Mensuration Notes

Introduction

You have already learnt that how to determine the surface area and volume of solids like cuboid, cube, right circular cylinder, right circular cone, sphere, hemisphere etc. in earlier classes. We come across various solids in our daily life. For example a circus tent with a cylindrical base is a combination of a right circular cylinder and a right circular cone. The shape of a bucket or a glass is not a complete cone but a part of a cone called a frustum of a cone. So in this section, we will learn how to determine the surface area and volume of such solids which are combination of two or more solids, frustum of a cone etc.

Review of surface areas and volumes of some solids

We give below the names and figures of the shapes of different solids and various formulae for determining the surface areas and volumes of these solids, which you already learnt in earlier classes, in tabular form for ready reference.

Table

Conversion of one solid into another solid

Sometimes it is necessary to melt a solid of a particular shape to convert it  in to a different shape. For example, an iron ball in the shape of a sphere is melted and recast into a thin wire of a cylindrical shape. In this type of problems, the volume of the material with the original shape is the same as the volume of the solid of converted shape.

Surface areas and volumes of combination of solids

            We already stated that in our daily-life problems, we come across with solids which are combinations of various solids like a cone, a cylinder, a hemisphere, etc. We shall now find the surface areas and volumes of such solids in this section.

Volume and surface area of frustum of a right circular cone

            If we cut a right circular cone OAB by a plane XY parallel to its base AB through a point E on the axis of the cone, then the section CMDN is a circle with centre at E [figure(i)].

Figure

            We now remove the portion OCD which contains the vertex O. The remaining portion ABDC which is in the shape of a bucket or a glass-tumbler for our daily use is called the frustum of the cone OAB shown separately in figure (ii).  

            Thus, we can define a frustum of a cone as,  if a right circular cone is cut off by a plane parallel to its base through a point on the axis of the cone and the portion containing the vertex of the cone is removed, then the remaining portion of the cone between the cutting plane and the base of the cone is called a frustum of the cone.

            The line-segment EF joining the centres of two circular bases (or ends) is called the height of the frustum. The line segments AC or BD is called its slant height.

Volume and Surface area of a frustum

            We shall now express the volume, the lateral surface area and the total surface area of a frustum of a right circular cone in terms of the radii of the circular ends, the vertical and the slant heights of the frustum.

            Let R and r be the radii of bigger and smaller circular ends with centres F and E respectively of the frustum, ‘h’ its height EF and ‘ l’ its slant height DB. Let H and L be respectively the height OF and slant height OB of the original cone OAB [Figure].

Figure

            We draw DP ⊥ AB. We shall now express H and L is terms of the known quantities r, R, h and l with respect to the frustum.

            Now,      Δ OCE ~ ΔOAF

∴        \frac{{OE}}{{OF}} = \frac{{OC}}{{OA}} = \frac{{CE}}{{AF}}

⇒        \frac{{OF-EF}}{{OF}} =\frac{{OA-AC}}{{OA}}  = \frac{{CE}}{{AF}}

⇒        \frac{{H-h}}{H} = \frac{{L-l}}{L} = \frac{r}{R}     [AC = DB = l ]

⇒        1 – \frac{h}{H} = 1 – \frac{l}{L} = \frac{r}{R}               ….. (i)

            From the first and the last ratios of (i), we get

 1 – \frac{h}{H} = \frac{r}{R}   ⇒ \frac{h}{H}= 1 –\frac{r}{R}  = \frac{{R-r}}{R}

⇒        H (R – r) = Rh

⇒        H = \frac{{Rh}}{{R-r}}                               …..(ii)

            From the second and the last ratios of (i), we get

      1 – \frac{l}{L} = \frac{r}{R}

⇒        \frac{l}{L} = 1 – \frac{r}{R} = \frac{{R-r}}{R}

⇒      L (R – r) = R l

⇒      L = \frac{{R l}}{{R-r}}                                                         …..(iii)

We can now express the slant height l in terms of r, R and h with respect to the frustum. Thus, we have from right-angled triangle DPB.

                        DB² = DP² + PB² = DP² + (FB – FP)²

                                = DP² + (FB – ED)²

⇒        l² = h² + (R – r)²

⇒        l = \sqrt {{h^2} + {{\left( {R - r} \right)}^2}}              ….. (iv)

                                 [Taking the positive square root only]

            Now, let V be the volume of the frustum. Then,

            V = Volume of the cone OAB – Volume of the cone OCD

=  \frac{1}{3}π R²H – \frac{1}{3} πr² (H – h)

 =  \frac{1}{3}π H (R² – r²) +  \frac{1}{3}π r² h

=  \frac{1}{3}π (R² – r²) ×  \frac{{Rh}}{{R-r}}+  \frac{1}{3}π r² h     [Substituting the value of H from (ii)]

=  \frac{1}{3}π (R + r) Rh + \frac{1}{3} πr² h

=  \frac{1}{3}πh (R² + Rr + r²)                                         …..(v)

            Let S be the lateral (or curved) surface area of the frustum. Then,

S = Lateral (or curved) surface area of the cone OAB – Lateral (or curved) surface area of the cone OCD

 = π RL – π r (L – l) = π (R – r) L + π rl

= π (R – r) × \frac{{Rl}}{{R-r}} + π r l         [Substituting the value of L from (iii)]

                        = π Rl + π r l                                                                

                        = π (R + r) l                                                       …..(vi)

            Total surface area of the frustum

                        = lateral surface area + sum of the areas of its circular ends       

                        = π(R + r) l + π (R² + r²)

                        = π [(R + r) l + R² + r²)]                          …..(vii)

ILLUSTRATIONS

Ex.1.      A plot of land in the form of a rectangle has a dimension 240 m × 180 m. A drainlet 10 m wide is dug all around it (on the outside) and the earth dug out is evenly spread over the plot, increasing its surface level by 25 cm. Find the depth of the drainlet.

Sol.        Let the depth of the drainlet be x metres.

We have, Width of the drainlet = 10 m.

∴ Volume of the drainlet

= (260 × 10 × x + 260 × 10 × x + 180 × 10 × x + 180 × 10 × x) m³

= (5200x + 3600x )m³ = 8800x m³

When earth dug out is evenly spread over the plot,

we get a cuboid whose base area is 240 × 180 m²

and height = 25 cm = 0.25 m.

∴ Volume of earth spread over the plot

= (240 × 180 × 0.25) m³

= 10800 m³.

Clearly,

Volume of earth spread over the plot = Volume of the drainlet    

Figure 

∴ 10800 = 8800x

⇒ x = \frac{{10800}}{{8800}} = 1.227 m

Ex.2.      A teak wood log is cut first in the form of a cuboid of length 2.3 m, width 0.75 m and of a certain thickness. Its volume is 1.104 m³. How many rectangular planks of size 2.3 m ×0.75 m × 0.04 m can be cut from the cuboid?

Sol.     Let the thickness of the log be x metres. Then,

Volume = 1.104 m³

⇒ 2.3 × 0.75 × x = 1.104

⇒  x = \frac{{1.104}}{{2.3 \times 0.75}} = 0.64 m

Since the length and breadth of each rectangular plank is the same as that of the cuboid.

No. of rectangular planks =\frac{{{\bf{Thickness of cuboid}}}}{{{\bf{Thickness of each plank}}}}  = \frac{{0.64}}{{0.04}} =\frac{{64}}{4}  = 16

ALITER : Number of rectangular planks = \frac{{{\bf{Volume of the cuboid}}}}{{{\bf{Volume of a plank}}}}

= \frac{{1.104}}{{2.3 \times 0.75 \times 0.04}} = \frac{{1.104}}{{0.069}} =\frac{{1104}}{{69}}  = 16

Ex.3.      A solid cylinder has total surface area of 462 square cm. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder. (Take = 22/7).

Sol.         Let r be the radius of the base and h be the height of the cylinder. Then,

Total surface area = 2 r (h + r) cm²

Curved surface area = 2 rh cm²

Now,  Curved surface area =\frac{1}{3} (Total surface area)

⇒ 2rh = \frac{1}{3}{2r(h + r)}

⇒ 6rh = 2rh + 2r²

⇒ 4rh = 2r²

⇒ 2h = r

∴ Total surface area = 462 cm²

⇒ 2r(h + r) = 462

⇒ 2r \left( {\frac{r}{2} + r} \right) = 462  [2h = r ∴ h =\frac{r}{2} ]

⇒ 2r ×\frac{{3r}}{2}  = 462

⇒ 2 ×\frac{{22}}{7}  × \frac{3}{2} × r² = 462  ⇒ r = 7 cm.

Now, 2h = r  ⇒ h = \frac{r}{2} = \frac{7}{2} cm

Hence, Volume of the cylinder = r²h = \left( {\frac{{22}}{7} \times {7^2} \times \frac{7}{2}} \right)cm³ = 539 cm³.

Ex.4.     The circumference of the base of a 10 m high conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 m. (Use = 22/7).

Sol.      Let r m be the radius of the base, h m be the height and l m be the slant height of the cone. Then,

Circumference = 44 metres

⇒  2r = 44

⇒  2 ×\frac{{22}}{7}  × r = 44   ⇒ r = 7 metres.

It is given that h = 10 metres.

∴ l² = r² + h²  ⇒ l =\sqrt {{r^2} + {h^2}}   = \sqrt {49 + 100} = \sqrt {149} = 12.2 m

Now,  Surface area of the tent = r l m² =\frac{{22}}{7}  × 7 × 12.2 m² = 268.4 m²

∴ Area of the canvas used = 268.4 m².

It is given that the width of the canvas is 2 m.

∴ Length of the canvas used = \frac{{Area}}{{width}}=\frac{{268.4}}{2}  = 134.2 m.

Ex.5.   A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of cup.

Sol.     When the semi-circular sheet is bent into an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

∴  l = slant height of the conical cup = 14 cm    

Figure (i)

Let r cm be the radius and h cm the height (depth) of the conical cup. Then, Circumference of the base of the conical cup = Circumference of the sheet

⇒  2r = × 14

⇒ r = 7 cm

Now, l² = r² + h²

⇒  h = \sqrt {{l^2}-{r^2}}

=   \sqrt {{{14}^2}-{7^2}} = 7\sqrt 3 cm

=  (7 × 1.732)cm = 12.12 cm

∴  Depth of the cup = 12.12 cm                                                                                                              

Now,

Capacity of the cup = Volume of the cup  

Figure (ii)

= \left( {\frac{1}{3}{\rm{\pi }} {r^2}h} \right) cm³

=  \left( {\frac{1}{3} \times \frac{{{\rm{22}}}}{{\rm{7}}} \times {\rm{7}} \times {\rm{7}} \times {\rm{12}}{\rm{.12}}} \right)cm3 = 622.16 cm³

Ex.6.      The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost to paint 1 cm² surface is Rs. 0.05. Find the total cost to paint the vessel all over. (Use= 22/7).

Sol.     Let the external and internal radii of the hemispherical vessel be R cm and r cm respectively. Then, R = 12.5 cm and r = 12 cm.

Now,

Area of outer surface = 2R²

Area of the inner surface = 2r²

Area of the ring at the top = R² – r²

∴ Total area to be painted = (2R² + 2 r² + R² – r²)

= (3R² + r²)

=  \frac{{22}}{7}× [3 × (12.5)² + (12)²] cm² 

Figure 

=  \frac{{22}}{7} × [468.75 + 144] cm²

=  \frac{{22}}{7}× 612.75 cm² = \frac{{13480.5}}{7} cm²                                                                                     

=  1925.78 cm²

∴  Cost of painting = Rs. (1925.78 × 0.05) = Rs. 96.28

Ex.7.      An iron rod of length 1 m and diameter 4 cm is melted and cast into thin wires of length 20 cm each. If the number of such wires be 2000, find the radius of each thin wire.

Sol.         Let the radius of each thin wire be r cm. Then, the sum of the volumes of 2000 thin wires will be equal to the volume of the iron rod. Now, the shape of the iron rod and each thin wire is cylindrical.

Hence, the volume of the iron rod of radius \frac{4}{2} cm

= 2 cm is × 2² × 1 × 100 cm³

Again, the volume of each thin wire = r² × 20

Hence, we have

× 2² × 100 = 2000 × r² × 20

⇒ 400 r² = 4  ⇒ r² = \frac{1}{{100}}

⇒  r = \frac{1}{{10}}  [Taking positive square root only]

Hence, the required radius of each thin wire is \frac{1}{{10}} cm or 0.1 cm.

Ex.8.      The height of a right circular cylinder is equal to its diameter. If it is melted and recast into a sphere of radius equal to the radius of the cylinder, find the part of the material that remained unused.

Sol.        Let h be the height of the cylinder. Then, its diameter is h and so its radius is\frac{h}{2} . Hence, its volume is V₁ = {\left( {\frac{h}{2}} \right)^2}.
h =\frac{{{\rm{\pi  }}{h^3}}}{4} .

Again, the radius of the sphere = \frac{h}{2}

Hence, the volume of the sphere is

V₂ =\frac{4}{3}  {\left( {\frac{h}{2}} \right)^3} = \frac{{{\rm{\pi  }}{h^3}}}{6}

∴  The volume of the unused material

=  V₁ – V₂ = \frac{{{\rm{\pi  }}{h^3}}}{4} – \frac{{{\rm{\pi  }}{h^3}}}{6} = \frac{{{\rm{\pi  }}{h^3}\left( {3 - 2} \right)}}{{12}}

=  \frac{{{\rm{\pi  }}{h^3}}}{{12}} =\frac{1}{3} × \frac{{{\rm{\pi  }}{h^3}}}{4} = \frac{1}{3} V₁

Hence, the required volume of the unused material is equal to\frac{1}{3} rd of the volume of the cylinder.

Ex. 9.     The base diameter of a solid in the form of a cone is 6 cm and the height of the cone is 10 cm. It is melted and recast intospherical balls of diameter 1 cm. Find the number of balls, thus obtained.

Sol.     Let the number of spherical balls be n. Then, the volume of the cone will be equal to the sum of the volumes of the spherical balls.

The radius of the base of the cone =\frac{6}{2} cm = 3 cm

and the radius of the sphere =\frac{1}{2} cm.

Now, the volume of the cone =\frac{1}{3}  × 3² × 10 cm³ = 30cm³

and the volume of each sphere =\frac{4}{3}  {\left( {\frac{1}{2}} \right)^3} cm³ =\frac{{\rm{\pi }}}{6} cm³

Hence, we have n\frac{{\rm{\pi }}}{6} = 30 ⇒ n = 6 × 30 = 180

Hence, the required number of balls = 180

Ex.10.    Water flows at the rate of 10 m per minute through a cylindrical pipe having its diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of the base is 40 cm and depth 24 cm?

Sol.         Diameter of the pipe = 5 mm = \frac{{\rm{5}}}{{10}} cm = \frac{{\rm{1}}}{2} cm

∴  Radius of the pipe = \frac{{\rm{1}}}{2}× \frac{{\rm{1}}}{2} cm =\frac{{\rm{1}}}{4}  cm.

In 1 minute, the length of the water column in the cylindrical pipe = 10 m = 1000 cm.

∴  Volume of water that flows out of the pipe in 1 minute = × \frac{{\rm{1}}}{4} × \frac{{\rm{1}}}{4}1000 cm³.

Also, volume of the cone = \frac{{\rm{1}}}{3} × × 20 × 20 × 24 cm³.

Hence, the time needed to fill up this conical vessel

= (\frac{{\rm{1}}}{3} × 20 × 20 × 24 ÷ × \frac{{\rm{1}}}{4} × \frac{{\rm{1}}}{4} × 1000) minutes

=\left( {\frac{{20 \times 20 \times 24}}{3} \times \frac{{4 \times 4}}{{1000}}} \right) minutes

=\frac{{4 \times 24 \times 16}}{{30}} minutes =\frac{{256}}{5} minutes = 51.2 minutes.

Hence, the required time is 51.2 minutes.

Ex.11.    A hemispherical tank of radius1\frac{3}{4}  m is full of water. It is connected with a pipe which empties it at the rate of 7 litres per second. How much time will it take to empty the tank completely?

Solution

Sol.         Radius of the hemisphere = \frac{7}{4} m =\frac{7}{4}  × 100 cm = 175 cm

∴  Volume of the hemisphere =\frac{2}{3}  × × 175 × 175 × 175 cm³

The cylindrical pipe empties it at the rate of 7 litres i.e., 7000 cm³ of water per second.

Hence, the required time to empty the tank =\left( {\frac{2}{3} \times \frac{{22}}{7} \times 175 \times 175 \times 175 \div 7000} \right) s

= \frac{2}{3}  ×\frac{{22}}{7} ×\frac{{175 \times 175 \times 175}}{{7000 \times 60}}min

= \frac{{11 \times 25 \times 7}}{{3 \times 2 \times 12}} min = \frac{{1925}}{{72}}min = 26.74 min nearly.

Ex.12.    A well of diameter 2 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 5 m to form anembankment. Find the height of the embankment.

Sol.       Let h m be the required height of the embankment.

The shape of the embankment will be like the shape of a cylinder of internal radius 1 m and external radius (5 + 1) m = 6 m [in Figure].

The volume of the embankment will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth

= volume of the cylindrical well

= × 1² × 14 m³ = 14 m³

Also, the volume of the embankment

= (6² – 1²)h cm³ = 35 p h m³

Hence, we have 35 h = 14

⇒ h =\frac{{14}}{{35}} = \frac{2}{5} = 0.4 m

Hence, the required height of the embankment = 0.4 m.     

Figure

Ex.13.    Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

Sol.         If the sphere exactly fits inside a cube i.e., if the sphere is inscribed in a cube, then the side of the cube will be equal to the diameter of the sphere. Let d be the diameter of the sphere. Then, the side of the cube = d.

∴  Volume of the cube = d³ And the volume of the sphere

=\frac{4}{3}  × {\left( {\frac{{\rm{d}}}{{\rm{2}}}} \right)^{\rm{3}}} = \frac{{{\rm{\pi  }}{{\rm{d}}^{\rm{3}}}}}{{\rm{6}}}

The required ratio   = d³ : \frac{{{\rm{\pi  }}{{\rm{d}}^{\rm{3}}}}}{{\rm{6}}}  = 6 :

Ex.14.    Water in a canal, 30 dm wide and 12 dm deep, is flowing with a speed of 10 km/hour. How much area will it irrigate in 30 minutes., if 8 cm of standing water is required for irrigation.

Sol.     Speed of water in the canal = 10 km/h

= 10000 m/60 min

=  \frac{{{\rm{500}}}}{{\rm{3}}}m/min.

∴  In 1 min, the volume of water which flows out of the canal

= volume of a cuboid of length \frac{{{\rm{500}}}}{{\rm{3}}} m,

breadth 30 dm = 3 m and height 12 dm = \frac{{{\rm{12}}}}{{{\rm{10}}}} m =\frac{{\rm{6}}}{{\rm{5}}} m

∴ The volume of the water flowing out of the canal in 1 minute = \left( {\frac{{500}}{3} \times 3 \times \frac{6}{5}} \right) m³ = 600 m³

∴ In 30 min, the amount of water flowing out of the canal = (600 × 30) m³ = 18000 m³

If the required area of the irrigated land is x m², then the volume of water to be needed to irrigate the land

=  \left( {x \times \frac{8}{{100}}} \right)m³ =  \frac{{2x}}{{25}}m³

Hence, \frac{{2x}}{{25}} = 18000 ⇒x = 18000 ×\frac{{25}}{2}  = 225000

Hence, the required area is 225000 m²

Ex.15.    If the diameter of the cross-section of a wire is decreased by 5%, how much per cent will the length be increased so that the volume remains the same?

Sol.      Let d be the original diameter and l be the original length of the wire. Then, the original volume of the wire = \frac{{{{\rm{d}}^{\rm{2}}}}}{{\rm{4}}} l

Now, if the diameter is decreased by 5% then the new diameter = d –\frac{{{\rm{5d}}}}{{{\rm{100}}}}  = d – \frac{{\rm{d}}}{{{\rm{20}}}} = \frac{{{\rm{19d}}}}{{{\rm{20}}}}

Let the length be increased by x%. Then, the new length is l +\frac{{lx}}{{{\rm{100}}}}  = l \left( {1 + \frac{x}{{{\rm{100}}}}} \right)= l\left( {\frac{{100 + x}}{{{\rm{100}}}}} \right)

Then, the new volume of the wire = {\left( {\frac{{{\rm{19d}}}}{{{\rm{40}}}}} \right)^{\rm{2}}}l\left( {\frac{{{\rm{100}} + x}}{{{\rm{100}}}}} \right)

By the condition of the problem, we have

{\left( {\frac{{{\rm{19d}}}}{{{\rm{40}}}}} \right)^{\rm{2}}}l \left( {\frac{{{\rm{100}} + x}}{{{\rm{100}}}}} \right)= \frac{{{\rm{\pi  }}{{\rm{d}}^{\rm{2}}}l}}{{\rm{4}}}

⇒ \frac{{361\left( {100 + x} \right)}}{{{\rm{160000}}}} = \frac{1}{{\rm{4}}}  ⇒ \frac{{361\left( {100 + x} \right)}}{{{\rm{40000}}}} = 1

⇒ 361 x = 40000 – 36100 = 3900

⇒   x = \frac{{3900}}{{{\rm{361}}}} = 10.8

Hence, the required per cent is 10.8% nearly.

 

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