Drinking
water is the most common activity. You must be wondering what does water have to
do with volume, well do you know how much water do you drink in a day? Have you
ever calculated the volume of the glass in which you drink water. Glass is an
example of the frustum of a cone. In the chapter below we shall help you
calculate the volume of a glass
Frustum of a Cone
We
have already learned in our previous classes to find the volume of perfectly
shaped 3-d structures. Cones, Cylinders etc have been under our consideration
for finding the respective volumes. These are the general shapes! In this
chapter, we shall deal with shapes that we use the most. A glass is the most
used shape. What do you find different in a glass tumbler? Which regular shape
does it resemble? A cone or a cylinder or none.
A
glass tumbler resembles a cone with its pointed part sliced. Yes, a glass is a
resemblance of the frustum of a cone. When the smaller end, parallel to the base
is cut from a cone, the shape we get is called the frustum of the cone. The case
of measurement here is different. So, how do we calculate the volume of the
frustum now?
In
the combination of solids, we added the volumes of two adjoining shapes which
gave us the total volume of any structure. But for frustum of the cone as we are
slicing the smaller end of the cone as shown in the figure, hence we need to
subtract the volume of the sliced part.
The Volume of the Frustum of a Cone
The
frustum as said earlier is the sliced part of a cone, therefore for calculating
the volume, we find the difference of volumes of two right circular cones.
From
the figure, we have, the total height H’ = H+h and the total slant height L
=l1 +l2.
The radius of the cone = R and the radius of the sliced cone = r. Now the volume
of the total cone = 1/3 π R2 H’
= 1/3 π R2 (H+h)
The
volume of the Tip cone = 1/3 πr2h. For
finding the volume of the frustum we calculate the difference between the two
right circular cones, this gives us
= 1/3 π
R2 H’
-1/3 πr2h
= 1/3π R2 (H+h) -1/3 πr2h
=1/3 π [ R2 (H+h)-r2 h ]
= 1/3π R2 (H+h) -1/3 πr2h
=1/3 π [ R2 (H+h)-r2 h ]
Now
on seeing the whole cone with the sliced cone, we come to know that the right
angle of the whole cone Δ QPS is similar to the sliced cone Δ QAB. This gives
us, R/ r = H+h / h ⇒ H+h = Rh/r . Substituting the value of H+h in the formula
for the volume of frustum we get,
=1/3
π [ R2 (Rh/r)-r2 h ]
=1/3 π [R3h/r-r2 h
)]
=1/3 π h (R3/r-r2 ) =1/3 π h (R3-r3 / r)
=1/3 π h (R3/r-r2 ) =1/3 π h (R3-r3 / r)
The
Volume of Frustum of Cone = 1/3 π h [(R3-r3)/
r]
Similar Triangles Property
Using
the same Similar triangles property lets find the value of h, R/ r = (H+h)/
h.
⇒ h= [r/(R-r)] H. Substituting the value of h this equation we get: =1/3 πH [r/(R-r)][(R3-r3)/ r)\]
=1/3 πH [(R3-r3)/(R-r)]
= 1/πH [(R-r)(R2 +Rr+r2 )/ (R-r) ]
=1/πH (R2 +Rr+r2 ).
Therefore, the volume (V) of the frustum of the cone is =1/3 πH (R2 +Rr+r2 ).
⇒ h= [r/(R-r)] H. Substituting the value of h this equation we get: =1/3 πH [r/(R-r)][(R3-r3)/ r)\]
=1/3 πH [(R3-r3)/(R-r)]
= 1/πH [(R-r)(R2 +Rr+r2 )/ (R-r) ]
=1/πH (R2 +Rr+r2 ).
Therefore, the volume (V) of the frustum of the cone is =1/3 πH (R2 +Rr+r2 ).
Curved Surface Area and Total Surface Area of the Frustum
The
curved surface area of the frustum of the cone = π(R+r)l1
The
total surface area of the frustum of the cone = π l1 (R+r)
+πR2 +πr2
The
slant height (l1)
in both the cases shall be = √[H2 +(R-r)2]
These
equations have been derived using the similarity of triangles property between
the two triangles QPS and QAB. Measurement of volume, surface area and curved
surface area like any other measurement depends on the understanding of the
subject.
For
the combination of solids, we add all the constituting shapes, here since we are
slicing a similar triangle from the cone, we find the difference between the two
shapes. These two parts of the measurements involve operations and depend highly
on the logic of understanding.
Sample Question for You
Q:
An open plastic drum of height 63 cm with radii of lower and upper ends as 15cm
and 25 cm respectively is filled with Milk. Find the cost of milk which can
completely fill the bucket at Rs. 45 per litre. Also find the surface area of
the drum, if it needs to be coloured with at the rate of .50ps/sq.cm.
Solution:
A plastic drum resembles a frustum of a cone. The Height (h) of the drum=63
cm. Upper Radius (R) of the drum = 25 cm. Lower Radius (r) of the drum= 15 cm.
Using the formula for the volume of the frustum of the cone the Volume (V) of
the drum shall be: 1/3πH (R2 +Rr+r2 )
=
1/3×22/7×63(252 +
25×15 +152 )
= 66(625+375+225)
= 80,850 cm3= 80 L 850 ml
One litre of milk costs Rs 45, and 80L & 850 ml shall cost = 80.850×45 = Rs 3,638.25
= 66(625+375+225)
= 80,850 cm3= 80 L 850 ml
One litre of milk costs Rs 45, and 80L & 850 ml shall cost = 80.850×45 = Rs 3,638.25
The
Drum stores milk of cost = Rs 3,638.25. Total Surface Area of the Drum
= π l (R+r)
+πR2 +πr2l
=√H2 +(R-r)2l
= 632 +
(25-15) 2=
√3969+100
= 63.79 cm.
= 63.79 cm.
Using
the formula π l (R+r)
+πR2 +πr2=22/7×63.79
(25+15) +22/7 (25)2 +22/7(15)2=
22/7×63.79×40+22/7×625+22/7×225
= 22/7[2551.6+625+225] = 10,690.74 cm2Cost of painting per sq. cm is .50ps. So for 10,690.74 sq.cm the cost of painting shall be 10,690.74×50 = Rs5,345 approximately.
= 22/7[2551.6+625+225] = 10,690.74 cm2Cost of painting per sq. cm is .50ps. So for 10,690.74 sq.cm the cost of painting shall be 10,690.74×50 = Rs5,345 approximately.
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