Part 3 - Area and Volume of Combined Solid Objects

Question 1

(a)

The solid object given in the figure contains a cone and a cylinder. The area of the base of cylinder is 100 cm² and its height is 3 cm. Find the total height of the solid if its total volume is 600 cm³.

Solution:

Here, area of the base (A) = 100 cm²

Height of the cylinder (h₁) = 3 cm

Total volume (V) = 600 cm³

Let the height of the cone = h₂

We know, Total volume (V) = volume of cylinder + volume of cone

$\text{or, } V = A h_1 + \frac{1}{3} A h_2$

$\text{or, } 600 = 100 \times 3 + \frac{1}{3} \times 100 \times h_2$

$\text{or, } 600 = 300 + \frac{100}{3} h_2$

$\text{or, } \frac{100}{3} h_2 = 300$

$\text{or, } h_2 = \frac{300 \times 3}{100} = 9 \text{ cm}$

∴ Height of the cone = 9 cm

Now, total height of the solid = h₁ + h₂ = 3 + 9 = 12 cm

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(b)

The upper part of a solid given aside is a pyramid with the slant height 5 cm. The lower part is a square based prism whose length of base is 8 cm. If the volume of solid is 448 cm³, find the height of the prism shaped portion.

Solution:

Here, length of the base (a) = 8 cm

Slant height of the pyramid (l) = 5 cm

Volume of solid (V) = 448 cm³

Area of base (A) = $a^2 = (8)^2 = 64$ cm²

Height of the pyramid (h₂) = $\sqrt{l^2 - \left(\frac{a}{2}\right)^2}$

$= \sqrt{(5)^2 - (4)^2}$

$= \sqrt{25 - 16}$

$= \sqrt{9} = 3 \text{ cm}$

Let the height of the prism = h₁

We know, Volume of solid (V) = volume of prism + volume of pyramid

$\text{or, } V = A h_1 + \frac{1}{3} A h_2$

$\text{or, } 448 = 64 \times h_1 + \frac{1}{3} \times 64 \times 3$

$\text{or, } 448 = 64 h_1 + 64$

$\text{or, } 64 h_1 = 384$

$\text{or, } h_1 = \frac{384}{64} = 6 \text{ cm}$

∴ Height of the prism shaped portion = 6 cm

Question 2

In the adjoining figure, a square based pyramid is placed on the top of a tower. The height of tower and pyramid are 6 ft. and the length of the base of tower is 1 ft.

1. Find the lateral surface area of the pyramid shaped portion.
2. Find the total surface area of the tower that can be painted.

Solution:

Here, length of the base (a) = 1 ft

Height of the tower (prism) (h₁) = 6 ft

Height of the pyramid (h₂) = 6 ft

(a) Slant height of the pyramid (l)

$= \sqrt{h_2^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(6)^2 + (0.5)^2}$

$= \sqrt{36.25}$

$= 6.02 \text{ ft}$

Lateral surface area of the pyramid shaped portion = 2al

$= 2 \times 1 \times 6.02$

$= 12.04 \text{ ft}^2$

(b) The paintable surface is the four walls of the tower and the four faces of the pyramid (the base rests on the ground and the top of the tower is covered by the pyramid).

Total surface area that can be painted = lateral surface of tower + lateral surface of pyramid

$= 4a h_1 + 2al$

$= 4 \times 1 \times 6 + 2 \times 1 \times 6.02$

$= 24 + 12.04$

$= 36.04 \text{ ft}^2$

Question 3

Ram received a spinning top (A toy) from his father as a gift on his sixth birthday. The lower part of the spinning top is conical and the upper part is hemispherical. The total height of the spinning top is 5 cm and diameter of hemisphere is 3.5 cm. Ram is planning to color it. Find the area of the surface that can be colored.

Solution:

Here, radius of the hemisphere (r) = $\frac{3.5}{2}$ = 1.75 cm

Total height of the spinning top = 5 cm

Height of the cone (h) = 5 − 1.75 = 3.25 cm

Slant height of the cone (l) = $\sqrt{h^2 + r^2}$

$= \sqrt{(3.25)^2 + (1.75)^2}$

$= \sqrt{10.5625 + 3.0625}$

$= \sqrt{13.625}$

$= 3.69 \text{ cm}$

The colorable surface is the curved surface of the cone and the curved surface of the hemisphere.

Area to be colored = $\pi r l + 2\pi r^2 = \pi r (l + 2r)$

$= \frac{22}{7} \times 1.75 \, (3.69 + 2 \times 1.75)$

$= 5.5 \times 7.19$

$= 39.55 \text{ cm}^2$

∴ The area of the surface that can be colored = 39.55 cm²