyz
2
Length of the bar = 25cm
Breadth of the bar = 18cm
Height of the bar = 6cm
∴
Volume of the iron bar = l × b × hcu unit
= ( 25 × 18 × 6 ) cu cm
= 2700cu cm
Weight of the bar = ( 2700 × 100 ) gm
= 270000gm
= 270kg
Ans 5 Let the dimensions of the cuboid be 4x, 3x and2x meters
Surface area of the cuboid = 2 ( 4 x × 3x × 2 x + 2 x × 4 x ) sq m
= 2 (12 x 2 + 6 x 2 + 8 x 2 ) sq m
= 52 x 2 sq m
Given surface area = 3328sq m
→
(i )
From (i) and (ii) we get
52 x 2 = 3328
3328
or
x2 =
= 64
52
or
x=6
∴
4 x = 32, 3 x = 24 and 2 x = 16
Thus the dimensions of the cuboid are 32m, 24m and 16m
∴
Volume of the cuboid = ( 32 × 24 × 16 ) m3
= 12288cu m
Ans 6. Let the length of the block be 3xdm
Width -2xdm 4height = xdm
Volume of the block = 10368 dm 2
∴
3 x × 2 x × x = 10368
10368
or
x3 =
6
= 1728
∴
x = 3 1728
= 3 12 × 12 × 12
= 12
also 2x = 24 and 3x = 36
Thus dimensions of the block are 36dm, 24dm and 12dm
Surface area of the block = 2 ( 36 × 24 + 24 × 12 + 36 × 12 ) dm 2
= 2(864+288+432) dm 2
= 2 × 1584dm2
= 3168dm2
Cost of polishing the surface = Rs ( 2 × 3168 )
= Rs6336
Ans 7.
42
m
10
35
Height of the drum =3.5m =
m
10
∴
Volume of the drum = π r 2 hcu units
Radius of the drum = 4.2m =
 22 42 42 35 
=  × × ×  cu m
 7 10 10 10 
21
Volume of wheat in each bags = 2.1cu m = cu m
10
volumeofdrum
∴
No.ofbags =
volumeofwheatineachbag
22 42 42 35
× × ×
7
10 10 10
=
[ from (i ) and (ii )]
21
10
924
=
= 92.4
10
= 92
Hence the number of full bags is 92
→
(i )
→
(ii )
Ans 8.
Inside diameter of the pipe = 24cm
Outside diameter of the pipe = 28cm
Length of the pipe = 35cm = (l says)
Outside radius of the pipe =
28
cm = 14cm = R( says )
2
Volume of the wood = External volume – Internal volume
= π R 2l − π 2l
= π × 35 (14 2 − 122 ) cu cm
22
× 35 (14 + 12 )(14 + 12 ) cu cm
7
= 5720cu cm
Mass of 1cu cm = 0.6g
∴ Mass of the pipe = ( 0.6 × 5720 ) g
=
= 3432g
= 3.432kg
Ans 9.
Let r1cm and r2cm can be the inner and outer radii respectively of the pipe
Area of the outside surface = 2π r2 hsq unit
Area of the inside surface = 2π r1hsq unit
∴ By the given condition
2π r2 h - 2π r1h = 44
2π h ( r2 − r1 ) = 44
or
∴
2×
or
∴
22
×14 × ( r2 − r1 ) = 44 (∵ h = 14cm )
7
88 ( r2 − r1 ) = 44
1
(i )
2
Again volume of the metal used in t he pipe = π ( r2 2 − r12 ) hcu units
( r2 − r1 ) =
22 2
r2 − r12 ) × 14 = 99
(
7
99 9
44 ( r2 2 − r12 ) =
=
44 4
∴
or
( given)
(ii )
Dividing (ii ) by (i ) we get
(r
2
or
∴
(r
r
2
2
− r12 )
r2 − r1
2
=
− r12 ) ( r2 + r1 )
( r2 − r1 )
9 1
÷
4 2
9 2
= ×
4 1
( r2 + r1 ) =
9
2
( r2 − r1 ) =
1
2
Also,
Adding
2π r = 5
5
r2 =
2
∴
and ,
∴
or
and
5
9
+ r1 =
2
2
9 5
r1 = −
2 2
r1 = 2
Thus outer radius = 25c
inner radius = 2cm
[ from(i )]