Part 2 - Mixed: Area, Volume & Cost Estimation

Question 1

There are two pillars with the base 6 ft × 6 ft and height 8 ft of each in a stadium. A pyramid of height 4 ft is placed on the top of each pillar.

1. Draw two figures as per the information given in question.
2. Write the formula to find the lateral surface area of a pyramid.
3. Find the slant height of the pyramid.
4. Find the surface area of the pyramid that need to be calculated for the painting purpose.
5. What will be the total cost to paint the pillars if the rate of cost of painting is Rs. 110 per square feet?

Solution:

Here, length of the base (a) = 6 ft

Height of the pillar (prism) (h₁) = 8 ft

Height of the pyramid (h₂) = 4 ft

(a)

(b) The formula to find the lateral surface area of a pyramid (A) = $2al$

(c) Slant height of the pyramid (l) = $\sqrt{h_2^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(4)^2 + (3)^2}$

$= \sqrt{16 + 9}$

$= \sqrt{25} = 5 \text{ ft}$

∴ The slant height of the pyramid = 5 ft

(d) The paintable surface of one pillar = lateral surface area of the pillar (prism) + lateral surface area of the pyramid. The base lies on the ground and the top of the pillar is covered by the pyramid, so they are not painted.

Lateral surface area of pillar (A₁) = 4a × h₁ = 4 × 6 × 8 = 192 ft²

Lateral surface area of pyramid (A₂) = 2al = 2 × 6 × 5 = 60 ft²

Paintable surface area of one pillar = A₁ + A₂ = 192 + 60 = 252 ft²

Total paintable surface area of two pillars = 2 × 252 = 504 ft²

(e) Rate of painting (R) = Rs. 110 per square feet

Total cost to paint the pillars (T) = R × A = 110 × 504 = Rs. 55,440

Question 2

There are two cuboid shaped pillars in the gate of a house, each of which has length 1 ft, breadth 1 ft and height 6 ft. On the top of each pillar, a pyramid of the same base and height 1 ft is placed.

1. Write the formula to calculate the lateral surface area of a prism.
2. What is the perimeter of the base of prism?
3. Find the slant height of the pyramid.
4. Find the area of pillars that need to be calculated for the painting purpose.
5. What will be the total cost to paint the pillars if the rate of cost of painting is Rs. 52 per square feet?

Solution:

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

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

Height of the pyramid (h₂) = 1 ft

(a) The formula to find the lateral surface area of a prism (A) = perimeter of base × height = $p h$ (where p = 4a)

(b) Perimeter of the base of prism (p) = 4a = 4 × 1 = 4 ft

(c) Slant height of the pyramid (l) = $\sqrt{h_2^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(1)^2 + (0.5)^2}$

$= \sqrt{1 + 0.25}$

$= \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \text{ ft} \approx 1.12 \text{ ft}$

∴ The slant height of the pyramid = $\frac{\sqrt{5}}{2}$ ft

(d) The paintable surface of one pillar = lateral surface area of the pillar (prism) + lateral surface area of the pyramid. The base lies on the ground and the top of the pillar is covered by the pyramid, so they are not painted.

Lateral surface area of pillar (A₁) = 4a × h₁ = 4 × 1 × 6 = 24 ft²

Lateral surface area of pyramid (A₂) = 2al = $2 \times 1 \times \frac{\sqrt{5}}{2} = \sqrt{5}$ ft²

Paintable surface area of one pillar = A₁ + A₂ = $(24 + \sqrt{5})$ = 26.24 ft²

Total paintable surface area of two pillars = $2(24 + \sqrt{5})$ = 52.47 ft²

(e) Rate of painting (R) = Rs. 52 per square feet

Total cost to paint the pillars (T) = R × A = 52 × 52.47 = Rs. 2728.55

Question 3

The diameter of concrete ring used in a well is 3.5 ft and its height is 1 ft. A ring costs Rs. 1200. If it requires 32 rings to construct the well,

1. Find the total cost to construct the well.
2. Find the volume of water.
3. If the water level is seen at the eighteenth ring, find the volume of well above the water level.

Solution:

Here, diameter of the ring (d) = 3.5 ft

Radius of the ring (r) = $\frac{d}{2} = \frac{3.5}{2}$ = 1.75 ft

Height of one ring = 1 ft

Number of rings = 32

Total height of the well (h) = 32 × 1 = 32 ft

(a) Cost of one ring = Rs. 1200

Total cost to construct the well = 32 × 1200 = Rs. 38,400

(b) Volume of water = volume of the well (cylinder) = $\pi r^2 h$

$= \frac{22}{7} \times (1.75)^2 \times 32$

$= \frac{22}{7} \times 3.0625 \times 32$

$= 308 \text{ ft}^3$

∴ The volume of water = 308 ft³

(c) If the water level is at the eighteenth ring, the number of empty rings above the water = 32 − 18 = 14 rings.

Height of the well above the water level (h') = 14 × 1 = 14 ft

Volume of the well above the water level = $\pi r^2 h'$

$= \frac{22}{7} \times (1.75)^2 \times 14$

$= \frac{22}{7} \times 3.0625 \times 14$

$= 134.75 \text{ ft}^3$

∴ The volume of the well above the water level = 134.75 ft³

Question 4

A solid object is formed combining two square based pyramids with the same base. The length of its base is 12 cm and the total height is 20 cm.

1. Write the formula to calculate the lateral surface area of a pyramid.
2. What is the volume of that combined solid object?

Solution:

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

Total height of the combined solid (H) = 20 cm

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

(a) The formula to calculate the lateral surface area of a pyramid (A) = $2al$

(b) The solid is formed by joining two pyramids at their common base, so the volume = volume of upper pyramid + volume of lower pyramid.

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

$= \frac{1}{3} \times 144 \times 20$

$= 960 \text{ cm}^3$

∴ The volume of the combined solid object = 960 cm³

Question 5

Nandakishor has established an agricultural farm of cows and goats in his village after he returned from foreign employment. There is a water tank with combined shape of a cylinder and cone. The inner diameter and height of the tank are 1.4 m and 2.1 m respectively. The vertical height of the cone is 0.36 m.

1. Write the formula to find the total surface area of a cone.
2. Find the perimeter of base of the tank.
3. How much water can be contained in the tank? Calculate in liters.

Solution:

Here, inner diameter of the tank (d) = 1.4 m

Radius of the tank (r) = $\frac{d}{2} = \frac{1.4}{2}$ = 0.7 m

Height of the cylindrical part (h) = 2.1 m

Vertical height of the cone (h₁) = 0.36 m

(a) The formula to find the total surface area of a cone (T.S.A) = $\pi r l + \pi r^2 = \pi r (l + r)$

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

$= 2 \times \frac{22}{7} \times 0.7$

$= 4.4 \text{ m}$

(c) The water is contained in both the cylindrical and conical parts.

Volume of the cylindrical part = $\pi r^2 h$

$= \frac{22}{7} \times (0.7)^2 \times 2.1$

$= 1.54 \times 2.1 = 3.234 \text{ m}^3$

Volume of the conical part = $\frac{1}{3} \pi r^2 h_1$

$= \frac{1}{3} \times \frac{22}{7} \times (0.7)^2 \times 0.36$

$= \frac{1}{3} \times 1.54 \times 0.36 = 0.1848 \text{ m}^3$

Total volume of the tank = 3.234 + 0.1848 = 3.4188 m³

We know, 1 m³ = 1000 liters

∴ Amount of water contained = 3.4188 × 1000 = 3418.8 liters

Question 6

A combined solid object with volume 240π cm³ is shown in the figure. The ratio of the slant height of the conical part and radius is 5 : 3.

1. What are the different shapes contained in the solid?
2. Write the formula to calculate the curved surface area of the cone.
3. Find the total surface area of the given solid.
4. Are the total surface and curved surface area of the solid equal? Justify with reason.

Solution:

Here, volume of the combined solid (V) = 240π cm³

Ratio of slant height to radius = l : r = 5 : 3

Let the slant height (l) = 5k and radius (r) = 3k.

(a) The solid contains a cone and a hemisphere.

(b) Curved surface area of the cone = $\pi r l$

(c) First, we find the value of k.

Height of the cone (h) = $\sqrt{l^2 - r^2} = \sqrt{(5k)^2 - (3k)^2} = \sqrt{16k^2} = 4k$

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

$\text{or, } 240\pi = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$

$\text{or, } 240\pi = \frac{1}{3}\pi (3k)^2 (4k) + \frac{2}{3}\pi (3k)^3$

$\text{or, } 240\pi = 12\pi k^3 + 18\pi k^3$

$\text{or, } 240\pi = 30\pi k^3$

$\text{or, } k^3 = 8$

$\text{or, } k = 2$

∴ Radius (r) = 3k = 6 cm, slant height (l) = 5k = 10 cm

Total surface area of the solid = curved surface area of cone + curved surface area of hemisphere

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

$= \pi \times 6 \times 10 + 2\pi \times (6)^2$

$= 60\pi + 72\pi$

$= 132\pi = 414.86 \text{ cm}^2$

∴ The total surface area of the given solid = 132π = 414.86 cm²

(d) Yes, the total surface area and the curved surface area of the solid are equal. This is because the combined solid has no plane (flat) surface on its exterior — the circular face where the cone and hemisphere meet is inside the solid. So the whole surface of the solid is made up only of curved surfaces (the curved surface of the cone and the curved surface of the hemisphere). Hence its total surface area is entirely curved surface area, i.e. T.S.A = C.S.A = 132π cm².

Question 7

The total surface area of a combined solid given in the figure is 840π cm². The ratio of slant height and vertical height of the conical part is 13 : 12 and the height of cylindrical part is 24 cm.

1. What are the different shapes contained in the given solid?
2. Write the formula to find the perimeter of base of a cone.
3. Find the volume of the solid.
4. Is the total surface area of solid equal to its curved surface area? If not, calculate and compare the result.

Solution:

Here, total surface area of the solid (T.S.A) = 840π cm²

Ratio of slant height to vertical height of cone = l : h = 13 : 12

Height of the cylindrical part (H) = 24 cm

Let the slant height (l) = 13m and vertical height of the cone (h) = 12m.

(a) The solid contains a cylinder and a cone.

(b) The perimeter of base of a cone (C) = $2\pi r$

(c) First, we find the radius.

Radius of the cone (r) = $\sqrt{l^2 - h^2} = \sqrt{(13m)^2 - (12m)^2} = \sqrt{25m^2} = 5m$

The total surface area of the solid = area of one circular end of cylinder + curved surface area of cylinder + curved surface area of cone

$\text{or, } \pi r^2 + 2\pi r H + \pi r l = 840\pi$

$\text{or, } \pi (5m)^2 + 2\pi (5m)(24) + \pi (5m)(13m) = 840\pi$

$\text{or, } 25m^2 + 240m + 65m^2 = 840$

$\text{or, } 90m^2 + 240m - 840 = 0$

$\text{or, } 3m^2 + 8m - 28 = 0$

$\text{or, } 3m^2 + 14m - 6m - 28 = 0$

$\text{or, } (3m + 14)(m - 2) = 0$

$\text{or, } m = 2 ; (\text{taking the positive value})$

∴ Radius (r) = 5m = 10 cm, vertical height of cone (h) = 12m = 24 cm

Volume of the solid = volume of cylinder + volume of cone

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

$= \pi \times (10)^2 \times 24 + \frac{1}{3}\pi \times (10)^2 \times 24$

$= 2400\pi + 800\pi$

$= 3200\pi = 10057.14 \text{ cm}^3$

∴ The volume of the solid = 3200π = 10057.14 cm³

(d) No, the total surface area is not equal to the curved surface area. The solid has one flat circular end (the outer face of the cylinder), which is a plane surface, so the total surface area is greater than the curved surface area.

Curved surface area (C.S.A) = 2πrH + πrl

$= 2\pi \times 10 \times 24 + \pi \times 10 \times 26$

$= 480\pi + 260\pi = 740\pi \text{ cm}^2$

Total surface area (T.S.A) = 840π cm²

Difference = T.S.A − C.S.A = 840π − 740π = 100π cm², which is exactly the area of the flat circular end ($\pi r^2 = \pi \times 10^2 = 100\pi$ cm²).

Question 8

There are a cylinder and a cone with the same base in the first figure. A solid shown in the second figure is formed combining the cylinder and cone of the first figure.

1. Write the formula to find the area of a cone.
2. Find the surface area of solid objects of first figure separately and find their sum.
3. Is the sum of area of solids of the first figure equal to the total surface area of combined solid of the second figure?
4. Is the sum of volume of solids of the first figure equal to the volume of combined solid of the second figure? Justify with reasons.
5. Compare the total cost of painting the solids of the first and second figures if the rate of cost of each is Rs. 150 per square meter.

Solution:

Here, diameter of the base (d) = 14 cm

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

Height of the cylinder (h) = 39 cm

Vertical height of the cone (h₁) = 24 cm

Slant height of the cone (l) = $\sqrt{h_1^2 + r^2} = \sqrt{(24)^2 + (7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm

(a) The formula to find the total surface area of a cone (A) = $\pi r l + \pi r^2 = \pi r (l + r)$

(b) Taking each solid of the first figure as a complete (closed) solid:

Total surface area of the cylinder = $2\pi r (h + r)$

$= 2 \times \frac{22}{7} \times 7 \, (39 + 7)$

$= 44 \times 46 = 2024 \text{ cm}^2$

Total surface area of the cone = $\pi r (l + r)$

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

$= 22 \times 32 = 704 \text{ cm}^2$

Sum of the surface areas of the two solids = 2024 + 704 = 2728 cm²

(c) Total surface area of the combined solid (second figure) = area of one circular end of cylinder + curved surface area of cylinder + curved surface area of cone

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

$= \frac{22}{7} \times (7)^2 + 2 \times \frac{22}{7} \times 7 \times 39 + \frac{22}{7} \times 7 \times 25$

$= 154 + 1716 + 550$

$= 2420 \text{ cm}^2$

No, the sum of the areas of the solids of the first figure (2728 cm²) is not equal to the total surface area of the combined solid (2420 cm²). When the two solids are joined, the flat circular end of the cylinder and the circular base of the cone come together inside the solid and are no longer part of the surface. So the combined surface area is smaller by $2\pi r^2 = 2 \times 154 = 308$ cm² (2728 − 2420 = 308 cm²).

(d) Volume of the cylinder = $\pi r^2 h$

$= \frac{22}{7} \times (7)^2 \times 39 = 154 \times 39 = 6006 \text{ cm}^3$

Volume of the cone = $\frac{1}{3}\pi r^2 h_1$

$= \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24 = 154 \times 8 = 1232 \text{ cm}^3$

Sum of the volumes of the two solids = 6006 + 1232 = 7238 cm³

Yes, the sum of the volumes of the solids of the first figure is equal to the volume of the combined solid of the second figure (7238 cm³ each). This is because combining the solids does not add or remove any material — the volume is conserved.

(e) Rate of painting = Rs. 150 per square unit.

Cost of painting the solids of the first figure = 150 × 2728 = Rs. 4,09,200

Cost of painting the combined solid of the second figure = 150 × 2420 = Rs. 3,63,000

Difference = 4,09,200 − 3,63,000 = Rs. 46,200. So painting the two separate solids of the first figure costs Rs. 46,200 more than painting the combined solid of the second figure (because of the extra 308 cm² of the two hidden circular faces).

Question 9

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

1. Write the formula to find the volume of a cone.
2. What is the slant height of the conical part?
3. A new toy is formed by exchanging the measurement of diameter and conical height. Which toy costs more to color its surface at the same rate?

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

Height of the conical part (h) = 12 cm

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

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

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

$= \sqrt{144 + 25}$

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

∴ The slant height of the conical part = 13 cm

(c) The colorable surface of a toy = area of the circular base of the cylinder + curved surface area of the cylinder + curved surface area of the cone = $\pi r^2 + 2\pi r H + \pi r l$

Original toy (diameter = 10 cm → r = 5 cm, cone height = 12 cm, l = 13 cm):

$S_1 = \pi r^2 + 2\pi r H + \pi r l$

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

$= 78.57 + 440 + 204.29$

$= 722.86 \text{ cm}^2$

New toy (diameter and conical height exchanged → diameter = 12 cm so r = 6 cm, cone height = 10 cm, cylinder length = 14 cm):

New slant height (l') = $\sqrt{(10)^2 + (6)^2} = \sqrt{100 + 36} = \sqrt{136} = 11.66$ cm

$S_2 = \pi r^2 + 2\pi r H + \pi r l'$

$= \frac{22}{7} \times (6)^2 + 2 \times \frac{22}{7} \times 6 \times 14 + \frac{22}{7} \times 6 \times 11.66$

$= 113.14 + 528 + 219.92$

$= 861.06 \text{ cm}^2$

Since $S_2 = 861.06 \text{ cm}^2 > S_1 = 722.86 \text{ cm}^2$, at the same rate of coloring the new toy (with the larger diameter of 12 cm) costs more to color its surface.

Difference in area = 861.06 − 722.86 = 138.2 cm²

Question 10

A solid object is formed with the combination of a cylinder and a cone with the same radius. The height of the cylinder and slant height of the cone are 28 cm and 17 cm respectively. The total cost of coloring the solid at the rate of Rs. 1.40 per square centimeter is Rs. 2851.20.

1. Draw a figure using the information given in the question.
2. What is the shape of the cylinder?
3. Find the height of the cone.

Solution:

Here, height of the cylinder (h) = 28 cm

Slant height of the cone (l) = 17 cm

Rate of coloring (R) = Rs. 1.40 per square cm

Total cost of coloring = Rs. 2851.20

(a)

(b) The shape of the base (cross-section) of the cylinder is a circle.

(c) First, we find the total surface area to be colored.

Total surface area (T.S.A) = $\frac{\text{Total cost}}{\text{Rate}}$

$= \frac{2851.20}{1.40}$

$= 2036.57 \text{ cm}^2$

The colored surface = area of the circular base of cylinder + curved surface area of cylinder + curved surface area of cone

$\text{or, } \pi r^2 + 2\pi r h + \pi r l = 2036.57$

$\text{or, } \pi r (r + 2h + l) = 2036.57$

$\text{or, } \frac{22}{7} \times r , (r + 56 + 17) = 2036.57$

$\text{or, } r (r + 73) = 648$

$\text{or, } r^2 + 73r - 648 = 0$

$\text{or, } (r - 8)(r + 81) = 0$

$\text{or, } r = 8 \text{ cm} ; (\text{taking the positive value})$

Now, height of the cone (H) = $\sqrt{l^2 - r^2}$

$= \sqrt{(17)^2 - (8)^2}$

$= \sqrt{289 - 64}$

$= \sqrt{225} = 15 \text{ cm}$

∴ The height of the cone = 15 cm

Note: The rate is taken as Rs. 1.40 per square cm. With a rate of Rs. 100 per square cm the total surface area would work out to only 28.5 cm², which is impossible for a cone of slant height 17 cm, so the rate "Rs. 100" appears to be a misprint.

Question 11

A water tank contains cylindrical and hemispherical parts. The total height of the tank is 20 m and the area of its base is 154 m². If the tank is filled with water at the rate of 45 paisa per liter,

1. How much water is contained in the volume of 1 m³?
2. What is the radius of its base?

Solution:

Here, total height of the tank = 20 m

Area of the base of the tank (A) = 154 m²

(a) The volume of 1 m³ contains 1000 liters of water.

(b) The base of the tank is circular, so area of base (A) = $\pi r^2$

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

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

$\text{or, } r^2 = \frac{154 \times 7}{22} = 49$

$\text{or, } r = \sqrt{49} = 7 \text{ m}$

∴ The radius of the base of the tank = 7 m