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 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 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