1.  They have two circular planes (base and top)

     (Yellow color-AP 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 =2r

Lateral surface are of cylinder = area of rectangle =

Length*Breadth = P*h = 2 r *h  =2rh

Total surface area of cylinder =

Area of circular plane of one side (top)                                             

+ Lateral surface area + Area of circular plane of another side (bottom)

= r²+2rh+r²= 2r²+2rh=2r(r+h) sq units

For easy calculation we use approximate value for  =22/7 in all our problems.

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

The radius of roller = 35cms or .35 meters ( d=2r =70)

Lateral surface area of roller = 2rh

= 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

Since tanker is closed on all sides,

we are required to calculate  total surface area = 2r(r+h)=

2*(22/7)*1.4*(1.4+2.6) = 2*22*.2*4 =    35.2  Sq. meters

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

= (r²)*h = r²h  cubic units

Note that volume is always represented in cubic units

First we need to find the height of the tins

Since diameter of the tin 14cm its radius(r)= 7cm

Volume of  tin = r²h  = (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

Since the diameter of the well is 10 meters. Its radius = 5 meters

Volume of well = r²h   = (22/7)*5²*7 = 22*25 = 550 cubic meters = 5, 50,000 liters

(  1cu meter = 100*100*100 cc and 1000cc = 1 liter )

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

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 (AX₁X₂, AX₂X₃,,AX₃X₄, …..)

as shown in the figure on the right hand side.

Though the sectors X₁X₂, X₂X₃, X₃X₄ 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 AX₁X₂=  (1/2)*base*height =(1/2)*base*l

Lateral surface area of the cone = Sum of areas of several small triangles

= (1/2)*B₁l +(1/2)*B₂l+(1/2)*B₃l+ …….+(1/2)*B₁l

= (1/2)l [B₁+ B₂+ B₃+ ………+ B_n]

But [B₁+ B₂+ B₃+ ………+ B_n] = perimeter of the base of the cone = 2r

Lateral (curved) surface area of the cone =(1/2) l *2r= rl

Total surface area of cone  = area of circular plane of  base + lateral surface area = r² +rl = r² +rl=r(r+l)

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

l²= h²+r²

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

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

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)* (r² )h cubic units ( we have seen volume of cylinder =r²h)

= (1/3)* (area of base) *h

The measurements of cylindrical rod are (r=3.5, h=100cm)

 Volume of cylinder = r²h = (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)* r²h

= (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

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 = 2r,  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= 1225-441=784 =(28)²

 height = h = 28 feet

Volume of the heap(cone) = (1/3)* (r² )h = (1/3)*(22/7)*21*21*28=22*21*28 =12,936 cubic feet

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

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.

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

 2r = 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)*(r² )*h= (1/6)*3*10²*6 = 300

Volume of heap with circumference of  15  

= {1/4}*(1/3)*(r² )*h= (1/12)*3*10²*6 = 150

Volume of heap with circumference of  45  

= {3/4}*(1/3)*(r² )*h= (3/12)*3*10²*6 = 450

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 H-h =5

 (H-5)/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)* (R² )H = (1/3)(6²)15

Volume of smaller cone = (1/3)* (r² )H = (1/3)(4²)10

 Volume of frustum =

Volume of larger cone - Volume of smaller cone

= (1/3) (36*15-16*10)

= (1/3) (380) cubic cms

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.

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 r² square units

(area of circular plane = area of circle = r²)

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)

We know Circumference = 2r

  r = circulferenc/2 = (1/2)*44*(7/22) = 7 meters

Surface area of sphere = 4r² = 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.

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 B₁r

Volume of 2^nd cone= 1/3 B₂r

…

Volume of nth cone =1/3 B_nr

Volume of sphere = Sum of volume of small cones 

= (1/3) B₁r+(1/3) B₂r.......... (1/3) B_nr

= (1/3)*r(B₁+B₂.......... +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*4r² = 4/3 r³(Surface area of sphere= 4r²)

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.

Here r = 21 cm. Note hemisphere is half of sphere

 Volume of bowl which is a hemisphere = 2/3 r³= 2/3 *22/7*14*14*14 = 5749.3 cc = 5.75 liters  ( 1000 cc = 1 liter)

Volume of 1 marble = 4/3 r³= (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 r³=(4/3)*(22/7)r³ = 88/21r³

 r³ = 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 r³= (4/3)*(22/7)*5.52*5.52*5.52 = 704(Calculator was used to get the approximate value)

Volume of shot-put = 1437cc

4/3 r³=1437cc

i,e (4/3)*(22/7)r³ =1437

r³ = 1437*21/(4*22)  =343

 r=7 cm

Therefore diameter of the shot-put is 14 cm

Verification:

Volume of shot-put of radius 7cm = 4/3  r³ = (4/3)*(22/7)*7*7*7 = (4/3)*22*7*7 =1437