Mixed: Area, Volume & Cost Estimation

Question 1

A student of grade 10 has submitted three paper made models of square based pyramids to his mathematics teacher as a project work. The length of base and slant height of each pyramid is 12 cm and 10 cm respectively.

1. Write the formula to find out the volume of a pyramid.
2. Find the vertical height of the pyramid.
3. How much space is occupied by the given pyramids?
4. Find the total cost of coloring each pyramid excluding the base if the rate of coloring is Rs. 80 per square cm.

Solution:

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

Slant height of the pyramid (l) = 10 cm

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

(a) The formula to find the volume of a pyramid (V) = $\frac{1}{3} A h = \frac{1}{3} a^2 h$

(b) Vertical height of the pyramid (h)

$= \sqrt{l^2 - \left(\frac{a}{2}\right)^2}$

$= \sqrt{(10)^2 - \left(\frac{12}{2}\right)^2}$

$= \sqrt{100 - 36}$

$= \sqrt{64} = 8 \text{ cm}$

(c) Volume of one pyramid (V) = $\frac{1}{3} a^2 h$

$= \frac{1}{3} \times 144 \times 8$

$= 384 \text{ cm}^3$

Space occupied by the three given pyramids = 3 × V = 3 × 384 = 1152 cm³

(d) Lateral surface area of one pyramid (excluding the base) (A) = 2al

$= 2 \times 12 \times 10$

$= 240 \text{ cm}^2$

Lateral surface area of the three pyramids = 3 × 240 = 720 cm²

Rate of coloring (R) = Rs. 80 per square cm

Total cost of coloring the pyramids excluding the base (T) = R × A = 80 × 720 = Rs. 57,600

Question 2

If the length of the side of a square based pyramid is 16 cm and volume is 1280 cm³,

1. Write the formula to find out the total surface area of the pyramid.
2. Find the perimeter of the base.
3. Find the slant height of the pyramid.
4. Find the total surface area and the area of triangular surfaces. Which area is greater and by how much? Calculate the area and compare the result.

Solution:

Here, length of the base (a) = 16 cm

Volume of the pyramid (V) = 1280 cm³

Area of the base (A) = $a^2 = (16)^2 = 256$ cm²

(a) The formula to find the total surface area of the pyramid (T.S.A) = $a^2 + 2al$

(b) Perimeter of the base (P) = 4a = 4 × 16 = 64 cm

(c) First, we find the vertical height of the pyramid.

We know, Volume (V) = $\frac{1}{3} a^2 h$

$\text{or, } 1280 = \frac{1}{3} \times 256 \times h$

$\text{or, } h = \frac{1280 \times 3}{256}$

$\text{or, } h = 15 \text{ cm}$

Now, slant height of the pyramid (l) = $\sqrt{h^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(15)^2 + (8)^2}$

$= \sqrt{225 + 64}$

$= \sqrt{289} = 17 \text{ cm}$

(d) Total surface area of the pyramid (T.S.A) = $a^2 + 2al$

$= (16)^2 + 2 \times 16 \times 17$

$= 256 + 544$

$= 800 \text{ cm}^2$

Area of the triangular surfaces (lateral surface area) = 2al

$= 2 \times 16 \times 17$

$= 544 \text{ cm}^2$

Comparing the two, the total surface area (800 cm²) is greater than the area of the triangular surfaces (544 cm²).

Difference = 800 − 544 = 256 cm²

Hence, the total surface area is greater than the area of the triangular surfaces by 256 cm² (which is equal to the area of the square base).

Question 3

The lateral surface area of a square based pyramid is 540 cm² and its slant height is 15 cm.

1. Write the formula to find out the total surface area of a pyramid.
2. Find the length of its base.
3. Find its volume.

Solution:

Here, lateral surface area (L.S.A) = 540 cm²

Slant height of the pyramid (l) = 15 cm

(a) The formula to find the total surface area of the pyramid (T.S.A) = $a^2 + 2al$

(b) We know, Lateral surface area (L.S.A) = 2al

$\text{or, } 540 = 2 \times a \times 15$

$\text{or, } 540 = 30a$

$\text{or, } a = \frac{540}{30} = 18 \text{ cm}$

∴ Length of the base = 18 cm

(c) Vertical height of the pyramid (h) = $\sqrt{l^2 - \left(\frac{a}{2}\right)^2}$

$= \sqrt{(15)^2 - (9)^2}$

$= \sqrt{225 - 81}$

$= \sqrt{144} = 12 \text{ cm}$

Now, Volume of the pyramid (V) = $\frac{1}{3} a^2 h$

$= \frac{1}{3} \times (18)^2 \times 12$

$= \frac{1}{3} \times 324 \times 12$

$= 1296 \text{ cm}^3$

∴ Volume of the pyramid = 1296 cm³

Question 4

The figure is a toy formed by combining a cylinder and hemisphere.

1. What is the diameter of its base?
2. Find the circumference of its base.
3. Find the total surface area.
4. Find the volume of the toy.

Solution:

Here, diameter of the base (d) = 7 cm

Radius of the base (r) = $\frac{d}{2} = \frac{7}{2}$ = 3.5 cm

Total length of the toy = 10 cm

Height of the cylinder (h) = total length − radius of hemisphere = 10 − 3.5 = 6.5 cm

(a) The diameter of the base = 7 cm

(b) Circumference of the base (C) = $2\pi r$

$= 2 \times \frac{22}{7} \times 3.5$

$= 22 \text{ cm}$

(c) The total surface area of the toy = curved surface area of cylinder + area of one circular (flat) end + curved surface area of hemisphere

$\text{T.S.A} = 2\pi r h + \pi r^2 + 2\pi r^2$

$= 2 \times \frac{22}{7} \times 3.5 \times 6.5 + \frac{22}{7} \times (3.5)^2 + 2 \times \frac{22}{7} \times (3.5)^2$

$= 143 + 38.5 + 77$

$= 258.5 \text{ cm}^2$

(d) Volume of the toy = volume of cylinder + volume of hemisphere

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

$= \frac{22}{7} \times (3.5)^2 \times 6.5 + \frac{2}{3} \times \frac{22}{7} \times (3.5)^3$

$= 250.25 + 89.83$

$= 340.08 \text{ cm}^3$

∴ Volume of the toy = 340.08 cm³

Question 5

A solid object contains a square based pyramid of slant height 5 cm on the top and a square based prism with length of base 8 cm on its bottom. The volume of object is 1024 cm³,

1. Draw the figure as per the information given in the question.
2. Find the vertical height of the pyramid shaped part.
3. Find by how much the vertical height of the pyramid shaped part is greater or less or equal to the height of prism shaped part.
4. Find the total surface area of the object.

Solution:

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

Slant height of the pyramid (l) = 5 cm

Volume of the object (V) = 1024 cm³

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

(a)

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

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

$= \sqrt{25 - 16}$

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

∴ Vertical height of the pyramid shaped part = 3 cm

(c) Let the height of the prism = h₁

Volume of pyramid (V₂) = $\frac{1}{3} A h_2 = \frac{1}{3} \times 64 \times 3 = 64$ cm³

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

$\text{or, } 1024 = A h_1 + 64$

$\text{or, } 64 h_1 = 1024 - 64$

$\text{or, } 64 h_1 = 960$

$\text{or, } h_1 = \frac{960}{64} = 15 \text{ cm}$

∴ Height of the prism shaped part = 15 cm

Difference = h₁ − h₂ = 15 − 3 = 12 cm

Hence, the vertical height of the pyramid shaped part (3 cm) is less than the height of the prism shaped part (15 cm) by 12 cm.

(d) The total surface area of the object = area of the base of prism + lateral surface area of prism + lateral surface area of pyramid

Area of the base of prism = $a^2 = 64$ cm²

Lateral surface area of prism = 4a × h₁ = 4 × 8 × 15 = 480 cm²

Lateral surface area of pyramid = 2al = 2 × 8 × 5 = 80 cm²

Now, total surface area (T.S.A) = 64 + 480 + 80 = 624 cm²

∴ The total surface area of the object = 624 cm²

Question 6

The lower part of a tent is cylindrical and the upper part is hemispherical. The radius of both parts is the same. If the total height of the tent is 54 m and the height of the cylindrical part is 40 m,

1. draw a figure as per the information given in the question.
2. find the radius of the cylindrical part.
3. find the total surface area of the tent.
4. The cost of cloth per meter is Rs. 500, find the total cost to construct the tent.

Solution:

Here, total height of the tent = 54 m

Height of the cylindrical part (h) = 40 m

(a)

(b) Since the height of the hemispherical part is equal to its radius,

Radius (r) = total height − height of cylindrical part = 54 − 40 = 14 m

∴ Radius of the cylindrical part = 14 m

(c) The total surface area of the tent (cloth required) = curved surface area of cylinder + curved surface area of hemisphere

$\text{T.S.A} = 2\pi r h + 2\pi r^2 = 2\pi r (h + r)$

$= 2 \times \frac{22}{7} \times 14 , (40 + 14)$

$= 88 \times 54$

$= 4752 \text{ m}^2$

∴ The total surface area of the tent = 4752 m²

(d) Rate of cloth per square meter (R) = Rs. 500

Total cost to construct the tent (T) = R × T.S.A = 500 × 4752 = Rs. 23,76,000

Question 7

A solid object is formed with a hemisphere and cone combined together. The radius of the base of both parts is 7 cm. The total cost to color the object is Rs. 5148 at the rate of Rs. 6 per square cm.

1. Find the total surface area of the combined solid.
2. What is the slant height of the cone?
3. Find the vertical height of the cone.
4. What is the total height of the solid?

Solution:

Here, radius of the base (r) = 7 cm

Total cost to color the object = Rs. 5148

Rate of coloring (R) = Rs. 6 per square cm

(a) Total surface area of the combined solid (T.S.A) = $\frac{\text{Total cost}}{\text{Rate}}$

$= \frac{5148}{6}$

$= 858 \text{ cm}^2$

∴ The total surface area of the combined solid = 858 cm²

(b) The total surface area = curved surface area of cone + curved surface area of hemisphere

$\text{or, } \pi r l + 2\pi r^2 = 858$

$\text{or, } \frac{22}{7} \times 7 \times l + 2 \times \frac{22}{7} \times (7)^2 = 858$

$\text{or, } 22 l + 308 = 858$

$\text{or, } 22 l = 550$

$\text{or, } l = \frac{550}{22} = 25 \text{ cm}$

∴ The slant height of the cone = 25 cm

(c) Vertical height of the cone (h) = $\sqrt{l^2 - r^2}$

$= \sqrt{(25)^2 - (7)^2}$

$= \sqrt{625 - 49}$

$= \sqrt{576} = 24 \text{ cm}$

∴ The vertical height of the cone = 24 cm

(d) Total height of the solid = height of cone + radius of hemisphere

$= h + r = 24 + 7 = 31 \text{ cm}$

∴ The total height of the solid = 31 cm

Question 8

A toy is formed combining a cylinder and a cone with the same base of diameter 10 cm. The length of the cylindrical part is 14 cm and vertical height of the conical part is 12 cm.

1. Find the slant height of the cone.
2. If the length of cylindrical part and the height of conical part are exchanged to form a new toy, which one costs more under the same rate of cost per square meter?

Solution:

Here, diameter of the base (d) = 10 cm

Radius of the base (r) = $\frac{d}{2} = \frac{10}{2}$ = 5 cm

Length of the cylindrical part (H) = 14 cm

Vertical height of the conical part (h) = 12 cm

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

$= \sqrt{(12)^2 + (5)^2}$

$= \sqrt{144 + 25}$

$= \sqrt{169} = 13 \text{ cm}$

∴ The slant height of the cone = 13 cm

(b) The surface area of a toy = area of the circular base of the cylinder + curved surface area of the cylinder + curved surface area of the cone

$\text{S} = \pi r^2 + 2\pi r H + \pi r l = \pi r \, (r + 2H + l)$

Original toy (cylinder = 14 cm, cone height = 12 cm, l = 13 cm):

$S_1 = \pi r \, (r + 2H + l)$

$= \frac{22}{7} \times 5 , (5 + 2 \times 14 + 13)$

$= \frac{22}{7} \times 5 \times 46$

$= 722.86 \text{ cm}^2$

New toy (cylinder = 12 cm, cone height = 14 cm):

Slant height of the new cone (l') = $\sqrt{(14)^2 + (5)^2} = \sqrt{196 + 25} = \sqrt{221} = 14.87$ cm

$S_2 = \pi r \, (r + 2H' + l')$

$= \frac{22}{7} \times 5 , (5 + 2 \times 12 + 14.87)$

$= \frac{22}{7} \times 5 \times 43.87$

$= 689.31 \text{ cm}^2$

Since $S_1 = 722.86 \text{ cm}^2 > S_2 = 689.31 \text{ cm}^2$, under the same rate of cost per unit area, the original toy (with the longer cylindrical part of 14 cm) costs more.

Difference in area = 722.86 − 689.31 = 33.55 cm²

Question 9

A pyramid of vertical height 12 cm is placed on the top of a square based cuboid. If the area of base of cuboid shaped solid is 100 cm² and its height is 9 cm,

1. What is the length of the base of cuboid shaped object?
2. Find the volume of the combined solid.

Solution:

Here, area of the base (A) = 100 cm²

Height of the cuboid (h₁) = 9 cm

Vertical height of the pyramid (h₂) = 12 cm

(a) Since the base is square, area of base (A) = $a^2$

$\text{or, } a^2 = 100$

$\text{or, } a = \sqrt{100} = 10 \text{ cm}$

∴ The length of the base of the cuboid shaped object = 10 cm

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

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

$= 100 \times 9 + \frac{1}{3} \times 100 \times 12$

$= 900 + 400$

$= 1300 \text{ cm}^3$

∴ The volume of the combined solid = 1300 cm³

Question 10

A cylindrical tank with radius 14 cm and height 40 cm is full of cement. A conical shaped pile of height 30 cm is formed while it is poured on the ground.

1. Find the radius of the conical shaped pile of cement.
2. Find the surface area of the pile of cement excluding the base.
3. If the weight of 1 cm³ cement is 2.5 gm, find the total weight of the cement.

Solution:

Here, radius of the cylindrical tank (r) = 14 cm

Height of the cylindrical tank (h) = 40 cm

Height of the conical pile (H) = 30 cm

Volume of cement = volume of the cylindrical tank (V) = $\pi r^2 h$

$= \frac{22}{7} \times (14)^2 \times 40$

$= \frac{22}{7} \times 196 \times 40$

$= 24640 \text{ cm}^3$

(a) When poured on the ground, the volume of the conical pile is equal to the volume of cement.

$\text{or, } \frac{1}{3} \pi R^2 H = 24640$

$\text{or, } \frac{1}{3} \times \frac{22}{7} \times R^2 \times 30 = 24640$

$\text{or, } \frac{220}{7} R^2 = 24640$

$\text{or, } R^2 = \frac{24640 \times 7}{220} = 784$

$\text{or, } R = \sqrt{784} = 28 \text{ cm}$

∴ The radius of the conical shaped pile = 28 cm

(b) Slant height of the cone (l) = $\sqrt{R^2 + H^2}$

$= \sqrt{(28)^2 + (30)^2}$

$= \sqrt{784 + 900}$

$= \sqrt{1684} = 41.04 \text{ cm}$

Surface area of the pile excluding the base = curved surface area of cone = $\pi R l$

$= \frac{22}{7} \times 28 \times 41.04$

$= 88 \times 41.04$

$= 3611.2 \text{ cm}^2$

∴ The surface area of the pile excluding the base = 3611.2 cm²

(c) Weight of 1 cm³ of cement = 2.5 gm

Total weight of the cement = volume × weight per cm³

$= 24640 \times 2.5$

$= 61600 \text{ gm} = 61.6 \text{ kg}$

∴ The total weight of the cement = 61600 gm (61.6 kg)