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Composite Solids

Definition: A Composite Solid is a combination of basic Solids put together. Consider the diagram below, the composite solids are made as follows:

(i) A cube and a square-based pyramid

(ii) A cylinder and a cone

Addition and Subtraction of Volumes

Hollow Shapes

Definition: These are shapes that have space inside. The volume of material in a hollow object is found by subtracting the volume of the shape inside from the volume of the shape as if it were solid; examples of such shapes are boxes and pipes.

hollow shapes

Frustum of Cone and Pyramid

 Introduction: If a cone or pyramid, standing on a horizontal table is cut through, parallel to the table, the top part is a smaller cone or pyramid. The other part is called Frustum, see the diagram below.

frustrum-of-a-cone-and-pyramid.c76634c

In order to find the volume or surface area of a frustum, it is necessary to consider the frustum as a complete cone (or pyramid) with the smaller cone removed.

Example 1:

A water tank is 1.2m square and 1.35m deep. It is half full of water. How many times can a 9-litre bucket be filled from the tank?

Solution:

Volume = base area x height

= 1.2 x 1.2 x 1.35

= 1.944m3

Recall, 1,000litres = 1m3

∴ Volume of tank in litres = 1000 x 1.944 = 1,944 litres

∴ Half Volume = \(\frac {1944}{2} \)

= 972 litres

Number of times 9 Litres will fill the half of tank

Number of times = \( \frac{972}{9} \\ \scriptsize = 108 \; times \)

Example 2:

The figure below shows the cross-section of a steel rail, dimensions being given in cm. Calculate the mass, in tonnes, of a 20metre length of the rail if the mass of 1cm3 of the steel is 7.5g.

Screen Shot 2021 02 09 at 12.37.05 PM

Solution

Screen Shot 2021 11 26 at 1.29.44 PM

Total base area = 40 + 32 + 24 = 96cm2

Length = 20m = 20 x 100 = 2,000cm

∴ Volume = base area x height

= 2,000 x 96 = 192,000cm3

If 1cm3  \( \scriptsize \equiv 7.5g \)

∴ \( \scriptsize 192,000cm^3 \equiv 7.5 \: \times \: 192000 \)

= 1,440,000g

= 1,440kg

1tonne = 1000kg

∴ Mass = 1.44 tonnes

Example 3:

A cylindrical container 30cm in diameter holds appropriately 30litres of oil. How far does the oil level reach after 1 litre of oil has been used?

Solution:

Recall 1 litre = 1,000cm3

i.e. Volume = 1,000 = πr2h which suggests the ‘h’ to be fall in height

Diameter = 30cm

radius = 15cm

1000 = \( \frac{22}{7}\scriptsize \: \times \: 15^2 \: \times \: h \)

i.e h = \( \frac{1000 \: \times \: 7}{22 \: \times \: 225} \\ = \frac{20 \: \times \: 7}{11 \: \times \: 9} \)

∴ h = \( \frac{140}{99} \scriptsize = 1.41414141 \)

i.e. h = 1.4cm

Example 4: 

A right pyramid on a base 4cm square has a slanted edge of 6cm. Calculate the volume of the pyramid.

Screen Shot 2021 02 09 at 12.56.53 PM

Consider triangle ABC

Using Pythagoras Theorem

AC = \( \scriptsize \sqrt {4^2 \: + \: 4^2} \)

AC = \( \scriptsize \sqrt {16 \: + \: 16} \\ \scriptsize = \sqrt {32} \\ \scriptsize = 4 \sqrt {2} \)

Note that AE =  \( \frac{AC}{2} \)

i.e AE = \( \frac {4 \sqrt {2}}{2} \\ \scriptsize = 2\sqrt{2} cm \)

Consider \( \scriptsize \Delta \; ADE \)

Using Pythagoras theorem

DE = \( \scriptsize \sqrt {36 \: – \: (2 \sqrt{2})^2} \)

DE = \( \scriptsize \sqrt {36 \: – \: 8} \)

DE = \( \scriptsize \sqrt {28} \)

DE = \( \scriptsize 2 \sqrt {7} \)

Volume of Pyramid

= \( \frac{1}{3} \: \scriptsize \times \: 4^2 \: \times \: 2 \sqrt {7} \\ = \frac{32}{3} \scriptsize \sqrt{7} \)

= 28.22134731

Therefore, Volume = 28.2cm3

Example 5:

A cone of height 9cm has a volume of ncm3 and a curved surface area of ncm2. If the cone is cut and opened out into the sector of a circle. Find the:

a. Vertical angle of the cone

b. Angle of the sector.

Solution:

Screen Shot 2021 11 26 at 6.02.32 PM

a.

Consider figure (i)

Volume = \( \frac{1}{3} \scriptsize \: \times \: \pi r^2h = ncm^3 \)

Curved Surface Area = \( \scriptsize \pi r l = ncm^2 \)

∴ \( \frac{1}{3}\scriptsize \pi r^2 h = \pi rl = n \)

Divide both sides by \( \scriptsize \pi r \)

\( \frac{1}{3}\scriptsize r \: \times \: h = l \)

Substitute for h = 9

i.e \( \frac{1}{3}\scriptsize r \: \times \: 9 = l \)

\(\scriptsize 3r = l \)

Screen Shot 2021 11 26 at 5.30.55 PM

Sinθ = \( \frac{r}{3r} \)

i.e Sinθ = \( \frac{1}{3} \scriptsize = 0.3333 \)

i.e θ = sin -1 (0.333)

θ = 19.47122°

Therefore, vertical angle = 2θ°

= 2 x 19.47122

= 38.942441°

2θ = 38.9°

vertical angle = 38.9°

b.

Screen Shot 2021 11 26 at 5.33.25 PM 1

tan 19.47122 = \( \frac{r}{9} \)

r = 9 x tan 19.47122

r = 3.1819804

r = 3.18

r = 3.2cm

Length of circular base = \( \scriptsize 2 \pi r \\ \scriptsize 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: 3.1819804 \)

= 20.001

= 20cm

Recall length of circular base = Arc length of sector

i.e \( \frac{\theta}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: l = 20\)

Recall l =3r

i.e. l = 3 x 3.1819804

therefore, l = 9.5459412

\( \frac{\theta}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: 9.546 = 20\)

\( \frac{\theta}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{44 \: \times \: 9.546}{7}\scriptsize = 20\)

θ = \( \frac{20 \: \times \: 360 \: \times \: 7}{44 \: \times \: 9.546} \)

θ = \( \frac{50,400}{420.021413} \\ \scriptsize = 119.993882 \\ \scriptsize \theta = 120^o\)

Example 6

An iron pipe has a cross-section as shown in the diagram below, the iron being 1cm thick. The mass of 1cm3 of cast iron is 7.2g. Calculate the mass of a 2 metre length of the pipe.

Screen Shot 2021 02 09 at 1.52.30 PM

Solution:

For external, L = 12cm, B = 10cm, H = 2m = 200cm

Volume = L x B x H

= 12 x 10 x 200

= 24,000

For Internal , L = 10, B= 8, H =200

Volume = 10 x 8 x 200

=18,000

Volume of Iron = 24,000 – 16,000

= 8,000cm3

Therefore, Mass of pipe = 8,000 x 7.2

= 57600g

= 57.6kg

Example 7

A cylindrical pipe, made of metal, is 3cm thick. If the internal radius of the pipe is 10cm, find the volume of the metal used in making 3m of the pipe. (Answer in cm3 in terms in terms of  π)           (JAMB)

Solution

Screen Shot 2021 02 10 at 7.12.32 AM

Consider the diagram above

R = 13cm, r= 10cm, h = 3m = 300cm

External:

Volume = πr2h

Volume = \( \frac{22}{7}\scriptsize \: \times \: 13^2 \: \times \: 300 \)

Volume = \( \frac{22}{7}\scriptsize \: \times \: 169 \: \times \: 300 \)

= 159,342.857

Internal:

r = 10cm, h = 300cm

Volume = πr2h

Volume = \( \frac{22}{7}\scriptsize \: \times \: 10^2 \: \times \: 300 \)

Volume = \( \frac{22}{7}\scriptsize \: \times \: 100 \; \times \: 300 \)

Volume = \( \frac{22}{7}\scriptsize \: \times \: 30,000 \)

= 94,285.7143

Volume of metal used = Volume of External – Volume of Internal

= πR2h – πr2h (where h is the length of the pipe)

πh(R2 – r2)

πh(R – r)(R + r)

=π x 300(13 – 10)(13 + 10)

= 300π(3)(23)

= 300 x 69 x π

= 20700π

= 20,700πcm3 (in terms of π)

Example 8:

 A solid cube of side 8cm is dropped into a cylindrical tank of radius 7cm. Calculate the new water level in the tank. (WAEC)

Solution

Screen Shot 2021 02 10 at 7.25.20 AM

Volume of cube = l3 = 83 = 512cm3

Therefore, equivalent height in the cylinder is given by 

512 = πr2h (r = 7cm)

i.e 512 = \( \frac{22}{7} \scriptsize \: \times \: 7 \: \times \: 7 \: \times \: h \)

512 = 154h

h = \( \frac{512}{514} \)

= 3.32467532

Therefore, new water level = 9 + 3.3246753

= 12.3246753

= 12.3cm

Example 9:

Screen Shot 2021 02 10 at 7.34.06 AM

In the diagram above an iron rod 8cm high and 6cm in diameter stands in a cylindrical tin 12cm in diameter. Water is poured into the tin until its depth is 8cm. How far would the level drop when the rod is removed?

Solution:

Volume of rod = πr2h

= π x 32 x 8 = 72πcm3

Corresponding height of 72  πcm3 in cylinder of radius 6cm is given as

72π = πR2h

72 =  62h

h = \( \frac{72}{36} \)

h = 2

Therefore, the drop in water level is 2cm.

Example 10

A frustum of a pyramid is 16cm square at the bottom, 6cm square at the top, and 12cm high. Find the volume of the frustum

Solution

Screen Shot 2021 02 10 at 7.39.23 AM

JG = \( \frac{EG}{2} \\ = \frac{\sqrt{6^2 \: + \: 6^2}}{2} \\ =\frac{\sqrt{72}}{2} \\ =\frac{6\sqrt{2}}{2} \\= \scriptsize 3 \sqrt{2}\)

KC = \( \frac{AC}{2} \\ = \frac{\sqrt{16^2 \: + \: 16^2}}{2} \\ =\frac{\sqrt{512}}{2}\\ =\frac{16\sqrt{2}}{2}\\ = \scriptsize 8 \sqrt{2}\)

Applying the principle of similarity on  \( \scriptsize \Delta IKC \; and \; IJG \)

\( \frac{x}{3\sqrt{2}} = \frac{x \: + \: 12}{8\sqrt{2}} \)

\( \scriptsize x \: \times \: 8\sqrt{2} = 3 \sqrt{2(x \: + \: 12)} \)

i.e. 8x = 3(x + 12)

8x – 3x =36

5x = 36

x = \( \frac{36}{5} \)

x = 7.2

Therefore, Volume of frustum = Volume of Big – Volume of small 

= \( \normalsize \frac{1}{3} \scriptsize \: \times \: 16^2 \: \times \: 19.2 \: – \: \normalsize \frac{1}{3} \scriptsize \: \times \: 6^2 \: \times \: 7.2 \)

= 1638.4 – 86.4

= 1,552cm3

Example 11

The volume of a right circular cone is 5litres. Calculate the volumes of the two parts into which the cone is divided by a plane parallel to the base one-third of the way down from the vertex to the base. Give your answers to the nearest ml. (WAEC)

Solution:

Recall, 1Litre = 1,000ml

Therefore, 5litres = 5,000ml

Consider the diagram below

Screen Shot 2021 02 10 at 7.55.31 AM

Comparing Volumes and height of the corresponding cones in terms of volumes

\( \left( \frac{x}{\frac{x}{3}}\right)^3 = \frac{Vbig}{Vsmall} \)

\( \left( \frac{x}{\frac{x}{3}}\right)^3 = \frac{5,000}{Vsmall} \)

\( \scriptsize x^3 \: \times \: \frac{27}{x^3} = \frac{5,000}{Vsmall} \)

\( \scriptsize 27 =\normalsize \frac{5,000}{Vsmall} \)

\( \scriptsize Vsmall = \frac{5,000}{27} \)

= 185.185185ml

Therefore, Volume of cone = 185ml (nearest ml)

∴ the volume of the \( \frac{2}{3} \) part is given by

5,000 – 185.185185 = 4,814.81481

∴ Volume of Frustum = 4,815ml (nearest ml)

Example 12

The cone in the diagram below is exactly half full of water by Volume. How deep is the water in the cone?

Screen Shot 2021 11 26 at 6.53.21 PM

 

Solution:

Screen Shot 2021 02 10 at 8.06.52 AM

Recall, V small = \( \frac{1}{2} \scriptsize \: \times \: Vbig \)

By proportion,

\( \frac{(16)^3}{x^3} \\= \frac {VBig}{\frac{VBig}{2}} \\ = \frac{2VBig}{VBig} \)

i.e. \( \frac{(16)^3}{x^3} = \frac{2}{1} \)

i.e 2x3 = 163 = 4096

2x3 = 4096

Therefore, x3 = \( \frac{4096}{2} \)

x3 = 2048

x = \(\scriptsize \sqrt [3] {2048} \)

x = 12.6992084157456

x = 12.7cm

Example 13

A bucket is 20cm in diameter at the open end, 12cm in diameter at the bottom, and 16cm deep. To what depth would the bucket fill a cylindrical tin 28cm in diameter? Calculate the internal surface area of the bucket.

Solution:

Screen Shot 2021 02 10 at 8.29.27 AM

By similar triangles

\( \frac{10}{16 \: + \: x} = \frac{6}{x} \)

10x =6(16 + x)

10x = 96 +6x

4x = 96

X = 28

16 + x = 44

Volume of Bucket = Volume of Big Cone – Volume of Small Cone

= \( \frac {1}{3}\scriptsize \pi (10^2 \: \times \: 44 \: – \; 6^2 \: \times \: 28) \)

= \( \frac {1}{3}\scriptsize \pi (4400 \: – \: 1008) \)

= \( \frac {1}{3}\scriptsize \pi (3,392) \)

= 3,553.5

3,554cm3

When the bucket is emptied into a cylinder of 28cm diameter, we have

Volume of a bucket = πr2h

i.e \(\scriptsize \pi \: \times \: 14^2 \: \times \: h = \frac{1}{3}\scriptsize \pi \: \times \: 3,392 \)

196h x 3 = 3392

h = \( \frac{3392}{588} \)

h = 5.76870748

Screen Shot 2021 02 10 at 8.38.57 AM

Internal Area of bucket =  π(RL – rl)

Screen Shot 2021 02 10 at 8.40.28 AM

l = \(\scriptsize \sqrt {10^2 \; + \; 44^2} \\ \scriptsize \sqrt {100 \: + \: 1936}\\ \scriptsize = \sqrt {2,036} \)

l = 45.12205669071391

i.e. l = 45.1cm

l = \( \scriptsize \sqrt {6^2 \: + \: 28^2}\\ \scriptsize \sqrt {36 \: + \: 784}\\ \scriptsize = \sqrt {820} \)

l = 28.6356

i.e. l = 28.7

Area of bucket (internal) = π(RL – rl)

= π(10 x 45.1 – 6 x 28.7)

= π(451 – 172.2)

π(278.8)

= 876.228571

= 876cm2

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