6.15 Solid figures:
In our daily life,
we come across several challenging questions such as
1. What is
the capacity of a water/petrol tanker, or a tanker
carrying milk?
2. How to
measure the heap of grains/sand stacked on the floor, against wall, at the
corner of wall?
3. What is
the requirement of paint to paint a pillar?
4. How much
cloth is required to cover a tent?
5. What is
the volume/mass of earth?
6.15.1
Cylinder
Surface
Area of Cylinder
In your daily life
you must have come across objects like water pipes, iron rods and Road roller.
They are all cylindrical in shape. They may be hollow like pipes, or solids
like rods and road rollers. They all have circular planes at two ends.
In the figure given
below, AB is the axis of cylinder. PQ is the height of the cylinder. AP and BQ
(AP=BQ) are the radius of the two circular planes.
Features of the
right circular cylinders are:
1. They have two circular planes (base and
top) (Yellow colorAP and BQ) 2. These circular planes are parallel to each
other and have same radius(r) 3. The line joining the centers of circular
planes (AB) is the axis of the cylinder 4. The curved surface joining the circular
planes is the lateral surface (Green color) 5. All the points on the lateral surface are
equidistant from the axis 6. The distance between the circular planes is
the height of the cylinder 

Imagine
we cut the outer surface of this cylinder across and if
we spread across, we get a rectangle
similar to ABCD. Breadth
of the rectangle is equal to the length of the cylinder (AB=h). Length
of the rectangle = circumference of the circular plane=P =2_{}r Lateral
surface are of cylinder = area of rectangle = Length*Breadth
= P*h = 2 _{}r *h =2_{}rh Total
surface area of cylinder = Area of circular
plane of one side (top) +
Lateral surface area + Area of circular plane of another side (bottom) =
_{}r^{2}+2_{}rh+_{}r^{2}= 2_{}r^{2}+2_{}rh=2_{}r(r+h) sq units For easy calculation we use approximate value for _{} =22/7 in all our
problems. 

6.15.1 Problem 1: A mansion has 12 cylindrical
pillars 0f 3.5 meters height and of circumference of 50cms. Find the cost of
painting the lateral surface of the pillars at Rs 25 per Sq.meter.
Solution:
Let
us use all the values in single unit of meters Lateral
surface of one pillar = Ph = .5*3.5
= 7/4 Sq. meter Since
there are 12 pillars, total area to be painted =
12*7/4 = 21 sq. meters Painting
cost = (area*rate) =21*25 = Rs 525 

6.15.1 Problem 2: A roller having 70cms as diameter
and length of 1 meter is used to level the playground in your school. If it
makes 200 complete revolutions what is the area of playground?
Solution:
The
radius of roller = 35cms or .35 meters ( _{}d=2r =70) Lateral
surface area of roller = 2_{}rh =
2*(22/7)*.35*1 = 44*.05 = 2.2 sq.
meters Since
roller makes 200 rounds, total
area of playground = 200*2.2 = 440 sq. meters 

6.15.1 Problem 3:
A Petrol tanker is made of sheet metal. If the tank’s length is 2.6
meters and the radius of the cylindrical tank is 140cm. How many sq meters of
sheet metal is required to make the tank?
Solution:
Since
tanker is closed on all sides, we
are required to calculate total surface
area = 2_{}r(r+h)= 2*(22/7)*1.4*(1.4+2.6)
= 2*22*.2*4 = 35.2 Sq. meters 

Volume
of cylinder
We
have seen that volume of cube = length*breadth*height =(area of the
base)*height Similarly
volume of a cylinder = (area of
base)*height =(area of the circular plane)*height =
(_{}r^{2})*h = _{}r^{2}h cubic
units Note that volume is always represented in cubic units 

6.15.1 Problem 4:
You must have observed that on the petrol tanker they mention the
capacity of the tanker. If the capacity of tanker mentioned on a petrol tanker
= x liters and
length of the tanker is =y
meters. Find out the diameter of the tanker in meters
Solution: Do it
yourself by taking measurements of a tanker(x, y)
6.15.1 Problem 5:
A Paint manufacturer sells paints in 1 litre cylindrical tins of
diameter 14cm. How many such tins can he store one above
the other in a
store room of height 3.245 meters
Solution:
First
we need to find the height of the tins Since
diameter of the tin 14cm its radius(r)= 7cm Volume
of tin = _{}r^{2}h =
(22/7)7*7*h = 154h It
is given that tin’s capacity is 1 liter We
know 1 liter = 1000cc _{} 154h =1000 i,e
h =6.49cm Since
the store room is of height 3.245 meters we can stack 50
(=3.245*100/6.49) tins one above the
other 

6.15.1 Problem 6: The height of water level in a circular
well is 7 meters and its diameter is 10 meters. Calculate the volume of water stored
in the well.
Solution:
Since
the diameter of the well is 10 meters. Its radius = 5 meters Volume
of well = _{}r^{2}h =
(22/7)*5^{2}*7 = 22*25 = 550 cubic meters = 5, 50,000 liters (_{} 1cu meter =
100*100*100 cc and 1000cc = 1 liter ) 

6.15.2
Circular cones
Surface Area of
Cone
The heap of grains,
heap of sand, ice cream introduce us the concept of cones in our daily life.
Like in the case of cylinders we need to know
the methods of calculating surface area and the volume of cones
In the adjacent
figure we have a right circular cone
Its properties are
1. A cone has only one circular plane and is
of radius OC =r 2. It has one vertex (A) which is the
intersection point of cone’s
axis (OA) and the slant height (CA=l) 3.
The line joining the centre of circular plane to the vertex is the height
(OA=h) 4.
The curved surface which connects the vertex and the circular edge of the
circular base is the lateral surface area 

If we cut a cone
along the edge and spread across, we get a sector of a circle as shown below.

The
section (APQ in the figure on the left) is formed by an arc of radius equal to
slant height l of the cone. The
sector APQ can be imagined to be made up of very small triangles (AX_{1}X_{2},
AX_{2}X_{3},,AX_{3}X_{4}, …..) as
shown in the figure on the right hand side. Though
the sectors X_{1}X_{2}, X_{2}X_{3}, X_{3}X_{4}
are not straight lines, they
tend to become straight lines when the circular section APQ is divided into
small portions. They
form the base of small triangles. Note
area of AX_{1}X_{2}=
(1/2)*base*height =(1/2)*base*l Lateral
surface area of the cone = Sum of areas of several small triangles =
(1/2)*B_{1}l +(1/2)*B_{2}l+(1/2)*B_{3}l+ …….+(1/2)*B_{1}l =
(1/2)l [B_{1}+ B_{2}+ B_{3}+ ………+ B_{n}] But
[B_{1}+ B_{2}+ B_{3}+ ………+ B_{n}] = perimeter
of the base of the cone = 2_{}r _{}Lateral (curved) surface area of the cone =(1/2) l *2_{}r= _{}rl Total
surface area of cone = area of
circular plane of base + lateral
surface area = _{}r^{2} +_{}rl = _{}r^{2} +_{}rl=_{}r(r+l) 

The
relationship between base, height and slant height in a cone:
As
in the figure on the right hand side Let
base radius of the cone =r Let
height of the cone =h Let
slant height of the cone =l From
Pythagoras theorem l^{2}=
h^{2}+r^{2} 

6.15.2 Problem 1:
A conical tent was is put up for a show supported by a pole of height 28
meters at the center. The diameter of the base is 42 meters. Find the cost of
the canvas used at the rate of Rs.20/ per sq mts
Solution:
Here
r=21( _{}d=2r =42): h=28 We
need to find the curved surface area. For
that we need to know the slant height of the tent The
slant height is the hypotenuse of the triangle formed by the radius as
the base and height By
Pythagoras theorem (hypotenuse)^{2}=
(base)^{2}+(height)^{2}= (21)^{2}+(28)^{2}=
441+784 =1225 =(35)^{2} _{} slant height = l =
35 meters Lateral
(curved) surface area of the cone =_{}rl =(22/7)*21*35 = 22*3*35 =2310
Sq. meters Cost
of the canvas = area*rate = 2310*20 = Rs 46,200 

6.15.2 Problem 2:
A factory was asked to use sheet metal to make a conical object of slant
height 8 meters with diameter of the base as 12 meters. How much sheet metal is
required?
Solution:
Here
r=6( _{}d=2r =12), l=8 Total
surface area of cone =_{}r(r+l) =
(22/7)*6*(6+8) = (22/7)*6*14 = 22*6*2 =264 Sq. meters 

Volume
of Cone
By observations and actual measurements we
notice that if a cylinder and a cone have same circular base and same height
then, volume
of a cylinder is equal to three times the volume of the cone. (observe
the adjoining figure) _{}Volume of cone =
(1/3)*volume of cylinder =
(1/3)* (_{}r^{2 })h cubic units ( _{}we have seen volume of cylinder =_{}r^{2}h) =
(1/3)* (area of base) *h 

6.15.2 Problem 3:
A worker in a factory was given a meter long cylindrical rod of radius
3.5cm. He was asked to melt the material and make from that, cones of radius
1cm and height 2.1 cm. How many cones can he produce?
Solution:
The
measurements of cylindrical rod are (r=3.5, h=100cm) _{} Volume of cylinder =
_{}r^{2}h = (22/7)*3.5*3.5*100 = 22*3.5*.5*100=3850cc After
melting this, he is asked to produce cones of sizes(r=1, h=2.1) The
volume of the cone to be manufactured = (1/3)* _{}r^{2}h =
(1/3)*(22/7)*1*1*2.1 = 22*.1 =2.2cc Number
of cones the worker can produce =
(Volume of the melted rod)/Volume of Cone to be made =
3850/2.2 =
1750 pieces 

6.15.2 Problem 4:
A heap is farmed when a farmer pours food grains on a ground. The
slanted height of heap is 35 feet. The circumference of the base is 132ft. He sells the food grains based on the volume.
Name the solid formed and find the volume.
Solution:
The
heap formed is of cone shape. We
are required to find the volume of the cone. We are given circumference of
the base and its slant height. In
order to calculate the volume We need to find the radius and the height. Since circumference = 2_{}r,^{ }r = (circumference)/ 2_{} = 132*7/(2*22) = 3*7=21^{} The
slant height is the hypotenuse of the triangle formed by the radius as the
base and height. By
Pythagoras theorem, (hypotenuse)^{2}= (base)^{2}+(height)^{2} _{}(height)^{2}=(hypotenuse)^{2}(base)^{2}= (35)^{2}(21)^{2}= 1225441=784 =(28)^{2} _{} height = h = 28 feet Volume
of the heap(cone) = (1/3)* (_{}r^{2 })h = (1/3)*(22/7)*21*21*28=22*21*28 =12,936
cubic feet 

Finding
the area and volume of an object similar to the shape of a bucket (Frustum):

Looking
at the above figures we notice 1.
Curved surface area of APBDQC(Frustum) = Curved surface area of the cone APBDOCA
– Curved surface area of the cone CQDOC 2.
Volume of APBDQC(Frustum) = Volume of
the cone APBDOCA – Volume of the cone CQDOC 
Relationship
between measures of original cone and the cone cut off:
Observe
the figure on the right hand side Let
base radius, height and the slant height of large cone be R,H,L
and that of small cut off cone be r,h,l respectively. Since two triangles in the figure are similar, the
sides opposite to equal angles are proportional Hence
r/R = h/H=l/L Thus, If
from the top of a cone, a smaller cone is cut off by slicing parallel to
the base, then the base radius, height and the slant height of two
cones are proportional. 

6.15.2 Frustum:
In the first figure radius of
circular base is r_{1. }If we cut_{ }the
cone somewhere in the middle we get a frustum as shown in second figure. The height of frustum
is h, slant height is l, radius of circular base is r_{1 }and radius
of circle at the top is r_{2}
Then: l= Volume of frustum = ^{ }h(r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) Curved surface area of frustum = l(r_{1} + r_{2}) Total surface area of frustum = Curved
surface area of frustum +Area of circular base +Area
of circular portion at the top = [l(r_{1}
+ r_{2})+ (r_{1}^{2}+ r_{2}^{2})] = [l(r_{1} + r_{2})+ (r_{1}^{2}+ r_{2}^{2})] 
Figure 1 Figure 2 
6.15.2 Problem 5: A heap of grains is stored against a wall, in
the inner corner of a wall and outer corner of wall having circumference of 30,
15 and 45 ‘Hasta’(Unit
of measurement) respectively. If the height of heap is 6 Hasta, tell the size of the heap (Lilavati Shloka 237)
Solution:
The heap is of the cone shape.
The base of the heap is part of a circle and hence
we need to find the radius. Note that their base is part of the same
circle having base of half of circle, quarter of circle and 3/4^{th} of a circle _{} 2_{}r = 60 , For
simplicity let us assume _{}=3, then r = 60/6 = 10 Volume
of cone = (1/3)* (area of base) *h ( It is
given that height = 6) Volume
of heap with circumference of 30
=
{1/2}*(1/3)*(_{}r^{2 })*h= (1/6)*3*10^{2}*6 = 300 Volume
of heap with circumference of 15 =
{1/4}*(1/3)*(_{}r^{2 })*h= (1/12)*3*10^{2}*6 = 150 Volume
of heap with circumference of 45 =
{3/4}*(1/3)*(_{}r^{2 })*h= (3/12)*3*10^{2}*6 = 450 

6.15.2 Problem 6: The Radius of the top circle of a
frustum is 4 cm and the radius of its bottom is 6cm. The height of frustum is
5cm. compute its volume.
Solution:
Let
the height of larger cone and smaller cone be H and h respectively. Since
the radii and heights are proportional h/H
= 4/6 =2/3 From
the given data and from the figure we note Hh =5 _{} (H5)/H = 4/6 =2/3 _{}1(5/H) = 2/3 _{} (5/H) = (2/3)1 =
(1/3) _{} H = 15 _{} h = 10 _{}Volume of larger cone = (1/3)* (_{}R^{2 })H = (1/3)(_{}6^{2})15^{} _{}Volume of smaller cone = (1/3)* (_{}r^{2 })H = (1/3)(_{}4^{2})10^{} _{} Volume of frustum = Volume
of larger cone  Volume of smaller cone =
(1/3) _{}(36*1516*10) =
(1/3) _{}(380) cubic cms 

6.15.3
Sphere
The
objects such as Football, cricket ball. and marbles introduce us the concept of sphere The
properties of sphere are 1.
Sphere has a centre. 2.
All the points on the surface of the sphere are at equidistance from centre
of the sphere. 3.
The equidistance is the radius of the sphere. 4.
A plane passing through the centre of the sphere divides the sphere in to two
equal parts called hemispheres. 

Surface area of
sphere:
By observation and actual measurement we
conclude that surface area of the sphere is four times the area of circular
plane surface passing thorough the diameter (the
plane which cuts sphere in two equal parts) _{} Surface area of
sphere = 4 _{}r^{2} square units (_{}area of circular plane = area of circle = _{}r^{2}) 

It
is interesting to note that the surface area of a sphere is equal to curved
surface area of
a cylinder just containing it(Refer Adjacent figure) 

6.15.3 Problem 1: A
building has hemispherical dome whose circumference is 44mtrrs. Calculate the
cost of painting at Rs. 200 per square meters
Solution:
We
know Circumference = 2_{}r _{} r = circulferenc/2_{} = (1/2)*44*(7/22) = 7 meters Surface
area of sphere = 4_{}r^{2} = 4*(22/7)*7*7 = 4*22*1*7 = 616 Sq meters Surface
area of dome (hemisphere) = half of surface area of sphere = 308 sq meters Cost
of painting = area*rate = 308*200 = 61,600 Rs. 

Volume
of sphere:
Who
has not heard of water melon which is rich in Vitamin A, C , Iron and
Calcium. Have
you observed how a vendor cuts a water melon before you decide to buy one? He
makes an incision and cuts through the water melon to take a piece out(cone
shape) similar to what is shown in the adjacent figure. We
shall introduce similar concept while working with sphere. 

The
sphere can be concluded to be consisting of small cones of height equal to
the radius of the sphere with circular base as can be inferred from the
adjoining figure. We
have learnt that volume of small cone =
1/3 (area of circular base of cone)*height
(_{}volume of cone = 1/3*Bh where B is area of base of cone and
h is the height of the cone) Volume
of 1^{st} cone = 1/3 B_{1}r Volume
of 2^{nd} cone= 1/3 B_{2}r … Volume
of nth cone =1/3 B_{n}r _{}Volume of sphere = Sum of volume of small cones =
(1/3) B_{1}r+(1/3) B_{2}r.......... (1/3) B_{n}r =
(1/3)*r(B_{1}+B_{2}.......... +B_{n}) =
(1/3)*r*(Surface area of sphere) (_{} If we join all the small cones together side by side, the
bases of small cones become the surface area of sphere) =
(1/3)*r*4_{}r^{2} = 4/3 _{}r^{3}(_{}Surface area of sphere= 4_{}r^{2}) 

Note
that if a cone is inscribed within the sphere as in the adjacent figure, then
the volume of the
sphere is four times the volume of the cone. 

6.15.3 Problem 2: A
hemisphere bowl of radius of 14 cm is used as a measure to prepare sweets in a
shop. . How many liters of ghee can it
hold?
Solution:
Here
r = 21 cm. Note hemisphere is half of sphere _{} Volume of bowl which
is a hemisphere = 2/3 _{}r^{3}= 2/3 *22/7*14*14*14 = 5749.3 cc = 5.75
liters (_{} 1000 cc = 1 liter) 

6.15.3 Problem 3: 21 lead
marbles having 2cm radius are melted to make a big sphere. Find the volume of
the sphere and its radius
Solution:
Volume
of 1 marble = 4/3 _{}r^{3}= (4/3)*(22/7)*2*2*2 = 32*22/21 _{}Volume of 21 marble = 21*32*22/21 = 32*22 = 704cc Since
marbles were melted and a sphere is made, Volume
of sphere = 4/3 _{}r^{3}=(4/3)*(22/7)r^{3} = 88/21r^{3} _{} r^{3} =
704*(21/88) (_{} the volume of sphere is given to be 704cc) =168
_{} r =5.52cm
(approximate value got using a calculator) Verification: Volume
of sphere of radius 5.52cm = 4/3 _{}r^{3}= (4/3)*(22/7)*5.52*5.52*5.52 = 704(Calculator
was used to get the approximate value) 

6.15.3 Problem 4 :
When a iron shotput ball was immersed in a
jar it spilled out approximately 1437cc of water. Find out the diameter of the
shotput.
Note, the water
spilled out is equal to volume of shotput.
Solution:
Volume
of shotput = 1437cc _{}4/3 _{}r^{3}=1437cc i,e
(4/3)*(22/7)r^{3} =1437 _{}r^{3} = 1437*21/(4*22) =343 _{} r=7 cm Therefore
diameter of the shotput is 14 cm Verification: Volume
of shotput of radius 7cm = 4/3 _{} r^{3} =
(4/3)*(22/7)*7*7*7 = (4/3)*22*7*7 =1437 

6.15.3 Problem 5: An
object has been created with a cone of
height 120 cm on top of a semi sphere of
radius 60 cm. Find out the volume of water in cubic meters remaining in the cylinder, when this object is immersed in a cylinder having radius of 60cm and height of 120cm.
Solution:
Radius of cylinder= Radius of semi sphere= Radius of cone = 60
cm=r Height of Cylinder = h1= 180 cm ; Height of cone =h2=120 cm Water remaining in cylinder = volume of cylinder volume of semi sphere volume of cone = r^{2 }h1  ^{ } r^{3}^{ }^{ }r^{2}h2
= r^{2}(h1
r h2) = r^{2}( 180 *60 *120) = *60*60*(1804040) = *60*60*100= *10000 113*10000 Cubic cm = 1.13 cubic
meter. 

6.15.3
Problem 6: A utensil
is made from a sheet of metal in the shape of a bucket with top open. The
height of bucket is 16cm. The radii of bottom and top portions are 8cm and 20
cm respectively. The bucket needs to be filled with milk. Find out the buying
cost of milk in the bucket if the rate of milk is Rs.20 per liter. Also find
out the cost of metal sheet, if the rate of metal sheet is Rs 8 per 100 sq.cm.( Take = 3.14).
Solution:
Write a rough sketch as shown in the figure. Here
h h=16, r_{1}=
8 and r_{2}=20 l== =
20 Volume of utensil =^{ }h(r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) =*3.14^{ }*16(20^{2}+ 8^{2}+ 20*8)= *3.14*16*(624) =
3.14*16*208 10449
cc=10.449 liter Cost of milk 10.449*20 209 Rs. 

Total Surface area of utensil = Curved surface area of utensil + Area of circular bottom = l(r_{1} + r_{2})+ r_{1}^{2}^{ }= [ l(r_{1} + r_{2}) + r_{1}^{2} ] =3.14[ 20*28+64] = 3.14*624= 1959.36 sq.cm.^{} Cost of metal sheet = 1959.36*8* Rs. 156 
6.15 Summary of learning
If l, b, h are
sides of cuboid and a is the side of cube h and l are the height and slant
height of geometrical figure then
Additional
points:
6.15.1
Hollow Cylinder
Have you observed
water pipes made of cement or cylindrical water tanks built on houses ? They
all have one thing in common that is they are all hollow and have one internal
and another external curved surface.
Let R be the
external radius, r be the internal radius (Note R>r) and h be the height of
the cylinder
No. 
Explanation 
Expression 

1 
Thickness
of the hollow cylinder 
Rr 

2 
Area
of cross section = (Area
of outer circular plane – Area of inner circular plane) 
_{}R^{2}  _{}r^{2} = _{}(R^{2}  r^{2}) 

3 
External
curved surface area 
2_{}Rh 

4 
Internal
curved surface area 
2_{}rh 

5 
Total
surface Area = External+
internal+ top plane+ bottom plane 
2_{}Rh + 2_{}rh + 2_{}(R^{2}  r^{2}) 

6 
Volume
of solid portion = (
External volume – inner volume) 
_{}R^{2}h  _{}r^{2}h = _{}(R^{2}  r^{2})h 
6.15.1 Problem 7: A hose pipe of cross sectional area
of 2 sq.cm. delivers 1500 litres in 5 minutes. What is the speed of water in meters/second
through the pipe?
Solution:
Cross section can be thought of as the area of a circular plane = _{}r^{2} = 2sq.cm.
Water discharged in
5 minutes = 1500 liters = 1500*1000 cc (_{}1 liter = 1000cc)
This can be thought
of as the volume of water collected for 5 minutes in a pipe of length h whose
cross section is 2sq.cm.
_{} _{}r^{2}h = 1,500,000 = 2h (_{}_{}r^{2} = 2)
_{}h = 750,000cm = 7500 meters (_{}1m = 100cm)
Since this is the
water collected in 5minutes (=300Sec), water collected in 1 second = 7500/300 =
25meters
6.15.1 Problem 8: A rectangular water tank measuring
80cmx60cmx60cm is filled from a pipe of cross sectional area of 1.5sq cm,
the water emerging
at 3.2mts/sec. How long does it take to fill the tank?
Hint:
1. Calculate
the volume of tank(l*b*h)
2. Calculate
the volume of water flowing through a pipe of circular plane area of 1.5sq cm
in one minute at 3.2mts/sec.
3. Divide
the volume of the tank (arrived at step 1) by the volume of water (arrived at
step 2)
6.15.1 Problem 9: A metal pipe has a bore (inner
diameter) of 5cm. The pipe is 5mm thick throughout. Find the weight (in kg) of
2 meters of the pipe if 1 cubic cm of metal weighs 7.7gm.
Solution:
r = 2.5cm, R = 2.5+0.5 cm = 3cm, h = 200cm
Volume of solid
portion =_{}(R^{2}  r^{2})h = 22/7*(96.25)*200 =
1728.57 cu cm
Weight in kg =
(volume in cc * weight in gram /cc)/1000 = 1728.57*7.7/1000 = 13.31 kg
We divided the
result by 1000 to get the answer in KG