Surface Areas of a Combination of Solids

In our daily life we come across many combined solids like,

a cylinder over a hemisphere,

a capsule which is combination of a cylinder attached with two hemispheres on both the ends,

a toy shaped with a cone over a hemispherical bottom etc.

Some real life examples are given below for a immediate reference:

While calculating the surface area of the combined figures, we should only calculate the areas that are visible to our eyes. For example, if a cone is surmounted by a hemisphere and if we have to find out the total surface area of this solid, we need to just find out the curved surface area of the hemisphere and curved surface area of the cone separately and add them together. Note that we are leaving the base area of both the cone and the hemisphere in this case since both the bases are attached together and are not visible to our eyes.

A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs.45 per m^{2}.

Radius of the cylindrical base = = 15 m

Height of the cylindrical portion = 5.5 m

Height of the conical portion = (8.25 – 5.5)m = 2.75m

Slant height (l) =

=

=

= 15.25 m

Surface area of the tent = Curved Surface area of the cylindrical portion + Curved Surface area of the conical portion

= 2πrh + πrl

** = + = =**

= 1237.5 m

^{2}

Therefore the cost of the canvas of the tent at the rate of Rs. 45

The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.

Radius of the cylindrical and conical base = 12 m

Height of the cylinder = 3.5 m

Slant height of the cone = 12.5 m

Height of the cone =

=

=

= 3.5m

Internal curved surface area = Curved surface area of the cone + Curved surface area of the cylinder

= πrl + 2πrh

=+

=

= 735.43 m^{2}

= π r^{2}h + πr^{2}h

=+

=

=

= 2112 m^{3}.

Capacity of the building = 2112 m^{3}

A godown building is in the form as in figure. The vertical cross-section parallel to the width side of the building is a rectangle of size 7 m

× 3 m mounted by a semicircle of radius 3.5 m. The inner measurements of the cuboidal portion are 10 m × 7 m × 3 m. Find the volume of the godown and the total internal surface area excluding the floor.

Volume of the godown = Volume of the cuboid + Volume of the half cylinder

= lbh + × π r^{2}h

= +

×= 210 + 192.5

= 402.5 m

^{3}

Total surface area of the godown

= Area of the four walls + Surface area of the two semi circle + curved surface area of the half cylinder

= 2(l + b) h + (2 × π r^{2}) + × 2 × π rh

= [2 × 3 (10 + 7)] + [2

× × × 3.5 × 3.5] + [× 2 × × 3.5 × 10]= 102 + 38.5 + 110

= 250.5 m

^{2}.

A rocket is in the form of a cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of the radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area.

Radius of the cylindrical portion = r = 2.5 m

Height of the cylindrical portion = h = 21 m

∴ Surface area of the cylindrical portion = 2πrh

= 2 × × 2.5 × 21 m^{2 }= 330 m^{2}

Radius of the conical portion = r = 2.5 m

Slant height of the conical portion = l = 8 m

∴ Curved surface area of the conical part = π r l

= × 2.5 × 8 m^{2}

= 62.86 m^{2}

Area of the base = π r^{2}

= × 2.5 × 2.5 m^{2}

= 19.64

∴ Total surface area of the rocket = Curved surface area of the cylindrical portion + Curved surface area of the conical portion + Area of the base

= (330 + 62.86 + 19.64) m^{2}

**= **412.5 m^{2}.

Unlike the surface area of the combination of solids, while calculating the volume of the combined solids we need to find the volume of each solid and add them together.

Given below are few example problems based on finding the volume of the combined solids.

Circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20m. The heights of the cylindrical and conical portions are 4.2 m and 2.1m, respectively. Find the volume of the tent.

Radius of the cylindrical base = 20m

Height of the cylindrical portion = 4.2m

Height of the conical portion = 2.1m

Volume of the tent = Volume of the cylindrical portion + volume of the conical portion

**= + = = **

= 6160 m

^{3}

Volume of the tent = 6160 m

^{3}.

Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose total length is 2.7m and the diameter of each hemispherical end is 0.7 m.

Radius of the hemispherical end = 0.7m/2 = 0.35m

Length of the cylinder = 2.7 – (0.35 + 0.35) = 2m

Volume of a hemisphere = = 0.0898 m^{3}

Volume of a right circular cylinder = = 0.77m^{3}

Volume of a solid = 2

= 2× 0.0898 + 0.77

= 0.1796 + 0.77= 0.9496 m

^{3}

= 0.95 m

^{3}

Volume of the solid = 0.95 m^{3}.

An iron pole consisting of a cylindrical portion 110 m high and of base diameter 12 m is surmounted by a cone 9 m high. Find the mass of the pole, given that 1 m^{3} of iron has 8 g mass (approx). (Use **π = ).**

Radius of the cone = = 6 m

Height of the cone = 9 m

Height of the cylindrical portion = (110 - 9) m = 101 m

Volume of the cone =

Volume of the cylindrical portion =

Volume of the iron pole = Volume of the cone + Volume of the cylindrical portion

=**+ **

=

=

= 12785.14 m^{3}

The mass of the pole = 12785.14 × 8 = 102281.14 g = 102.281 kg.

A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimetres of cork dust will be required?

Radius of inner cylindrical vessel = 7cm

Radius of outer cylindrical vessel = 8cm

Height of both inner and outer cylindrical vessel = 42cm

The total space between the two vessels is filled with cork dust for heat insulation

= Volume of outer cylindrical vessel - Volume of inner cylindrical vessel

= -

=

=

= 1980cm^{3}

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.

Diameter of the base of the cylinder = 21 cm ∴ Radius of the base of the cylinder = 10.5 cm

Length of the cylinder = 18 cm

Radius of the conical end = 10.5 cm

Height of the conical end = 9 cm ∴ Volume of the cylindrical part = π r^{2}h

= × 10.5 × 10.5 × 18 cm^{3}

= 6237 cm^{3}** ∴ **Volume of one conical end = π r^{2}h

= × × 10.5 × 10.5 × 9 cm^{3 }= 1039.5 cm^{3}** **

**∴ **Volume of two conical ends = 2 × 1039.5 cm^{3 }= 2079 cm^{3}

∴ Volume of the petrol tank = Volume of the cylindrical part + 2 × Volume of the conical portion

= (6237 + 2079) cm^{3}

= 8316 cm^{3}.

Volume of a Combination of Solids

Unlike the surface area of the combination of solids, while calculating the volume of the combined solids we need to find the volume of each solid and add them together.

Given below are few example problems based on finding the volume of the combined solids.

Circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20m. The heights of the cylindrical and conical portions are 4.2 m and 2.1m, respectively. Find the volume of the tent.

Radius of the cylindrical base = 20m

Height of the cylindrical portion = 4.2m

Height of the conical portion = 2.1m

Volume of the tent = Volume of the cylindrical portion + volume of the conical portion

**= + = = **

= 6160 m

^{3}

Volume of the tent = 6160 m

^{3}.

Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose total length is 2.7m and the diameter of each hemispherical end is 0.7 m.

Radius of the hemispherical end = 0.7m/2 = 0.35m

Length of the cylinder = 2.7 – (0.35 + 0.35) = 2m

Volume of a hemisphere = = 0.0898 m^{3}

Volume of a right circular cylinder = = 0.77m^{3}

Volume of a solid = 2

= 2× 0.0898 + 0.77

= 0.1796 + 0.77= 0.9496 m

^{3}

= 0.95 m

^{3}

Volume of the solid = 0.95 m^{3}.

An iron pole consisting of a cylindrical portion 110 m high and of base diameter 12 m is surmounted by a cone 9 m high. Find the mass of the pole, given that 1 m^{3} of iron has 8 g mass (approx). (Use **π = ).**

Radius of the cone = = 6 m

Height of the cone = 9 m

Height of the cylindrical portion = (110 - 9) m = 101 m

Volume of the cone =

Volume of the cylindrical portion =

Volume of the iron pole = Volume of the cone + Volume of the cylindrical portion

=**+ **

=

=

= 12785.14 m^{3}

The mass of the pole = 12785.14 × 8 = 102281.14 g = 102.281 kg.

A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimetres of cork dust will be required?

Radius of inner cylindrical vessel = 7cm

Radius of outer cylindrical vessel = 8cm

Height of both inner and outer cylindrical vessel = 42cm

The total space between the two vessels is filled with cork dust for heat insulation

= Volume of outer cylindrical vessel - Volume of inner cylindrical vessel

= -

=

=

= 1980cm^{3}

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.

Diameter of the base of the cylinder = 21 cm

∴ Radius of the base of the cylinder = 10.5 cm

Length of the cylinder = 18 cm

Radius of the conical end = 10.5 cm

Height of the conical end = 9 cm

∴ Volume of the cylindrical part = π r^{2}h

= × 10.5 × 10.5 × 18 cm^{3}

= 6237 cm^{3}** ∴ **Volume of one conical end = π r^{2}h

= × × 10.5 × 10.5 × 9 cm^{3 }= 1039.5 cm^{3}** **

**∴ **Volume of two conical ends = 2 × 1039.5 cm^{3 }= 2079 cm^{3}

∴ Volume of the petrol tank = Volume of the cylindrical part + 2 × Volume of the conical portion

= (6237 + 2079) cm^{3}

= 8316 cm^{3}.

You must have seen in our daily life that some solids are melted and recast into another solid for various purposes. For example a logwood is cut to make small pencils, which are in the shape of a cylinder surmounted by a cone. Also, for your school activities you must have converted a cylindrical shaped candle to form a spherical ball or conical lamp etc. by melting them. Now, in this topic we are going to learn as to how to find the volume and surface area of the converted solid from the original solid.

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is of the radius of the original ball, how many such balls are made? Compare the surface area, of all smaller balls combined together with that of the original ball.

Radius of the spherical ball = R

Radius of the smaller spherical ball (r) = R

Volume of a spherical ball =

Volume of a smaller spherical ball =

Number of balls =

== = 64

= = 16

Therefore the surface area of the spherical ball is 16 times bigger than surface area of the smaller spherical balls.

An inverted conical vessel of radius of 6 cm and height 8 cm is filled with water. A sphere is lowered into the vessel. Find

(i) the radius of the sphere, given that when it touches the sides, the highest point of the sphere is in level with the base of the cone.

(ii) volume of water that flows out of conical vessel consequent to lowering the sphere in it.

(i) In Δ AOC and Δ ABD,

∠ OAC = ∠ DAB (Common)∠ OCA = ∠ ABD = 90

^{o}

Δ AOC ∼ Δ ADB (AA Similarity)

…………..(i)

In Δ ADB,

AD^{2} = AB^{2} + BD^{2}

AD =

=

= = 10 cm

AO = AB – OB = 8 - OC

Substituting in (i)

10 OC = 48 – 6 OC

16 OC = 48

OC = 3 cm

(ii) Volume of a sphere =

Volume of a sphere = 56. 571 cm^{3}

Volume of water that flows out of conical vessel consequent to lowering the sphere in it = Volume of the Sphere =56.571 cm^{3}.

From a sphere of radius 10 cm, a right circular cylinder diameter of whose base is 12 cm, is carved out. Calculate the volume of the right circular cylinder correct to 2 decimal places.

From the figure, OB = OC = 10 cm (radius of the sphere)

In Δ AOB,

OB^{2} = OA^{2} + AB^{2}

10^{2} = OA^{2} + 6^{2}

OA =

=

== 8 cm

Height of the cylinder = 2 × OA

Volume of a right circular cylinder =

^{2}h

=

× 6^{2}× 16

= × 36 × 16

= 1810.29 cm^{3}

Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Radius of a coin r = 0.75 cm and height h = 0.2 cm.

Volume of a coin =πr^{2}

= cm^{3}

Radius of a cylinder r = 2.25 cm and height h = 10 cm.

Volume of the cylinder = cm^{3}

Therefore number of coins =

=

=

= 3× 3 × 50

Number of coins = 450A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.

Let the required thickness of the cylinder be x cm, the external and internal radius of the hollow cylinder are 5 cm and (5 - x) cm.

Radius of the sphere = 6 cm

Then, volume of sphere = volume of the solid in hollow cylinder

π × 6 × 6 × 6 = π × 32[5

^{2}- (5 - x)

^{2}]

4

× 6 × 6 × 6 = 3 × 32[(5 × 5) - (5 – x)(5 - x)]864 = 96 [25 - (5 – x)(5 - x)]

9 = [25 - (5 – x)(5 - x)

-16 = - (5 – x)(5 - x)

25 – 10x + x^{2} = 16

x^{2} – 10 x + 9 = 0

(x – 9)(x - 1) = 0

x = 9 or x = 1

Therefore the uniform thickness of the cylinder is 9 or 1cm.

The rainwater from a roof 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.

Length of the roof = 22 m

Breadth of the roof = 20 m ∴ Area of the roof = 22 × 20 = 440 m^{2} = 440 × 100 × 100 cm^{2}

Diameter of the base of the cylinder = 2 m ∴ Radius of the base of the cylinder = 1 m

Height of the cylinder = 3.5 m ∴ Volume of the cylinder = π r^{2} h

= × 1 × 1 × 3.5 m^{2}

= 11 m^{3}

= 11 × 100 × 100 × 100cm^{3}

Let the rainfall be h cm.

∴ Area of the roof × rainfall in cm = Volume of the cylinder

∴ 440 × 100× 100 × h = 11 × 100 × 100 × 100

∴ h = = 2.5 cm

∴ Rainfall = 2.5 cm.

Frustum of a Cone

**Bucket Flower Pot Drinking glass Table lamp**

**Definition **

Right circular cone VAB is cut by a plane A’O’B’ parallel to its circular base AOB. This gives the solid AA’O’B’ BOA which is the frustum of right circular cone VAB. The circular faces AOB and A’O’B’ are called the circular ends of the Frustum.

Let us now define height, lateral (slant) height of the frustum.

**Height**

The height or thickness of a frustum is the perpendicular distance between its two ends.

OO’ is the height of the frustum of a cone.

**Slant Height**

The slant height of a frustum of a right circular cone is the length of the line segment joining the extremities of two parallel radii, drawn in the same direction, of the circular ends.

AA’ = BB’ is the slant height of the frustum.

**Volume and Surface Area of a Frustum**

Let R and r be the radii of two circular ends and h be the height of a frustum of a right circular cone Let l be the slant height of the frustum. Clearly, triangles VOA and VO’A’ are similar.

= =

= =

= =

= =

= =

= =

H = and L =

From right angled triangle AMA’, we have

AA’^{2 }= AM^{2 }+ A’M^{2 }

⇒ l^{2} = (R - r)^{2 }+ h^{2 }

⇒ l =

Let V be the volume of the frustum. Then,

V = Volume of cone VAB – Volume of cone VA’B’

⇒ V = π R^{2}H - π r^{2}(H - h)

⇒ V = π {R^{2}H - r^{2}(H - h)}

⇒ V = π {(R^{2} ×) - (r^{2} ×)} [Q H = ∴ H - h= ]

⇒ V = π ()h

⇒ V = π (R^{2} +Rr+r^{2})h

Let S be the lateral surface area of the frustum. Then,

S = Lateral surface area of cone VAB – Lateral surface area of cone VA’B’

= π RL - π r(L - l)

= π R × - π r [ L = ; L - l = ]

= π ()l

= π (R + r)l

Total surface area of the frustum = Lateral surface area + Surface area of its circular ends

= π(R + r)l + πR^{2} + πr^{2}

= π [(R + r)l + R^{2 }+ r^{2}]

Thus, if R and r are the radii of the circular ends of a frustum of a cone, h is the height and l is the slant height, then

Volume of the frustum = π (R^{2} + Rr + r^{2})h

Lateral surface area = π (R + r)l

Total surface area = π [(R + r)l + R^{2 }+ r^{2}]

Slant height =

Let us now discuss some examples:

If the radii of the circular ends of a conical bucket are 28 cm and 7 cm, whose height is 45 cm. Find the capacity of the bucket (Use π = ).

Clearly, bucket forms a frustum of a cone such that the radii of its circular ends are R = 28 cm, r = 7 cm and height is = 45 cm.

Therefore,Capacity of the bucket = Volume of the frustum

= πh (R^{2} + Rr + r^{2})

= × π × 45(28^{2 }+ 7^{2 }+ 28 × 7)

= 22 × 15 × (28 × 4 + 7 + 28)

= 330 × 147 cm^{3}

= 48510 cm^{3}.

The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the lateral surface area and total surface area of the frustum.

R = 14 cm, r = 6 cm and h = 6 cm.

Let l be the slant height of the frustum. Then,

l =

⇒ l =

= 10 cm.

⇒ Lateral surface area = π (R + r)l

= π (14 + 6)10 cm^{2}

= 628.57 cm^{2}

Total surface area = π [(R + r)l + R^{2 }+ r^{2}]

= π (200 + 196 + 36) cm^{2}

= π × 432 cm^{2}

= 1357.71 cm^{2}.

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

Let the height of the bucket be h cm.

We have, R = 28 cm, r = 21 cm and,

V = Volume of the bucket = 28.490 litres = 28.490 × 1000 cm^{3} = 28490 cm^{3}.

V = 28490 cm^{3}.

⇒ π (R^{2} + Rr + r^{2})h = 28490

= 28490

⇒ h =

⇒ h = 15 cm.

Thus, height of the bucket = 15 cm.

**Example**

A friction clutch is in the form of a frustum of a cone the diameters of the ends being 32 cm and 20 cm and length 8 cm. Find its bearing surface and volume.

**Solution **

Let ABB’A’ be the friction clutch of slant height l cm.

We have,R = 16 cm, r = 10 cm and h = 8 cm.

∴ l^{2} = h^{2 }+ (R - r)^{2 }

⇒ l^{2} = 64 + (16 - 10)^{2 }

^{ }^{ }= 64 + (6)^{2}

^{ }= 64 + 36 = 100 cm

⇒ l = 10 cm

Bearing surface of the clutch = Lateral surface of the frustum

= π (R + r)l

= π (16 + 10)10 cm^{2 }

^{ } = (26)10

= 817.14 cm^{2}.

Volume = π (R^{2} + Rr + r^{2})h

= × π × 8(16^{2 }+ 16 × 10 + 10^{2}) cm^{3}

= × π × 8(256 + 160 + 100) cm^{3}

= 4324.57 cm^{3}.

==================================== Summary

# Summary

1. Volume of a cube = (side)^{3}cu. units

2. Surface area of a cube = 6(side)^{2} sq. units

3. Volume of a cuboid = lbh cu. units

4. Surface area of a cuboid = 2(lb + bh + hl) sq. units

5. Volume of a cylinder = πr^{2}h cu. units

6. Curved surface area of a cylinder = 2πr × h sq. units

7. Total surface area of a cylinder = 2πr(r + h) sq. units

8. Volume of a cone = π r^{2}h cu. units

9. Curved surface area of a cone = πrl

10. Total surface area of cone = πr^{2} + πrl = πr(r + l) sq. units

11. Volume of a sphere = πr^{3}

12. Surface area of a sphere = 4πr^{2}

13. Volume of a hemisphere = π r^{3} cu. units.

14. Total surface area of a hemisphere = 3πr^{2}

15. Curved surface area of a hemisphere = 2πr^{2}

^{2}+ Rr + r

^{2})h

17. Lateral surface area of a frustum = π (R + r)l

18. Total surface area of a frustum = π (R + r)l + πR

^{2}+ πr

^{2}.