25.VOLUME AND SURFACE AREA
IMPORTANT FORMULAE
I. CUBOID
Let
length = 1, breadth = b and height = h units. Then, 1. Volume = (1 x b x h)
cubic units.
2. Surface area= 2(lb + bh + lh)
sq.units.
3. Diagonal.=Öl2 +b2
+h2 unitsII. CUBE
Let
each edge of a cube be of length a. Then,
1. Volume = a3
cubic units.
2. Surface area = 6a2
sq. units.
3. Diagonal = Ö3 a units.
III. CYLINDER
Let
radius of base = r and Height (or length) = h. Then,
1. Volume = (P r2h) cubic
units.
2. Curved surface
area = (2P rh). units.
3. Total surface
area =2Pr (h+r) sq. units
IV. CONE
Let
radius of base = r and Height = h. Then,
1. Slant height, l =Ö h2+r2
2. Volume = (1/3) Pr2h cubic units.
3.
Curved surface area = (Prl) sq. units.
4. Total surface
area = (Prl + Pr2 ) sq. units.
V. SPHERE
Let
the radius of the sphere be r. Then,
1.
Volume = (4/3)Pr3 cubic units.
2. Surface area = (4Pr2) sq. units.
VI. HEMISPHERE
Let
the radius of a hemisphere be r. Then,
1.
Volume = (2/3)Pr3
cubic units.
2.
Curved surface area = (2Pr2)
sq. units.
3. Total surface area = (3Pr2) units.
Remember: 1 litre = 1000 cm3.
SOLVED EXAMPLES
.
Ex. 1. Find the volume and surface area of a
cuboid 16 m long, 14 m broad and
7 m high.
Sol.
Volume = (16 x 14 x 7) m3 = 1568 m3.
1
Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] cm2
= (2 x 434) cm2 = 868 cm2.
Ex. 2. Find the length of the longest pole that can be placed in a room
12 m long
8m broad and
9m high.
i
Sol. Length of
longest pole = Length of the diagonal of the room
=
Ö(122+82+92=
.Ö(289)= 17 m.
Ex. 3. Tbe volume of a wall, 5 times as high
as it is broad and 8 times as long as
it is high, is
12.8 cu. metres. Find the breadth of the wall.
Sol. Let the breadth
of the wall be x metres.
Then, Height = 5x
metres and Length = 40x metres.
\x * 5x * 40x = 12.8 Û x3=12.8/200 =
128/2000 = 64/1000
So, x = (4/10) m =((4/10)*100)cm = 40 cm
Ex. 4. Find the number of bricks, each measuring 24
cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm
thick, if 10% of the wall is filled with mortar?
Sol. Volume of the
wall = (2400 x 800 x 60) cu. cm.
Volume of bricks =
90% of the volume of the wall
=((90/100)*2400
*800 * 60)cu.cm.
Volume of 1 brick = (24 x
12 x 8) cu. cm.
\Number of
bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.
Ex. 5. Water flows into a tank 200 m x 160 m througb a rectangular pipe
of
1.5m x 1.25 m @ 20 kmph . In what time (in
minutes) will the water rise by 2
metres?
Sol. Volume required in the
tank = (200 x 150 x 2) m3 = 60000 m3.
.
.
Length
of water column flown in1 min =(20*1000)/60 m =1000/3 m
Volume flown per minute = 1.5 * 1.25 *
(1000/3) m3 = 625 m3.
\ Required time =
(60000/625)min = 96min
Ex. 6. Tbe
dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is
2 cm. If 1 cubic
cm of metal used in the box weighs 0.5 gms, find the weight of the box.
Sol. Volume of the metal used in the box = External Volume - Internal
Volume
=
[(50 * 40 * 23) - (44 * 34 * 20)]cm3
=
16080 cm3
\ Weight of the
metal =((16080*0.5)/1000) kg = 8.04 kg.
Ex. 7. The
diagonal of a cube is 6Ö3cm. Find its
volume and surface area.
Sol. Let the edge of the cube be a.
\Ö3a = 6../3 _ a = 6.
So,Volume = a3 = (6 x 6 x 6) cm3
= 216 cm3.
Surface area = 6a2 = (6 x
6 x 6) cm2 == 216 cm2.
Ex. 8. The
surface area of a cube is 1734 sq. cm.
Find its volume.
Sol. Let the edge of
the cube bea. Then,
6a2 = 1734 Þ a2 = 289 => a
= 17 cm.
\ Volume = a3
= (17)3 cm3 = 4913 cm3.
Ex. 9. A rectangular block 6 cm by 12 cm
by 15 cm is cut up into an exact number of equal cubes. Find the least possible
number of cubes.
Sol. Volume of the
block = (6 x 12 x 15) cm3 = 1080 cm3.
Side of the largest cube = H.C.F. of 6 cm, 12
cm, 15 cm = 3 cm.
Volume of this cube = (3 x 3 x 3) cm3
= 27 cm3.
Number
of cubes = 1080/27 = 40.
Ex.l0. A cube of edge 15 cm
is immersed completely in a rectangular vessel containing water . If the dimensions of the base of
vessel are 20 cm x 15 cm, find the rise in water level.
Sol. Increase in
volume = Volume of the cube = (15 x 15 x 15) cm3.
\Rise in water
level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.
Ex. 11. Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to
form a new
cube. Find the
surface area of the cube so formed.
Sol. Volume of new cube = (13 + 63 + 83)
cm+ = 729 cm3.
Edge of new cube = 3Ö729 cm = 9 cm.
\ Surface area of the new cube = (6 x 9 x 9) cm2
= 486 cm2.
Ex. 12. If each edge of a cube is increased by 50%, find the percentage increase in
Its surface area. "
Sol. Let original length of each edge = a.
Then, original surface area = 6a2.
New edge = (150% of a) = (150a/100) = 3a/2
New surface area = 6x (3a/2)2 = 27a2/2
Increase percent in surface area = ((15a2)
x ( 1 ) x 100)% = 125%
2
6a2
Ex. 13. Two cubes have their volumes in the ratio 1 : 27. Find the
ratio of their
surface areas.
Sol. Let their edges be a and b. Then,
a3/b3
= 1/27 (or) (a/b)3 = (1/3)3 (or) (a/b) = (1/3).
\Ratio of their surface area
= 6a2/6b2 = a2/b2 = (a/b)2
= 1/9, i.e. 1:9.
Ex.14.Find the volume , curved surface area and the total surface area
of a cylinder with diameter of base 7 cm and height 40 cm.
Sol. Volume = ∏r2
h = ((22/7)x(7/2)x(7/2)x40) = 1540 cm3. .
Curved surface area = 2∏rh =
(2x(22/7)x(7/2)x40)= 880 cm2 .
Total surface area = 2∏rh + 2∏r2
= 2∏r (h + r)
= (2 x
(22/7) x (7/2) x (40+3.5)) cm2
= 957 cm2
Ex.15. If the capacity of a cylindrical tank
is 1848 m3 and the diameter of its base
is 14 m, then
find the depth of the tank.
Sol. Let the depth of the tank
be h metres. Then,
∏ x 72
x h = 1848 Þ h = (1848 x (7/22) x (1/49) = 12 m
Ex.16. 2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm
diameter. Find
the length of the wire in metres.
Sol. Let the length of the wire
be h metres. Then,
∏ (0.50/(2 x 100))2 x h = 2.2/1000
Þ
h = ( (2.2/1000) x (100 x 100)/(0.25 x 0.25) x
(7/22) ) = 112 m.
Ex. 17. How many iron rods, each of length 7 m and diameter 2 cm
can be made
out of 0.88 cubic metre of
iron?
Sol. Volume of 1 rod
= (( 22/7) x (1/100) x (1/100) x 7 ) cu.m = 11/5000 cu.m
Volume
of iron = 0.88 cu. m.
Number
of rods = (0.88 x 5000/11) = 400.
Ex. 18. The radii of two cylinders are in the
ratio 3: 5 and their heights are in tbe
ratio of 2 : 3.
Find the ratio of their curved surface areas.
Sol. Let the radii of the cylinders be 3x, 5x and their heights be 2y, 3y
respectively. Then
Ratio of their curved
surface area = 2∏ X 3x X 2y = 2/5 = 2.5
2∏ X 5x X 3y
Ex. 19. If 1 cubic cm of cast iron weighs 21 gms, then find the eight
of a cast iron
pipe of length 1 metre with a
bore of 3 cm and in which thickness of the metal is 1 em.
Sol. Inner radius =
(3/2) cm = 1.5 cm, Outer radius = (1.5 + 1) = 2.5 cm.
\Volume of iron =
[∏ x (2.5)2 x 100 - ∏ x (1.5)2 x 100] cm3
= (22/7) x
100 x [(2.5)2 - (1.5)2] cm3
= (8800/7)
cm3
Weight
of the pipe = ((8800/7) x (21/1000))kg = 26.4 kg.
Ex. 20. Find the slant height, volume, curved
surface area and the whole surface
area of a cone of radius 21 cm and height 28 cm.
Sol. Here, r = 21 cm and h = 28 cm.
\ Slant height, l = Ör2 + h2 = Ö (21)2 + (28)2 = Ö 1225 = 35cm
Ex. 21. Find the length of canvas 1.25 m wide required to build a conical tent of base radius 7 metres and height 24 metres.
Sol. Here, r = 7m and
h = 24 m.
So,l
= Ö(r2 +
h2) = Ö(72 +
242) = Ö (625) = 25 m.
Area of canvas = Õrl=((22/7)*7*25)m2
= 550 m2 .
Length of canvas = (Area/Width) = (550/1.25) m = 440 m.
Ex. 22. The heights of two right circular cones are in the ratio 1 : 2 and the
perimeters of
their bases are in the ratio 3 : 4. Find the ratio of their volumes.
Sol. Let the radii
of their bases be r and R and their heights be h and 2h respectively.
Then,(2Õr/2ÕR)=(3/4) Þ R=(4/3)r.
\ Ratio of volumes
= (((1/3)Õ r2h)/((1/3)Õ(4/3r)2(2h)))=9 : 32.
Ex. 23. The radii of the bases of a
cylinder and a cone are in the ratio of 3 : 4 and It heights are in the ratio 2
: 3. Find the ratio of their volumes.
Sol. Let the radii
of the cylinder and the cone be 3r and 4r and their heights be 2h and
3h
respectively.
:. Volume
of cylinder = Õ x (3r)2
* 2h = 9/8 = 9 : 8.
Volume of cone (1/3)Õr2 * 3h
Ex. 24. A conical vessel, whose
internal radius is 12 cm and height 50 cm, is full of liquid. The contents are emptied into a
cylindrical vessel with internal radius 10 cm. Find the height to which the
liquid rises in the cylindrical vessel.
Sol.
Volume of the liquid in the cylindrical vessel
=
Volume of the conical vessel
=
((1/3)* (22/7)* 12 * 12 * 50) )cm3 = (22 *4 *12 * 50)/7 cm3.
Let the height of
the liquid in the vessel be h.
Then (22/7)*10*10*h =(22*4*12*50)/7 or h = (4*12*50)/100 = 24 cm
Ex. 25. Find the volume and surface area of a sphere of radius 10.5 cm.
3
3
Sol. Volume = (4/3)Õr3 =(4/3)*(22/7)*(21/2)*(21/2)*(21/2)
cm3 = 4851 cm3.
Surface area = 4Õr 2
=(4*(22/7)*(21/2)*(21/2)) cm2 = 1386 cm2
Ex. 26. If the radius of a sphere is
increased by 50%, find the increase percent in
volume and the
increase percent in the surface area.
Sol. Let original radius = R. Then, new radius =
(150/100)R=(3R/2)
Original volume = (4/3)ÕR3,
New volume = (4/3)Õ(3R/2)3 =(9ÕR3/2)
Increase % in volume=((19/6)ÕR3)*(3/4ÕR3)*100))% =
237.5%
Original surface area =4ÕR2.
New surface area = 4Õ(3R/2)2=9ÕR2
Increase % in surface area =(5ÕR2/4ÕR2) * 100) % =
125%.
Ex. 27. Find the number of lead balls, each 1 cm in diameter that can be
a sphere of
diameter 12 cm.
Sol. Volume of larger
sphere = (4/3)Õ*6*6*6) cm3
= 288Õ cm3.
Volume of 1 small lead ball =
((4/3)Õ*(1/2)*(1/2)*(1/2))
cm3 = Õ/6 cm3.
\ Number of lead balls = (288Õ*(6/Õ)) = 1728.
Ex.28.How many spherical bullets can be made out of a lead cylinder
28cm high and with radius 6 cm, each bullet being 1.5 cm in diameter ?
Sol. Volume of cylinder = (∏ x 6 x 6
x 28 ) cm3 = ( 9∏/16)
cm3.
Number of bullet
= Volume of cylinder = [(36 x 28)∏ x 16] /9∏ = 1792.
Volume of each bullet
Ex.29.A copper sphere of diameter 18cm is drawn into a wire of diameter
4 mm Find the length of the wire.
Sol. Volume of sphere = ((4∏/3) x 9 x
9 x 9 ) cm3 = 972∏ cm3
Volume of sphere = (∏
x 0.2 x 0.2 x h ) cm3
\ 972∏ = ∏ x (2/10) x (2/10)
x h Þ h = (972 x 5 x 5
)cm = [( 972 x 5 x5 )/100 ] m
= 243m
Ex.30.Two metallic right circular cones having their heights 4.1 cm and
4.3 cm and the radii of their bases 2.1 cm each, have been melted together and
recast into a sphere. Find the diameter of the sphere.
Sol. Volume of sphere = Volume of 2 cones
= ( 1 ∏ x (2.102)
x 4.1 + 1 ∏ x (2.1)2 x 4.3)
3
3
Let the radius of sphere
be R
\(4/3)∏R3 =
(1/3)∏(2.1)3 or R = 2.1cm
Hence , diameter of the
sphere = 4.2.cm
Ex.31.A Cone and
a sphere have equal radii and equal volumes. Find the ratio of the sphere of
the diameter of the sphere to the height of the cone.
Sol. Let radius of
each be R and height of the cone be H.
Then, (4/3) ∏ R3 = (1/3) ∏ R2H
(or) R/H = ¼ (or) 2R/H = 2/4 =1/2
\Required ratio =
1:2.
Ex.32.Find
the volume , curved surface area and the total surface area of a hemisphere of
radius 10.5 cm.
Sol. Volume = (2∏r3/3)
= ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))cm3
= 2425.5 cm3
Curved surface area = 2∏r3 =
(2 x (22/7) x (21/2) x (21/2))cm2
=693 cm2
Total surface area = 3∏r3 =
(3 x (22/7) x (21/2) x (21/2))cm2
= 1039.5 cm2.
Ex.33.Hemispherical
bowl of internal radius 9 cm contains a liquid. This liquid is to be filled
into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How
many bottles will be needed to empty the bowl ?
Sol. Volume of bowl =
((2∏/3) x 9 x 9 x 9 ) cm3 =
486∏ cm3.
Volume of 1 bottle = (∏ x
(3/2) x (3/2) x 4 ) cm3 = 9∏ cm3
Number of bottles =
(486∏/9∏) = 54.
Ex34.A
Cone,a hemisphere and a cylinder stand on equal bases and have the same
height.Find ratio of their volumes.
Sol. Let R be the
radius of each
Height of the hemisphere = Its radius = R.
\Height of each =
R.
Ratio of volumes = (1/3)∏ R2 x
R : (2/3)∏R3 : ∏ R2 x R = 1:2:3
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