Surface Areas and Volumes

Area of circle $= π r^{2}$

Circumference of circle $= 2 π r$

Volume and capacity are sometimes vary, with capacity being used for how much a container can hold (in litres), and volume being how much space an object displaces (in cubic metres).

Area of square $= (side)^{2}$
Perimeter of square $= 4 \times a$

1.0Surface areas and volumes

Surface areas and volumes of a combination of solid

Two or more standard solids (Cube, Cuboid, Cylinder, Cone, Sphere and hemisphere) can be combined to form a new solid. Some of the examples are given below.

Combination of a cuboid and a right circular cylinder.

Combination of a right Circular cylinder and a right circular cone.

Combination of a cylinder and a hemisphere and a cylinder

Combination of a right circular cone and a hemisphere

Now, how do we find the surface area and volume of such a solid? The total surface area and volume of such a solid is the sum of surface areas and volumes of individual parts respectively.

Numerical Ability 1 An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m . If the diameter of the base is 168 m , find the quantity of canvas required to make the tent. Allow $20%$ extra for folds and for stitching. Give your answer to the nearest $m^{2}$ . Solution: Radius of the tent, $r = (\frac{168}{2}) m = 84 m$ . Height of the tent $= 85 m$ . Height of the cylindrical part, $H = 50 m$ . Height of the conical part, $h = (85 - 50) m = 35 m$ . Slant height of the conical part, $ℓ = h^{2} + r^{2} = (35)^{2} + (84)^{2}$ $= 8281 m = 91 m$ .

Quantity of canvas required i.e.,

\Rightarrow

Curved surface area of the tent = Curved surface area of the cylindrical part + Curved surface area of the conical part

\Rightarrow 2 π rH + π r ℓ = π r (2 H + ℓ) = [\frac{22}{7} \times 84 (2 \times 50 + 91)] m^{2}

\Rightarrow (22 \times 12 \times 191) m^{2} = 50424 m^{2}

. Area of canvas required for folds and stitching

= (20%

of 50424

m^{2}

= (\frac{20}{100} \times 50424) m^{2} = 10084.80 m^{2}

∴

Total quantity of canvas required to make the tent

= (50424 + 10084.80) m^{2} = 60508.80 m^{2}

= 60509 m^{2}

. (to the nearest

m^{2}

)

While solving question, change all the given quantities in same units also, while writing answer always write units.

Numerical Ability 2 A solid is in the form of a solid cylinder mounted on a solid hemisphere with same radius is made from a solid material. The diameter of the hemisphere is 21 cm and the total height of the solid is 24.5 cm . Determine the weight of the solid if the weight of $1 cm^{3}$ material is 6 gm. $($ Take $π = \frac{22}{7}$ ) Solution:

For the hemispherical part, radius

r = \frac{1}{2} \times

diameter

= \frac{1}{2} \times 21 cm = 10.5 cm

Volume of the hemispherical part of the solid

= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times (10.5)^{3} cm^{3}

= \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 cm^{3}

= 44 \times 0.5 \times 10.5 \times 10.5 cm^{3}

= 2425.5 cm^{3}

For the cylindrical part,

r = 10.5 cm, h = 14 cm

. Volume of the cylindrical part of the solid

= π r^{2} \times h = \frac{22}{7} \times (10.5)^{2} \times 14 cm^{3}

= 44 \times 110.25 cm^{3} = 4851 cm^{3}

Then, the total volume of the solid

=

volume of hemispherical part + volume of cylindrical part

= 2425.5 cm^{3} + 4851 cm^{3} = 7276.5 cm^{3}

Now, the weight of solid material at the rate of 6 gm per

cm^{3} = 7276.5 \times 6 gm

= 43659 gm = 43.659 kg

Volume of hemisphere is half of the volume of sphere.

Numerical Ability 3 A solid is in the form of a cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the hemisphere. Slant height of the cone is $ℓ$ and radius of the base of the cone is $\frac{1}{2} r$ , where $r$ is the radius of the hemisphere. Prove that the surface area of the solid is $\frac{π}{4} (11 r + 2 ℓ) r$ sq. units. Solution: Curved surface area of the hemispherical part $= 2 π r^{2}$ sq. For the conical part, radius $= \frac{1}{2} r$ and slant height $= ℓ$ Then, the curved surface area of the conical part $= π \times \frac{1}{2} r \times ℓ = \frac{π}{2} r ℓ$ sq. units. The exposed area of the upper base of the hemisphere $= π {r^{2} - (\frac{1}{2} r)^{2}}$ sq. units $= \frac{3}{4} π r^{2}$ . sq. units. Thus, the total surface area of the solid

= 2 π r^{2} + \frac{π}{2} r ℓ + \frac{3}{4} π r^{2}

sq. units

= \frac{11 π}{4} r^{2} + \frac{π}{2} r ℓ

sq. units.

= \frac{π}{4} (11 r + 2 ℓ) r

sq. units.

2.0Additional subtopic

Conversion of solid from one shape to another For public works and for industrial development activities, we need to convert a solid into another solid of different shape or more than one solid of similar shape but with reduced size. For example, solid metallic sphere is melted and recast into more than one spherical ball or recast into a wire of cylindrical shape, the earth taken out by digging a well and spreading it uniformly around the well to form an embankment taking the shape of a hollow cylinder from its original shape of right circular cylinder, etc.

A right circular cylinder is a solid generated by the revolution of a rectangle about its sides.

Numerical Ability 4 2.2 cu dm of brass is to be drawn into a cylindrical wire of diameter 0.50 cm . Find the length of the wire. Solution: Volume of brass $= 2.2 cu dm = (2.2 \times 10 \times 10 \times 10) cm^{3} = 2200 cm^{3}$ . Let the required length of wire be xcm . Then, its volume $= (π r^{2} x) cm^{3} = (\frac{22}{7} \times 0.25 \times 0.25 \times x) cm^{3}$ $∴ \frac{22}{7} \times 0.25 \times 0.25 \times x = 2200 \Rightarrow x = (2200 \times \frac{7}{22} \times \frac{1}{0.25 \times 0.25}) = 11200 cm = 112 m .$ Hence, the length of wire is 112 m .

If the line joining the centres of circular ends of a cylinder is not perpendicular to the circular ends, then the cylinder is not a right circular cylinder.

When we convert one solid into another. Volume of given solid will always be equal to volume of obtained solid after meeting.

Numerical Ability 5 A field is $80 m$ long and 50 m broad. In one corner of the field, a pit which is $10 m$ long, $7.5$ m broad and 8 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. Find the rise in the level of the field.

Solution: Area of the field

= (80 \times 50) m^{2} = 4000 m^{2}

Area of the pit

= (10 \times 7.5) m^{2} = 75 m^{2}

Area over which the earth is spread out

= (4000 - 75) m^{2} = 3925 m^{2}

Volume of earth dug out

= (10 \times 7.5 \times 8) m^{3} = 600 m^{3}

∴

Rise in level

= (\frac{Volume}{Area}) = (\frac{600}{3925}) m

= (\frac{600 \times 100}{3925}) cm = 15.3 cm

$1 ℓ = 1000 m^{3}$ $1000 cm^{3} = 1 ℓ$

Numerical Ability 6 The water in a rectangular reservoir having a base $80 m \times 60 m$ , is 6.5 m deep. In what time can the water be emptied by a pipe of having a cross section is a square of side 20 cm , if water runs through the pipe at the rate of $15 km / h$ ? Solution: Volume of water in the reservoir $= (80 \times 60 \times 6.5) m^{3} = 31200 m^{3}$ . Area of cross section of the pipe $= \frac{20 \times 20}{100 \times 100} m^{2} = 0.4 m^{2} = \frac{1}{25} m^{2}$ Volume of water emptied in 1 hour $= (\frac{1}{25} \times 15000) m^{3} = 600 m^{3}$ . Time taken to empty the reservoir $= (\frac{31200}{600})$ hrs $= 52 hrs$ .

While solving question, you need to know the concept of unit conversion. Because sometimes you need to convert one unit in to another.

Frustum of a cone

Frustum of a right circular cone

In our day-to-day life we come across a number of solids of the shape as shown in the figure. For example, a bucket or a glass tumbler. We observe that this type of solid is a part of a right circular cone and is obtained when the cone is cut by a plane parallel to the base of the cone. If a right circular cone is cut off by a plane parallel to its base, the portion of the cone between the plane and the base of the cone is called a frustum of the cone.

We can see this process from the figures given below: The lower portion in figure is the frustum of the cone. It has two parallel flat circular bases, mark as Base (1) and Base (2). A curved surface joins the two bases.

Fig. (a)

Fig. (b)

The line segment MN joining the centres of the two bases is called the height of the frustum. Diameter CD of Base (2) is parallel to diameter AB of base (1). Each of the line segments AC and $B D$ is called the slant height of the frustum. We observe from the fig. (a) and fig. (b) that,

Height of the frustum $=$ (the height of the cone $O A B$ ) - (the height of the cone OCD)
Slant height of the frustum $=$ (the slant height of the cone OAB) - (the slant height of the cone OCD)

A sphere is a solid generated by revolving a circle about any of its diameter.

Volume of a frustum of a right circular cone

Let $h$ be the height, $r_{1}$ and $r_{2}$ be the radii of the two bases ( $r_{1} > r_{2}$ ) of frustum of a right circular cone. [Fig.(c)] The frustum is made from the complete cone OAB by cutting off the conical part OCD. Let $h_{1}$ be the height of the cone OAB and $h_{2}$ be the height of the cone OCD. Here, $h_{2} = h_{1} - h$ . Since right angled triangles OND and OMB are similar, therefore, we have

$\frac{h _{2}}{h _{1}} = \frac{r _{2}}{r _{1}} \Rightarrow \frac{h _{1} - h}{h _{1}} = \frac{r _{2}}{r _{1}} \Rightarrow 1 - \frac{h}{h _{1}} = \frac{r _{2}}{r _{1}} \Rightarrow \frac{h}{h _{1}} = 1 - \frac{r _{2}}{r _{1}} = \frac{r _{1} - r _{2}}{r _{1}} \Rightarrow h_{1} = \frac{hr _{1}}{r _{1} - r _{2}} and h_{2} = h_{1} - h = \frac{hr _{1}}{r _{1} - r _{2}} - h \Rightarrow h_{2} = \frac{hr _{2}}{r _{1} - r _{2}}$

Fig. (c)

Volume V of the frustum of cone $=$ Volume of the cone $OAB -$ Volume of the cone OCD $= \frac{1}{3} π r_{1}^{2} h_{1} - \frac{1}{3} π r_{2}^{2} h_{2} = \frac{1}{3} π r_{1}^{2} \times \frac{h r _{1}}{( r _{1} - r _{2} )} - \frac{1}{3} π r_{2}^{2} \times \frac{h r _{2}}{( r _{1} - r _{2} )}$ $= \frac{1}{3} πh {\frac{r _{1}^{3} - r _{2}^{3}}{r _{1} - r _{2}}} = \frac{1}{3} π h {r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}}$ $∴ V = \frac{1}{3} π h {r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}}$

When 2 triangles are similar then their corresponding sides are in proportion.

Note : Volume of frustum $V = \frac{1}{3} πh {r_{1}^{2} + r_{1} r_{2} + r_{2}^{2}}$ $= \frac{h}{3} {π r_{1}^{2} + π r_{2}^{2} + π r_{1} r_{2}} = \frac{h}{3} {π r_{1}^{2} + π r_{2}^{2} + (π r_{1}^{2}) (π r_{2}^{2})}$ $= \frac{h}{3} {($ area of base 1 $) + ($ area of base 2 $) + (area of base 1) (area of base 2)$

Curved surface area of a frustum of a right circular cone

Let $h$ be the height, $ℓ$ be the slant height and $r_{1}, r_{2}$ be the radii of the bases where $r_{1} > r_{2}$ .

In figure, we observe $EB = r_{1} - r_{2}$ and $ℓ^{2} = h^{2} + (r_{1} - r_{2})^{2}$

$∴ = h^{2} + (r_{1} - r_{2})^{2} In fig, we have OAB as the co ABDC. Let ℓ_{1} be the slant heigh Δ OND and △ OMB are similar, \frac{ℓ _{2}}{ℓ _{1}} = \frac{r _{2}}{r _{1}} \Rightarrow \frac{ℓ _{1} - ℓ}{ℓ _{1}} = \frac{r _{2}}{r _{1}} \Rightarrow ℓ_{1} = \frac{ℓ r _{1}}{r _{1} - r _{2}} Now, ℓ_{2} = ℓ_{1} - ℓ = \frac{ℓ r _{1}}{r _{1} - r _{2}} - ℓ \Rightarrow ℓ_{2} = \frac{ℓ r _{2}}{r _{1} - r _{2}}$

In fig, we have OAB as the complete cone from which cone OCD is cut off to make the frustum ABDC. Let $ℓ_{1}$ be the slant height of the cone OAB and $ℓ_{2}$ be the slant height of the cone OCD. Since,

Curved surface area of frustum ABDC

= (

Curved surface area of cone OAB) - (Curved surface area of cone OCD)

= π r_{1} ℓ_{1} - π r_{2} ℓ_{2} = π r_{1} \times \frac{ℓ r _{1}}{r _{1} - r _{2}} - π r_{2} \times \frac{ℓ r _{1}}{r _{1} - r _{2}} = π ℓ {\frac{r _{1}^{2} - r _{2}^{2}}{r _{1} - r _{2}}}

Therefore, CSA of frustum

= π ℓ (r_{1} + r_{2})

Total surface area of a frustum of a solid right circular cone

Let h be the height, $ℓ$ be the slant height and $r_{1}, r_{2}$ the radii of the bases where $r_{1} > r_{2}$ as shown in figure.

Total surface area of this frustum $=$ Curved surface area + Area of Base $1 +$ Area of Base 2 $= π ℓ (r_{1} + r_{2}) + π r_{1}^{2} + π r_{2}^{2}$

∴

TSA of frustum

= π ℓ (r_{1} + r_{2}) + π r_{1}^{2} + π r_{2}^{2}

Area of the metal sheet used to make a bucket A bucket is in the shape of a frustum of a right circular hollow cone. Let

h

be the depth,

ℓ

be the slant height,

r_{1}

be the radius of the top and

r_{2}

be the radius of the bottom as shown in figure.

The area of the metal sheet used for making the bucket $=$ Outer (or inner) curved surface area + Area of bottom $= π ℓ (r_{1} + r_{2}) + π r_{2}^{2}$

Numerical Abilty 7 A bucket of height 16 cm and made up of metal sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as 3 cm and 15 cm respectively. Calculate (i) the height of the cone of which the bucket is a part. (ii) the volume of water which can be filled in the bucket. (iii) the slant height of the bucket. (iv) the area of the metal sheet required to make the bucket. Solution: Let ABCD be the bucket which is frustum of a cone with vertex 0 (as shown in figure). Let $ON = x cm$ $Δ ONB \sim Δ OMC$ $∴ \frac{x}{16 + x} = \frac{3}{15} (∵ \frac{ON}{OM} = \frac{NB}{MC}) \Rightarrow \frac{x}{16 + x} = \frac{1}{5} \Rightarrow 5 x = 16 + x \Rightarrow 4 x = 16 \Rightarrow x = 4 ∴ ON = 4 cm and OM = 4 + 16 = 20 cm$

∴

Height of the cone

= 20 cm

Volume of the bucket

= \frac{1}{3} π (15)^{2} \times 20 - \frac{1}{3} π (3)^{2} \times 4 cm^{3}

{i.e., Volume of the large cone - Volume of the small cone

= \frac{1}{3} π (225 \times 20 - 36) cm^{3} = π (75 \times 20 - 12) cm^{3} = 1488 π cm^{3}

∴

Slant height of cone of radius 15 cm

= (15)^{2} + (20)^{2} cm = 625 cm = 25 cm

∴

Slant height of cone of radius

3 cm = (4)^{2} + (3)^{2} = 5 cm

∴

Slant height of bucket

= (25 - 5) cm = 20 cm

, i.e,

ℓ = 20 cm

∴

The area of the metal sheet

= π ℓ (R + r) + π r^{2}

= π \times 20 \times (15 + 3) + π \times (3)^{2} cm^{2} = 360 π + 9 π cm^{2} = 369 π cm^{2}

Numerical Abilty 8 A container made up of a metal sheet is in the form of a frustum of a cone of height $16 cm$ with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container at the rate of $₹ 15$ per litre and the cost of the metal sheet used, if it costs ₹ 5 per $100 cm^{2}$ . (Take $π = 3.14$ ) Solution: $R = 20 cm, r = 8 cm, h = 16 cm$ $ℓ = h^{2} + (R - r)^{2} = 256 + 144 cm = 20 cm$ Volume of container $= \frac{1}{3} πh {R^{2} + r^{2} + Rr}$ $= \frac{1}{3} \times (3.14) \times 16 {400 + 64 + 160} cm^{3}$ $= 3.14 \times \frac{16}{3} {624} cm^{3}$

= 3.14 \times 16 \times 208 cm^{3} = 10449.92 cm^{3}

Therefore, the quantity of milk in the container

= \frac{10449.92}{1000}

litres

= 10.45

litres Cost of milk at the rate of

₹15

per litre

= ₹ (10.45 \times 15) = ₹156.75

Surface area of the metal sheet used to make the container

= π ℓ (R + r) + π r^{2} = π (ℓ (R + r) + r^{2})

= (3.14) \times {20 \times 28 + 64} cm^{2}

= (3.14) \times 624 cm^{2} = 1959.36 cm^{2}

Therefore, the cost of the metal sheet at the rate of ₹ 5 per

100 cm^{2}

= ₹ \frac{1959.36 \times 5}{100} = ₹97.97

approx.

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