Surface Area, Volume and Capacity

Q1: Multiple Choice Type:

i. The dimensions of a hall are 40m × 25m × 5m. The number of persons which can be accommodated in the hall is (each person requires 5 m³ of air):

Step 1: Find the volume of the hall: \[ \text{Volume} = \text{length} \times \text{breadth} \times \text{height} = 40 \times 25 \times 5 = 5000 \, \text{m}^3 \]Step 2: Each person requires 5 m³ of air.
Step 3: Find number of persons: \[ \text{Number of persons} = \frac{\text{Volume of hall}}{\text{Air required per person}} = \frac{5000}{5} = 1000 \]Answer: a. 1000

ii. The external dimensions of a closed rectangular box are 82 cm × 47 cm × 60 cm. If it is made of wood of 1 cm thickness, the internal dimensions of the box are:

Step 1: Thickness of wood = 1 cm.
Step 2: Internal length = External length − 2 × thickness = 82 − 2 × 1 = 80 cm
Internal breadth = 47 − 2 × 1 = 45 cm
Internal height = 60 − 2 × 1 = 58 cm
Answer: d. 80 cm × 45 cm × 58 cm

iii. The outer dimensions of a closed small box are 12 cm × 12 cm × 10 cm. If it is made of 1 cm thick walls, its capacity is:

Step 1: Find internal dimensions:
Internal length = 12 − 2 × 1 = 10 cm
Internal breadth = 12 − 2 × 1 = 10 cm
Internal height = 10 − 2 × 1 = 8 cm
Step 2: Calculate volume (capacity): \[ V = 10 \times 10 \times 8 = 800 \, \text{cm}^3 \]Answer: d. 800 cm³

iv. A room is 3 m long, 2 m broad and 2 m high. It has one door 2 m × 1 m; two windows each 1 m × 0.5 m; the remaining area of the walls is:

Step 1: Calculate total area of the four walls: \[ \text{Total wall area} = 2 \times \text{height} \times (\text{length} + \text{breadth}) = 2 \times 2 \times (3 + 2) = 4 \times 5 = 20 \, \text{m}^2 \]Step 2: Calculate area of door: \[ 2 \times 1 = 2 \, \text{m}^2 \]Step 3: Calculate area of two windows: \[ 2 \times (1 \times 0.5) = 2 \times 0.5 = 1 \, \text{m}^2 \]Step 4: Calculate remaining wall area: \[ \text{Remaining wall area} = 20 – (2 + 1) = 20 – 3 = 17 \, \text{m}^2 \]Answer: b. 17 m²

Q2: A room 5 m long, 4.5 m wide and 3.6 m high has one door 1.5 m by 2.4 m and two windows, each 1 m by 0.75 m. Find:

i. The area of its walls, excluding doors and windows.

Step 1: Calculate total wall area: \[ \text{Wall area} = 2 \times \text{height} \times ( \text{length} + \text{width} ) \\ = 2 \times 3.6 \times (5 + 4.5) = 7.2 \times 9.5 = 68.4 \, \text{m}^2 \]Step 2: Calculate area of door: \[ 1.5 \times 2.4 = 3.6 \, \text{m}^2 \]Step 3: Calculate area of two windows: \[ 2 \times (1 \times 0.75) = 2 \times 0.75 = 1.5 \, \text{m}^2 \]Step 4: Calculate wall area excluding doors and windows: \[ 68.4 – (3.6 + 1.5) = 68.4 – 5.1 = 63.3 \, \text{m}^2 \]Answer: The area of walls excluding doors and windows is 63.3 m².

ii. The cost of distempering its walls at the rate of ₹4.50 per m².

Step 5: Calculate cost: \[ \text{Cost} = 63.3 \times 4.50 = ₹284.85 \]Answer: The cost of distempering the walls is ₹284.85

iii. The cost of painting its roof at the rate of ₹9 per m².

Step 6: Calculate area of roof: \[ \text{Roof area} = \text{length} \times \text{width} = 5 \times 4.5 = 22.5 \, \text{m}^2 \]Step 7: Calculate cost of painting: \[ 22.5 \times 9 = ₹202.5 \]Answer: The cost of painting the roof is ₹202.5

Q3: The dining hall of a hotel is 75 m long, 60 m broad and 16 m high. It has five doors 4 m by 3 m each and four windows 3 m by 1.6 m each. Find the cost of:

i. Papering its walls at the rate of ₹12 per m²;

Step 1: Given dimensions of the dining hall:
Length \(l = 75\, m\)
Breadth \(b = 60\, m\)
Height \(h = 16\, m\)
Step 2: Find the area of four walls (lateral surface area): \[ \text{Area of four walls} = 2h(l + b) \\ = 2 \times 16 \times (75 + 60) \\ = 32 \times 135 = 4320\, m^2 \]Step 3: Area of doors:
Area of one door = \(4 \times 3 = 12\, m^2\)
Area of five doors = \(5 \times 12 = 60\, m^2\)
Step 4: Area of windows:
Area of one window = \(3 \times 1.6 = 4.8\, m^2\)
Area of four windows = \(4 \times 4.8 = 19.2\, m^2\)
Step 5: Net area to be papered: \[ = 4320 – (60 + 19.2) = 4320 – 79.2 = 4240.8\, m^2 \]Step 6: Cost of papering walls: \[ = 4240.8 \times 12 = ₹50889.6 \]Answer: Cost of papering the walls = ₹50,889.60

ii. Cost of carpeting the floor

Step 7: Area of the floor: \[ \text{Floor area} = l \times b = 75 \times 60 = 4500\, m^2 \]Step 8: Cost of carpeting: \[ = 4500 \times 25 = ₹112500 \]Answer: Cost of carpeting the floor = ₹1,12,500

Q4: Find the volume of wood required to make a closed box of external dimensions 80 cm, 75 cm and 60 cm, the thickness of walls of the box being 2 cm throughout.

Step 1: Let external dimensions be:
Length \(L = 80\) cm, Breadth \(B = 75\) cm, Height \(H = 60\) cm.
Step 2: Thickness of the walls \(t = 2\) cm.
Step 3: Calculate internal dimensions:
Internal length \(l = L – 2t = 80 – 2 \times 2 = 76\) cm
Internal breadth \(b = B – 2t = 75 – 2 \times 2 = 71\) cm
Internal height \(h = H – 2t = 60 – 2 \times 2 = 56\) cm
Step 4: Calculate external volume: \[ V_{\text{external}} = L \times B \times H = 80 \times 75 \times 60 = 360,000 \, \text{cm}^3 \]Step 5: Calculate internal volume: \[ V_{\text{internal}} = l \times b \times h = 76 \times 71 \times 56 \] Calculate: \[ 76 \times 71 = 5396 \\ 5396 \times 56 = 302,176 \, \text{cm}^3 \]Step 6: Calculate volume of wood used: \[ V_{\text{wood}} = V_{\text{external}} – V_{\text{internal}} = 360,000 – 302,176 = 57,824 \, \text{cm}^3 \]Answer: The volume of wood required is 57,824 cm³.

Q5: A closed box measures 66 cm, 36 cm and 21 cm from outside. If its walls are made of metal sheet, 0.5 cm thick, find:

i. The capacity of the box;

Step 1: Given external dimensions:
Length \(L = 66\) cm, Breadth \(B = 36\) cm, Height \(H = 21\) cm.
Thickness \(t = 0.5\) cm.
Step 2: Calculate internal dimensions: \[ l = L – 2t = 66 – 2 \times 0.5 = 66 – 1 = 65 \, \text{cm} \\ b = B – 2t = 36 – 1 = 35 \, \text{cm} \\ h = H – 2t = 21 – 1 = 20 \, \text{cm} \]Step 3: Calculate capacity (internal volume): \[ V_{\text{internal}} = l \times b \times h = 65 \times 35 \times 20 \] Calculate: \[ 65 \times 35 = 2275 \\ 2275 \times 20 = 45,500 \, \text{cm}^3 \]Answer: Capacity of the box = 45,500 cm³.

ii. Volume of metal sheet used

Step 4: Calculate external volume: \[ V_{\text{external}} = L \times B \times H = 66 \times 36 \times 21 \] Calculate: \[ 66 \times 36 = 2376 \\ 2376 \times 21 = 49,896 \, \text{cm}^3 \]Step 5: Volume of metal sheet used: \[ V_{\text{metal}} = V_{\text{external}} – V_{\text{internal}} = 49,896 – 45,500 = 4,396 \, \text{cm}^3 \]Answer: Volume of metal sheet used = 4,396 cm³.

iii. Weight of the box, if 1 cm³ of metal weighs 3.6 g.

Step 6: Calculate weight: \[ \text{Weight} = V_{\text{metal}} \times 3.6 = 4,396 \times 3.6 = 15,825.6 \, \text{g} \]Answer: Weight of the box = 15825.60 g.

Q6: The internal length, breadth and height of a closed box are 1 m, 80 cm and 25 cm respectively. If its sides are made of 2.5 cm thick wood, find:

i. The capacity of the box

Step 1: Convert all dimensions to the same unit (cm):
Length \(l = 1 \, \text{m} = 100 \, \text{cm}\)
Breadth \(b = 80 \, \text{cm}\)
Height \(h = 25 \, \text{cm}\)
Step 2: Calculate capacity (internal volume): \[ V_{\text{internal}} = l \times b \times h = 100 \times 80 \times 25 = 200,000 \, \text{cm}^3 \]Answer: Capacity of the box = 200,000 cm³.

ii. The volume of wood used to make the box

Step 3: Thickness of wood \(t = 2.5 \, \text{cm}\)
Step 4: Calculate external dimensions: \[ L = l + 2t = 100 + 2 \times 2.5 = 105 \, \text{cm} \\ B = b + 2t = 80 + 5 = 85 \, \text{cm} \\ H = h + 2t = 25 + 5 = 30 \, \text{cm} \]Step 5: Calculate external volume: \[ V_{\text{external}} = L \times B \times H = 105 \times 85 \times 30 \] Calculate: \[ 105 \times 85 = 8925 \\ 8925 \times 30 = 267,750 \, \text{cm}^3 \]Step 6: Calculate volume of wood used: \[ V_{\text{wood}} = V_{\text{external}} – V_{\text{internal}} = 267,750 – 200,000 = 67,750 \, \text{cm}^3 \]Answer: Volume of wood used = 67,750 cm³.

Q7: Find the area of metal sheet required to make an open tank of length = 10 m, breadth = 7.5 m and depth = 3.8 m.

Step 1: Since the tank is open at the top, the surface area required is the sum of the bottom area and the four sides.
Step 2: Calculate the area of the bottom: \[ \text{Area}_{\text{bottom}} = \text{length} \times \text{breadth} = 10 \times 7.5 = 75 \, \text{m}^2 \]Step 3: Calculate the area of the two longer sides (length × depth): \[ 2 \times (10 \times 3.8) = 2 \times 38 = 76 \, \text{m}^2 \]Step 4: Calculate the area of the two shorter sides (breadth × depth): \[ 2 \times (7.5 \times 3.8) = 2 \times 28.5 = 57 \, \text{m}^2 \]Step 5: Calculate total area of metal sheet required: \[ \text{Total area} = 75 + 76 + 57 = 208 \, \text{m}^2 \]Answer: The area of metal sheet required is 208 m².

Q8: A tank 30 m long, 24 m wide and 4.5 m deep is to be made. It is open from the top. Find the cost of iron sheet required, at the rate of ₹65 per m², to make the tank.

Step 1: Since the tank is open at the top, surface area required is sum of bottom area and four sides.
Step 2: Calculate the area of the bottom: \[ \text{Area}_{\text{bottom}} = \text{length} \times \text{width} = 30 \times 24 = 720 \, \text{m}^2 \]Step 3: Calculate the area of two longer sides (length × depth): \[ 2 \times (30 \times 4.5) = 2 \times 135 = 270 \, \text{m}^2 \]Step 4: Calculate the area of two shorter sides (width × depth): \[ 2 \times (24 \times 4.5) = 2 \times 108 = 216 \, \text{m}^2 \]Step 5: Calculate total area of iron sheet required: \[ \text{Total area} = 720 + 270 + 216 = 1206 \, \text{m}^2 \]Step 6: Calculate cost of iron sheet: \[ \text{Cost} = 1206 \times 65 = ₹78,390 \]Answer: The cost of iron sheet required to make the tank is ₹78,390.

Q9: The edges of three solid cubes are 6 cm, 8 cm and 10 cm. These cubes are melted and recasted into a single cube. Find the edge of the resulting cube.

Step 1: Calculate volumes of the three cubes: \[ V_1 = 6^3 = 216 \, \text{cm}^3 \\ V_2 = 8^3 = 512 \, \text{cm}^3 \\ V_3 = 10^3 = 1000 \, \text{cm}^3 \]Step 2: Total volume of the resulting cube: \[ V = V_1 + V_2 + V_3 = 216 + 512 + 1000 = 1728 \, \text{cm}^3 \]Step 3: Let the edge of the resulting cube be \(a\). Then, \[ a^3 = 1728 \]Step 4: Calculate the edge length: \[ a = \sqrt[3]{1728} = 12 \, \text{cm} \]Answer: The edge of the resulting cube is 12 cm.

Q10: The ratio between the lengths of the edges of two cubes is 3 : 2. Find the ratio between their:

i. Total surface area

Step 1: Let the edges of the two cubes be \(3x\) and \(2x\).
Step 2: Total surface area of a cube = \(6 \times (\text{edge})^2\).
Calculate total surface areas: \[ S_1 = 6 \times (3x)^2 = 6 \times 9x^2 = 54x^2 \\ S_2 = 6 \times (2x)^2 = 6 \times 4x^2 = 24x^2 \]Step 3: Ratio of total surface areas: \[ \frac{S_1}{S_2} = \frac{54x^2}{24x^2} = \frac{54}{24} = \frac{9}{4} \]Answer: Ratio of total surface areas = 9 : 4.

ii. Volume

Step 4: Volume of a cube = \((\text{edge})^3\).
Calculate volumes: \[ V_1 = (3x)^3 = 27x^3 \\ V_2 = (2x)^3 = 8x^3 \]Step 5: Ratio of volumes: \[ \frac{V_1}{V_2} = \frac{27x^3}{8x^3} = \frac{27}{8} \]Answer: Ratio of volumes = 27 : 8.

Q11: The length, breadth and height of a cuboid (rectangular solid) are 4 : 3 : 2.

i. If its surface area is 2548 cm², find its volume.

Step 1: Let the common ratio be \(x\). Then,
Length = \(4x\), Breadth = \(3x\), Height = \(2x\).
Step 2: Surface area of cuboid: \[ S = 2(lb + bh + hl) = 2[(4x \times 3x) + (3x \times 2x) + (2x \times 4x)] = 2[12x^2 + 6x^2 + 8x^2] = 2 \times 26x^2 = 52x^2 \]Step 3: Given surface area \(S = 2548\) cm², so \[ 52x^2 = 2548 \\ x^2 = \frac{2548}{52} = 49 \\ x = 7 \]Step 4: Calculate volume \(V\): \[ V = l \times b \times h = 4x \times 3x \times 2x = 24x^3 = 24 \times 7^3 = 24 \times 343 = 8232 \, \text{cm}^3 \]Answer: Volume of the cuboid = 8232 cm³.

ii. If its volume is 3000 m³, find its surface area.

Step 5: Given volume \(V = 3000 \, \text{m}^3 = 3000 \times 10^6 \, \text{cm}^3 = 3 \times 10^9 \, \text{cm}^3\) (since 1 m = 100 cm, \(1 m^3 = 10^6 cm^3\))
Step 6: Volume in terms of \(x\): \[ 24x^3 = 3 \times 10^9 \\ x^3 = \frac{3 \times 10^9}{24} = 1.25 \times 10^8 \\ x = \sqrt[3]{1.25 \times 10^8} = \sqrt[3]{125 \times 10^6} = 5 \times 100 = 500 \, \text{cm} \]Step 7: Calculate surface area: \[ S = 52x^2 = 52 \times (500)^2 = 52 \times 250,000 = 13,000,000 \, \text{cm}^2 \] Convert to \(m^2\): \[ 13,000,000 \, \text{cm}^2 = \frac{13,000,000}{10,000} = 1300 \, \text{m}^2 \]Answer: Surface area of the cuboid = 1300 m².