Part 2 - Area and Volume of Combined Solid Objects

Question 1

Answer the following questions on the basis of given information corresponding to the combined solid objects.

Solution:

(a)

Total volume of combined solid (V) = 1050 cm³

Volume of cylindrical part (V₁) = 748 cm³

Volume of conical part (V₂) = ?

Here, V = V₁ + V₂

$\text{or, } V_2 = V - V_1$

$= 1050 - 748$

$= 302$ cm³

∴ Volume of conical part (V₂) = 302 cm³

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(b)

Curved surface area of the conical part = 252 cm²

Curved surface area of the cylindrical part = 272 cm²

Area of the circular part = 154 cm²

Total surface area = ?

Total surface area = Curved surface area of conical part + Curved surface area of cylindrical part + Area of the circular part

$= 252 + 272 + 154$

$= 678$ cm²

∴ Total surface area = 678 cm²

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(c)

Area of the base (A) = 36 cm²

Volume of the prism shaped part (V₁) = 144 cm³

Height of the prism = h, height of the pyramid = 2h

Volume of the pyramid shaped part (V₂) = ?

We know, Volume of the prism shaped part (V₁) = A × h

$\text{or, } 144 = 36 \times h$

$\text{or, } h = \frac{144}{36} = 4 \text{ cm}$

So, height of the pyramid = 2h = 2 × 4 = 8 cm

Now, Volume of the pyramid shaped part (V₂) = $\frac{1}{3} \times A \times 2h$

$= \frac{1}{3} \times 36 \times 8$

$= 96$ cm³

∴ Volume of the pyramid shaped part (V₂) = 96 cm³

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(d)

Curved surface area of the conical shaped part = 308 cm²

Radius = r, slant height of cone (l) = 2r

Curved surface area of the hemispherical part = ?

We know, Curved surface area of the conical part = $\pi r l$

$\text{or, } 308 = \pi r \times 2r$

$\text{or, } 308 = 2\pi r^2$

$\text{or, } \pi r^2 = 154$

$\text{or, } \frac{22}{7} r^2 = 154$

$\text{or, } r^2 = 49$

$\text{or, } r = 7 \text{ cm}$

Now, Curved surface area of the hemispherical part = $2\pi r^2$

$= 2 \times \frac{22}{7} \times (7)^2$

$= 2 \times 154$

$= 308$ cm²

∴ Curved surface area of the hemispherical part = 308 cm²

Question 2

Find the volume of solid objects with given dimensions.

Solution:

(a)

Radius of the base (r) = $\frac{40}{2}$ = 20 cm

Height of the cylinder (h₁) = 70 cm

Height of the cone (h₂) = 10 cm

Volume of solid (V) = volume of cylinder + volume of cone

$= \pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2$

$= \pi r^2 \left(h_1 + \frac{1}{3} h_2\right)$

$= \frac{22}{7} \times (20)^2 \left(70 + \frac{1}{3} \times 10\right)$

$= \frac{22}{7} \times 400 \times \frac{220}{3}$

$= 92190.48$ cm³

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(b)

Length of the base (a) = 12 cm

Height of the prism (h₁) = 6 cm

Total height = 15 cm

Height of the pyramid (h₂) = 15 − 6 = 9 cm

Area of base (A) = $a^2 = (12)^2 = 144$ cm²

Volume of solid (V) = volume of prism + volume of pyramid

$= A \times h_1 + \frac{1}{3} A \times h_2$

$= A \left(h_1 + \frac{1}{3} h_2\right)$

$= 144 \left(6 + \frac{1}{3} \times 9\right)$

$= 144 \times 9$

$= 1296$ cm³

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(c)

Height of the cone (h) = 42 cm

Total height = 56 cm

Radius of the base (r) = 56 − 42 = 14 cm

Volume of solid (V) = volume of cone + volume of hemisphere

$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$

$= \frac{1}{3}\pi r^2 (h + 2r)$

$= \frac{1}{3} \times \frac{22}{7} \times (14)^2 (42 + 2 \times 14)$

$= \frac{1}{3} \times \frac{22}{7} \times 196 \times 70$

$= 14373.33$ cm³

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(d)

Radius of the base (r) = $\frac{14}{2}$ = 7 cm

Height of the left cone (h₁) = 6 cm

Height of the right cone (h₂) = 9 cm

Volume of solid (V) = volume of two cones

$= \frac{1}{3}\pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2$

$= \frac{1}{3}\pi r^2 (h_1 + h_2)$

$= \frac{1}{3} \times \frac{22}{7} \times (7)^2 (6 + 9)$

$= \frac{1}{3} \times \frac{22}{7} \times 49 \times 15$

$= 770$ cm³

Question 3

In the given figure of a crystal, the shaded part is square shaped of length 2.5 cm. The total height of the object is 4.5 cm. If the height of upper and lower pyramids are equal, find the total surface area and volume of the crystal.

Solution:

Here, length of the base (a) = 2.5 cm

Total height of the crystal = 4.5 cm

Since the heights of the two pyramids are equal,

Height of each pyramid (h) = $\frac{4.5}{2}$ = 2.25 cm

Volume:

Volume of the crystal (V) = 2 × volume of one pyramid

$= 2 \times \frac{1}{3} a^2 h$

$= \frac{2}{3} \times (2.5)^2 \times 2.25$

$= \frac{2}{3} \times 6.25 \times 2.25$

$= 9.375$ cm³

Total surface area:

Slant height of each pyramid (l)

$= \sqrt{h^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(2.25)^2 + (1.25)^2}$

$= \sqrt{5.0625 + 1.5625}$

$= \sqrt{6.625}$

$= 2.57$ cm

All the faces of the crystal are triangles (the square base is hidden), so

Total surface area (TSA) = 2 × (2al) = 4al

$= 4 \times 2.5 \times 2.57$

$= 25.74$ cm²

∴ Volume = 9.375 cm³ and Total surface area = 25.74 cm²

Question 4

Find the total surface area of the given combined solids.

Solution:

(a)

Radius of the base (r) = $\frac{6}{2}$ = 3 cm

Length of the cylinder (h₁) = 116 cm

Height of the cone (h₂) = 120 − 116 = 4 cm

Slant height of the cone (l) = $\sqrt{h_2^2 + r^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ cm

Total surface area (TSA) = $\pi r^2 + 2\pi r h_1 + \pi r l = \pi r (r + 2h_1 + l)$

$= \frac{22}{7} \times 3 \, (3 + 2 \times 116 + 5)$

$= \frac{22}{7} \times 3 \times 240$

$= 2262.86$ cm²

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(b)

Radius of the base (r) = $\frac{14}{2}$ = 7 cm

Length of the cylinder (h₁) = 32 cm

Slant height of the cone (l) = 25 cm

Total surface area (TSA) = $\pi r l + 2\pi r h_1 + \pi r^2 = \pi r (l + 2h_1 + r)$

$= \frac{22}{7} \times 7 \, (25 + 2 \times 32 + 7)$

$= 22 \times 96$

$= 2112$ cm²

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(c)

Height of the cone (h) = 24 cm

Radius of the base (r) = 31 − 24 = 7 cm

Slant height of the cone (l) = $\sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{625} = 25$ cm

Total surface area (TSA) = $2\pi r^2 + \pi r l = \pi r (2r + l)$

$= \frac{22}{7} \times 7 \, (2 \times 7 + 25)$

$= 22 \times 39$

$= 858$ cm²

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(d)

Height of the cone (h) = 48 cm

Slant height of the cone (l) = 50 cm

Radius of the base (r) = $\sqrt{l^2 - h^2} = \sqrt{50^2 - 48^2} = \sqrt{196} = 14$ cm

Total surface area (TSA) = $\pi r l + 2\pi r^2 = \pi r (l + 2r)$

$= \frac{22}{7} \times 14 \, (50 + 2 \times 14)$

$= 44 \times 78$

$= 3432$ cm²

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(e)

Length of the base (a) = 30 cm

Height of the pyramid (h₂) = 8 cm

Height of the prism (h₁) = 28 − 8 = 20 cm

Slant height of the pyramid (l) = $\sqrt{h_2^2 + \left(\frac{a}{2}\right)^2} = \sqrt{8^2 + 15^2} = \sqrt{289} = 17$ cm

Total surface area (TSA) = area of base + lateral surface of prism + lateral surface of pyramid

$= a^2 + 4a h_1 + 2al$

$= (30)^2 + 4 \times 30 \times 20 + 2 \times 30 \times 17$

$= 900 + 2400 + 1020$

$= 4320$ cm²

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(f)

Length of the base (a) = 6 cm

Height of the prism (h₁) = 6 cm

Total height = 29 cm

Height of the pyramid (h₂) = 29 − 6 = 23 cm

Slant height of the pyramid (l) = $\sqrt{h_2^2 + \left(\frac{a}{2}\right)^2} = \sqrt{23^2 + 3^2} = \sqrt{538} = 23.19$ cm

Total surface area (TSA) = area of base + lateral surface of prism + lateral surface of pyramid

$= a^2 + 4a h_1 + 2al$

$= (6)^2 + 4 \times 6 \times 6 + 2 \times 6 \times 23.19$

$= 36 + 144 + 278.3$

$= 458.3$ cm²

Question 5

A toy contains conical and hemispherical parts of radius 14 cm. Find its total surface area if its total height is 49 cm.

Solution:

Here, radius of the base (r) = 14 cm

Total height = 49 cm

Height of the cone (h) = 49 − 14 = 35 cm

Slant height of the cone (l) = $\sqrt{h^2 + r^2} = \sqrt{35^2 + 14^2} = \sqrt{1421} = 37.7$ cm

Total surface area (TSA) = $\pi r l + 2\pi r^2 = \pi r (l + 2r)$

$= \frac{22}{7} \times 14 \, (37.7 + 2 \times 14)$

$= 44 \times 65.7$

$= 2890.8$ cm²

∴ Total surface area = 2890.8 cm²

Question 6

A pyramid of height 12 cm is placed on the top of a solid of square based cuboid. If the base area of cuboid shaped solid is 100 cm² and its height is 10 cm, find (a) total volume and (b) total surface area of the combined solid.

Solution:

Here, base area (A) = 100 cm²

Length of the base (a) = $\sqrt{100}$ = 10 cm

Height of the cuboid (h₁) = 10 cm

Height of the pyramid (h₂) = 12 cm

(a) Volume of the combined solid (V) = volume of cuboid + volume of pyramid

$= A \times h_1 + \frac{1}{3} A \times h_2$

$= 100 \times 10 + \frac{1}{3} \times 100 \times 12$

$= 1000 + 400$

$= 1400$ cm³

(b) Slant height of the pyramid (l) = $\sqrt{h_2^2 + \left(\frac{a}{2}\right)^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$ cm

Total surface area (TSA) = area of base + lateral surface of cuboid + lateral surface of pyramid

$= a^2 + 4a h_1 + 2al$

$= (10)^2 + 4 \times 10 \times 10 + 2 \times 10 \times 13$

$= 100 + 400 + 260$

$= 760$ cm²

Question 7

A pyramid with vertical height 8 cm is placed on the top of a cubical solid. Find the total volume of the combined solid if the length of the base of cube is 12 cm.

Solution:

Here, length of the base of the cube (a) = 12 cm

Height of the pyramid (h) = 8 cm

Area of base (A) = $a^2 = (12)^2 = 144$ cm²

Volume of the combined solid (V) = volume of cube + volume of pyramid

$= a^3 + \frac{1}{3} A \times h$

$= (12)^3 + \frac{1}{3} \times 144 \times 8$

$= 1728 + 384$

$= 2112$ cm³