Class 8 Maths - Mensuration NCERT Solutions

Chapter 11: Mensuration (NCERT Solutions)

Exercise 11.1

Q1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
(Square side = 60 m, Rectangle length = 80 m)

Perimeter of square = 4 × side = 4 × 60 = 240 m.
Perimeter of rectangle = 2(length + breadth) = 2(80 + b) = 160 + 2b.
Given, both have the same perimeter:
160 + 2b = 240
2b = 240 - 160 = 80
b = 40 m.
Area of square = side × side = 60 × 60 = 3600 m².
Area of rectangle = length × breadth = 80 × 40 = 3200 m².
Thus, the square field has a larger area.

Q2. Mrs. Kaushik has a square plot with the measurement as shown in the figure (Outer square 25m, inner house 15m × 20m). She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Area of the square plot = side × side = 25 × 25 = 625 m².
Area of the house (rectangle) = length × breadth = 20 × 15 = 300 m².
Area of the garden = Area of plot - Area of house
= 625 - 300 = 325 m².
Cost of developing the garden = Area × Rate
= 325 × 55 = ₹ 17875.

Q3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 - (3.5 + 3.5) metres, width is 7m].

Diameter of semi-circle = 7 m ⇒ Radius (r) = 3.5 m.
Length of rectangular part = 20 - (3.5 + 3.5) = 20 - 7 = 13 m.
Area:
Area of rectangular part = 13 × 7 = 91 m².
Area of 2 semi-circles = 1 full circle = πr² = (22/7) × 3.5 × 3.5 = 38.5 m².
Total Area = 91 + 38.5 = 129.5 m².

Perimeter:
Perimeter includes two straight lengths and curves of two semi-circles.
Length of straight edges = 13 + 13 = 26 m.
Circumference of 2 semi-circles = 1 full circle = 2πr = 2 × (22/7) × 3.5 = 22 m.
Total Perimeter = 26 + 22 = 48 m.

Q4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners).

Area of one tile (parallelogram) = base × height = 24 cm × 10 cm = 240 cm².
Area of floor = 1080 m².
Convert m² to cm²: 1080 × 10000 = 10800000 cm².
Number of tiles required = (Area of floor) / (Area of one tile)
= 10800000 / 240
= 1080000 / 24 = 45000 tiles.

Q5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr.
(a) Semi-circle (r=1.4 cm, diameter straight edge 2.8cm)
(b) Semi-circle on a rectangle (height 1.5cm, width 2.8cm)
(c) Cone/Sector shape (straight edges 2cm, curved edge over 2.8cm diameter)

Radius (r) = 2.8 / 2 = 1.4 cm for all circular parts.

(a) Perimeter = Circumference of semi-circle + diameter
= πr + d = (22/7 × 1.4) + 2.8 = 4.4 + 2.8 = 7.2 cm.

(b) Perimeter = Circumference of semi-circle + sum of 3 straight sides
= πr + 1.5 + 2.8 + 1.5 = 4.4 + 5.8 = 10.2 cm.

(c) Perimeter = Circumference of semi-circle + 2 straight sides
= πr + 2 + 2 = 4.4 + 4 = 8.4 cm.

Therefore, the ant would have to take a longer round for food piece (b).

Exercise 11.2

Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Parallel sides: a = 1 m, b = 1.2 m.
Height: h = 0.8 m.
Area of trapezium = ½ × (a + b) × h
= ½ × (1 + 1.2) × 0.8
= ½ × 2.2 × 0.8
= 1.1 × 0.8 = 0.88 m².

Q2. The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Area = 34 cm². a = 10 cm, h = 4 cm. Let other side be b.
Area = ½ × (a + b) × h
34 = ½ × (10 + b) × 4
34 = 2 × (10 + b)
17 = 10 + b
b = 17 - 10 = 7 cm.

Q3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Perimeter = AB + BC + CD + AD = 120 m.
AB + 48 + 17 + 40 = 120
AB + 105 = 120
AB = 120 - 105 = 15 m.
Since AB is perpendicular to parallel sides AD and BC, AB is the height (h = 15 m).
Parallel sides: AD (a) = 40 m, BC (b) = 48 m.
Area = ½ × (a + b) × h
= ½ × (40 + 48) × 15
= ½ × 88 × 15 = 44 × 15 = 660 m².

Q4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Diagonal (d) = 24 m. Perpendiculars (h₁) = 8 m, (h₂) = 13 m.
Area of quadrilateral = ½ × d × (h₁ + h₂)
= ½ × 24 × (8 + 13)
= 12 × 21 = 252 m².

Q5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

d₁ = 7.5 cm, d₂ = 12 cm.
Area of rhombus = ½ × d₁ × d₂
= ½ × 7.5 × 12
= 7.5 × 6 = 45 cm².

Q6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Rhombus is a special parallelogram. Area = base × height (altitude).
Area = 5 × 4.8 = 24 cm².
We also know Area = ½ × d₁ × d₂.
24 = ½ × 8 × d₂
24 = 4 × d₂
d₂ = 24 / 4 = 6 cm.

Q7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4.

Area of one tile (rhombus) = ½ × d₁ × d₂
= ½ × 45 × 30 = 45 × 15 = 675 cm².
Area of floor = 3000 × 675 = 2025000 cm².
Convert to m²: 2025000 / 10000 = 202.5 m².
Cost of polishing = Area × Rate
= 202.5 × 4 = ₹ 810.

Q8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Let the side along the road be x meters.
Then, side along the river = 2x meters.
Height (h) = 100 m. Area = 10500 m².
Area = ½ × (a + b) × h
10500 = ½ × (x + 2x) × 100
10500 = 50 × (3x)
150x = 10500
x = 10500 / 150 = 70 m.
Side along the river = 2x = 2 × 70 = 140 m.

Q9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface. (Side=5m, distance between parallel opposite sides = 11m, height of trapezium part = 4m).

The regular octagon can be split into two identical trapeziums and one central rectangle.
Trapezium: Parallel sides = 11m and 5m (since regular octagon side is 5m), Height = 4m.
Area of one trapezium = ½ × (11 + 5) × 4 = ½ × 16 × 4 = 32 m².
Area of two trapeziums = 32 × 2 = 64 m².
Rectangle: Length = 11m, Breadth = 5m.
Area of rectangle = 11 × 5 = 55 m².
Total Area = 64 + 55 = 119 m².

Q10. There is a pentagonal shaped park. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways.
(Jyoti splits it vertically into two trapeziums. Kavita splits it horizontally into a triangle and a square). (Square side 15m, total height 30m).

Jyoti's way (Two trapeziums):
Parallel sides of one trapezium = 15m and 30m. Height = 15/2 = 7.5m.
Area of one trapezium = ½ × (15 + 30) × 7.5 = ½ × 45 × 7.5 = 168.75 m².
Total Area = 2 × 168.75 = 337.5 m².

Kavita's way (Triangle + Square):
Area of square part = 15 × 15 = 225 m².
Area of triangle part = ½ × base × height = ½ × 15 × (30 - 15) = ½ × 15 × 15 = 112.5 m².
Total Area = 225 + 112.5 = 337.5 m².

Q11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Width of each section = (24 - 16)/2 = 8/2 = 4 cm (or (28-20)/2 = 4 cm).
The frame has 4 sections, opposite pairs are equal in area. They are trapeziums.
Two vertical sections: Parallel sides = 28 cm and 20 cm, height = 4 cm.
Area of each = ½ × (28 + 20) × 4 = ½ × 48 × 4 = 96 cm².
Two horizontal sections: Parallel sides = 24 cm and 16 cm, height = 4 cm.
Area of each = ½ × (24 + 16) × 4 = ½ × 40 × 4 = 80 cm².

Exercise 11.3

Q1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
(a) Cuboid: l=60, b=40, h=50 cm.
(b) Cube: l=50, b=50, h=50 cm.

Amount of material = Total Surface Area (TSA).
(a) Cuboid: TSA = 2(lb + bh + hl)
= 2(60×40 + 40×50 + 50×60)
= 2(2400 + 2000 + 3000)
= 2(7400) = 14800 cm².
(b) Cube: TSA = 6a²
= 6(50×50) = 6(2500) = 15000 cm².
The cuboidal box (a) requires lesser amount of material.

Q2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

TSA of 1 suitcase = 2(lb + bh + hl)
= 2(80×48 + 48×24 + 24×80)
= 2(3840 + 1152 + 1920)
= 2(6912) = 13824 cm².
TSA of 100 suitcases = 13824 × 100 = 1382400 cm².
Area of tarpaulin cloth required = Area of suitcases.
Length × Width = 1382400
Length × 96 = 1382400
Length = 1382400 / 96 = 14400 cm.
Convert cm to meters = 14400 / 100 = 144 meters.

Q3. Find the side of a cube whose surface area is 600 cm².

TSA of cube = 6a²
6a² = 600
a² = 100
a = √100 = 10 cm.

Q4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.

l = 2 m, b = 1 m, h = 1.5 m. (assuming length is largest base dimension)
Area painted = Area of 4 walls + Area of top
= 2h(l + b) + lb
= 2 × 1.5 × (2 + 1) + (2 × 1)
= 3(3) + 2
= 9 + 2 = 11 m².

Q5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will she need to paint the room?

l = 15 m, b = 10 m, h = 7 m.
Area to be painted = Area of 4 walls + Area of ceiling
= 2h(l + b) + lb
= 2 × 7 × (15 + 10) + (15 × 10)
= 14(25) + 150 = 350 + 150 = 500 m².
Number of cans needed = Total Area / Area per can
= 500 / 100 = 5 cans.

Q6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
(Cylinder: d=7cm, h=7cm. Cube: side=7cm).

Alike: They both have the same height of 7 cm.
Different: One is a cylinder and the other is a cube.

LSA of Cylinder = 2πrh = 2 × (22/7) × 3.5 × 7 = 44 × 3.5 = 154 cm².
LSA of Cube = 4a² = 4 × 7² = 4 × 49 = 196 cm².
The cube has a larger lateral surface area.

Q7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

TSA of cylinder = 2πr(r + h)
= 2 × (22/7) × 7 × (7 + 3)
= 44 × 10 = 440 m².

Q8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet.

LSA = Area of rectangular sheet = 4224 cm².
Width (b) = 33 cm.
Area = l × b ⇒ l × 33 = 4224 ⇒ l = 4224 / 33 = 128 cm.
Perimeter = 2(l + b) = 2(128 + 33) = 2(161) = 322 cm.

Q9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Diameter = 84 cm ⇒ r = 42 cm = 0.42 m.
Height/Length (h) = 1 m.
In one revolution, the roller covers an area equal to its Curved Surface Area (CSA).
CSA = 2πrh = 2 × (22/7) × 0.42 × 1 = 44 × 0.06 = 2.64 m².
Area of road = 750 revolutions × CSA
= 750 × 2.64 = 1980 m².

Q10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container. If the label is placed 2 cm from top and bottom, what is the area of the label.

Diameter = 14 cm ⇒ r = 7 cm.
Height of the container = 20 cm.
Height of the label (h) = 20 - 2 (top) - 2 (bottom) = 16 cm.
The label forms a cylinder. Area corresponding to label = CSA of this cylindrical part.
CSA = 2πrh = 2 × (22/7) × 7 × 16
= 44 × 16 = 704 cm².

Exercise 11.4

Q1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

(a) We need to find the Volume to know its capacity.
(b) We need to find the Surface Area to plaster its surfaces.
(c) We need to find the Volume because it deals with capacity/quantity of water.

Q2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it.

Volume of cylinder = πr²h. The variable 'r' is squared. Cylinder B has a larger radius (7 cm compared to 3.5 cm), so its volume should be greater.
Verification:
Volume of A = (22/7) × 3.5 × 3.5 × 14 = (22/7) × 12.25 × 14 = 539 cm³.
Volume of B = (22/7) × 7 × 7 × 7 = 22 × 49 = 1078 cm³.
Yes, Volume of B is greater.

Q3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?

Volume = Base Area × Height
900 = 180 × Height
Height = 900 / 180 = 5 cm.

Q4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Number of cubes = (Volume of cuboid) / (Volume of one small cube)
= (60 × 54 × 30) / (6 × 6 × 6)
= 10 × 9 × 5 = 450 cubes.

Q5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?

r = 140/2 = 70 cm = 0.7 m.
Volume = πr²h
1.54 = (22/7) × 0.7 × 0.7 × h
1.54 = 1.54 × h
h = 1.54 / 1.54 = 1 m.

Q6. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?

Volume = πr²h = (22/7) × 1.5 × 1.5 × 7
= 22 × 2.25 = 49.5 m³.
1 m³ = 1000 Litres.
Capacity = 49.5 × 1000 = 49500 Litres.

Q7. If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?

Let initial edge be a.
(i) Initial SA = 6a². New edge = 2a. New SA = 6(2a)² = 24a² = 4(6a²). Surface area increases 4 times.
(ii) Initial Volume = a³. New Volume = (2a)³ = 8a³ = 8(a³). Volume increases 8 times.

Q8. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.

Volume in Litres = 108 × 1000 = 108000 L.
Time taken in minutes = 108000 / 60 = 1800 minutes.
Time taken in hours = 1800 / 60 = 30 hours.

Class 8 Maths - Mensuration Practice Questions

Chapter 11: Mensuration (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 8 cm.

Parallel sides, a = 12 cm, b = 20 cm.
Height, h = 8 cm.
Area of trapezium = ½ × (a + b) × h
= ½ × (12 + 20) × 8
= ½ × 32 × 8
= 16 × 8 = 128 cm².

Q2. The area of a rhombus is 240 cm² and one of the diagonals is 16 cm. Find the other diagonal.

Area = 240 cm², d₁ = 16 cm.
Area = ½ × d₁ × d₂
240 = ½ × 16 × d₂
240 = 8 × d₂
d₂ = 240 / 8 = 30 cm.

Q3. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Radius (r) = 7 m, Height (h) = 3 m.
Since the tank is closed, we need Total Surface Area (TSA).
TSA = 2πr(r + h)
= 2 × (22/7) × 7 × (7 + 3)
= 2 × 22 × 10
= 44 × 10 = 440 m².
Therefore, 440 m² of metal sheet is required.

Q4. Find the volume of a cube whose surface area is 150 cm².

Total Surface Area of cube = 6a²
6a² = 150
a² = 150 / 6 = 25
a = √25 = 5 cm (side of the cube).
Volume of cube = a³ = 5³ = 5 × 5 × 5 = 125 cm³.

Q5. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet?

LSA of cylinder = Area of rectangular sheet = 4224 cm².
Width of sheet (b) = 33 cm.
Area = l × b ⇒ 4224 = l × 33
l = 4224 / 33 = 128 cm.
Perimeter of rectangular sheet = 2(l + b)
= 2(128 + 33) = 2(161) = 322 cm.

Q6. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.

Volume of cuboid = Base Area × Height
900 = 180 × Height
Height = 900 / 180 = 5 cm.

Q7. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Volume of cuboid = 60 × 54 × 30 cm³.
Volume of one small cube = 6 × 6 × 6 cm³.
Number of small cubes = (Volume of cuboid) / (Volume of one small cube)
= (60 × 54 × 30) / (6 × 6 × 6)
= 10 × 9 × 5 = 450 cubes.

Q8. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm.

Volume = 1.54 m³.
Diameter = 140 cm = 1.4 m. So, Radius (r) = 1.4 / 2 = 0.7 m.
Volume of cylinder = πr²h
1.54 = (22/7) × 0.7 × 0.7 × h
1.54 = 22 × 0.1 × 0.7 × h
1.54 = 1.54 × h
h = 1.54 / 1.54 = 1 m.

Q9. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?

Radius (r) = 1.5 m, Height/Length (h) = 7 m.
Volume = πr²h
= (22/7) × 1.5 × 1.5 × 7
= 22 × 2.25 = 49.5 m³.
Capacity in Litres = 49.5 × 1000 (since 1 m³ = 1000 L)
= 49500 Litres.

Q10. If each edge of a cube is doubled, (i) how many times will its surface area increase? (ii) how many times will its volume increase?

Let original edge be 'a'.
New edge = 2a.
(i) Original Surface Area = 6a².
New Surface Area = 6(2a)² = 6(4a²) = 24a² = 4(6a²).
Surface area increases 4 times.
(ii) Original Volume = a³.
New Volume = (2a)³ = 8a³.
Volume increases 8 times.

Q11. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.

Volume of reservoir = 108 m³ = 108 × 1000 = 108000 Litres.
Rate of flow = 60 Litres/minute.
Time taken in minutes = 108000 / 60 = 1800 minutes.
Time taken in hours = 1800 / 60 = 30 hours.

Q12. The area of a quadrilateral is 252 cm² and the perpendiculars dropped on it from the opposite vertices are 7 cm and 11 cm. Find the length of its diagonal.

Area = ½ × d × (h₁ + h₂)
252 = ½ × d × (7 + 11)
252 = ½ × d × 18
252 = 9d
d = 252 / 9 = 28 cm.

Q13. Two cubes of side 5 cm are joined end to end. Find the surface area of the resulting cuboid.

When joined end to end, the length becomes 5 + 5 = 10 cm.
Breadth and height remain the same = 5 cm.
TSA of cuboid = 2(lb + bh + hl)
= 2(10×5 + 5×5 + 5×10)
= 2(50 + 25 + 50) = 2(125) = 250 cm².

Q14. The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the total cost of whitewashing all four walls of a room if the cost of white washing is ₹ 5 per m².

Area of four walls (LSA) = 2h(l + b)
= 2 × 4 × (12 + 8)
= 8 × 20 = 160 m².
Cost = Area × Rate = 160 × 5 = ₹ 800.

Q15. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold?

Circumference = 2πr = 132
2 × (22/7) × r = 132
(44/7)r = 132 ⇒ r = (132 × 7) / 44 = 3 × 7 = 21 cm.
Volume = πr²h = (22/7) × 21 × 21 × 25
= 22 × 3 × 21 × 25 = 34650 cm³.
Capacity in Litres = 34650 / 1000 = 34.65 Litres.

Q16. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Area of rhombus = ½ × d₁ × d₂
= ½ × 7.5 × 12
= 7.5 × 6 = 45 cm².

Q17. The shape of a garden is rectangular in the middle and semi circular at the ends (total length 20m, width 7m). Find the area of this garden.

Width = diameter of semi-circles = 7 m. Radius = 3.5 m.
Length of rectangle = Total length - 2(radius) = 20 - (3.5 + 3.5) = 20 - 7 = 13 m.
Area of rectangle = 13 × 7 = 91 m².
Area of 2 semi-circles = Area of 1 full circle = πr² = (22/7) × 3.5 × 3.5 = 22 × 0.5 × 3.5 = 38.5 m².
Total Area = 91 + 38.5 = 129.5 m².

Q18. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.

Perimeter = 2(l + b) = 130
2(l + 30) = 130
l + 30 = 65
l = 65 - 30 = 35 cm.
Area = l × b = 35 × 30 = 1050 cm².

Q19. Find the lateral surface area of a cube of side 8 cm.

Lateral Surface Area = 4a²
= 4 × (8)²
= 4 × 64 = 256 cm².

Q20. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

CSA = 2πrh = 88
2 × (22/7) × r × 14 = 88
44 × r × 2 = 88
88 × r = 88
r = 1 cm.
Diameter = 2r = 2(1) = 2 cm.

Class 8 Maths - Mensuration Summary

Chapter 11: Mensuration (Concepts & Formulas)

1. Plane Figures (2D Shapes)

Perimeter is the distance around a closed figure. Area is the region occupied by the closed figure.

Rectangle: Area = l × b ; Perimeter = 2(l + b)
Square: Area = a × a = a² ; Perimeter = 4a
Triangle: Area = ½ × base × height
Parallelogram: Area = base × height ; Perimeter = 2(sum of adjacent sides)
Circle: Area = πr² ; Circumference = 2πr (where π ≈ 22/7 or 3.14)

2. Area of Trapezium and General Quadrilateral

Trapezium: Area = ½ × (sum of parallel sides) × (perpendicular distance between them)
Area = ½ × (a + b) × h
General Quadrilateral: Area can be found by splitting it into two triangles by drawing a diagonal.
Area = ½ × d × (h₁ + h₂), where 'd' is the diagonal and 'h₁', 'h₂' are perpendiculars drawn on it from opposite vertices.
Rhombus: Area = ½ × d₁ × d₂ (half the product of its diagonals).

3. Surface Area of Solid Figures (3D Shapes)

Total Surface Area (TSA) is the sum of the areas of all faces. Lateral/Curved Surface Area (LSA/CSA) is the area of the side faces (excluding top and bottom).

Cuboid (length l, breadth b, height h):
TSA = 2(lb + bh + hl)
LSA = 2h(l + b)
Cube (side a):
TSA = 6a²
LSA = 4a²
Right Circular Cylinder (radius r, height h):
TSA = 2πr(r + h)
CSA = 2πrh

4. Volume and Capacity

Volume is the amount of space occupied by a 3D object. Capacity is the maximum amount that a container can hold.

Volume of Cuboid = l × b × h (or Area of base × height)
Volume of Cube = a × a × a = a³
Volume of Cylinder = πr²h (Area of base × height)

Important Conversions:

1 cm³ = 1 mL
1 L = 1000 cm³
1 m³ = 1000000 cm³ = 1000 L

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