Mensuration

Mensuration is the branch of mathematics that deals with the measurement of geometric figures and their parameters like length, area, and volume.

Key Topics Covered:

1. Perimeter and Area:
   • Rectangle: Perimeter = 2(l + b), Area = l × b
   • Square: Perimeter = 4a, Area = a²
   • Triangle: Perimeter = a + b + c, Area = ½ × base × height
2. Area of Special Quadrilaterals:
   • Parallelogram: Area = base × height
   • Rhombus: Area = ½ × d1 × d2 (where d1 and d2 are diagonals)
   • Trapezium: Area = ½ × (sum of parallel sides) × height
3. Circle:
   • Circumference = 2πr
   • Area = πr²
4. Volume and Surface Area:
   • Cuboid: Volume = l × b × h, Surface Area = 2(lb + bh + hl)
   • Cube: Volume = a³, Surface Area = 6a²
   • Cylinder: Volume = πr²h, Surface Area = 2πr(h + r)
5. Conversions:
   • Understanding and converting between different units of measurement for length, area, and volume.

The chapter emphasizes problem-solving and application-based learning to calculate these parameters for various geometrical shapes, helping students develop a solid understanding of practical geometry.

Exercise 9.1

1. 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.

Ans : 

• The parallel sides of the trapezium measure 1 m and 1.2 m.
• The perpendicular distance between the parallel sides is 0.8 m.

Formula for the area of a trapezium:

Area of a trapezium = (1/2) * (sum of parallel sides) * (perpendicular distance between them)

Calculation:

Area = (1/2) * (1 + 1.2) * 0.8

     = 0.5 * 2.2 * 0.8

     = 0.88 square meters

Therefore, the area of the top surface of the table is 0.88 square meters.

2. 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 sides.

Ans :

• Area of a trapezium = (1/2) * (sum of parallel sides) * height

Given:

• Area = 34 cm²
• One parallel side = 10 cm
• Height = 4 cm

Substituting the values in the formula:

• 34 = (1/2) * (10 + b) * 4

Simplifying the equation:

• 34 = 2 * (10 + b)
• 17 = 10 + b
• b = 17 – 10
• b = 7 cm

3. 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.

Ans : 

Calculate the length of AB:
Since the fence is the perimeter of the field, we have:
AB + BC + CD + DA = 120

AB + 48 + 17 + 40 = 120

AB = 120 – 48 – 17 – 40

AB = 15 m

Calculate the area of the trapezium:
The area of a trapezium is given by:
Area = (1/2) * (sum of parallel sides) * (perpendicular distance between them)

In this case, the parallel sides are AD and BC, and the perpendicular distance between them is AB.
Area = (1/2) * (AD + BC) * AB

     = (1/2) * (40 + 48) * 15

     = 660 square meters

4. 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.

Ans : 

Divide the quadrilateral into two triangles:

The diagonal of the quadrilateral divides it into two triangles: triangle ABE and triangle CDE.

Calculate the area of each triangle:

Area of triangle ABE:

Area = (1/2) * base * height

     = (1/2) * 24 * 8

     = 96 square meters

Area of triangle CDE:

Area = (1/2) * base * height

     = (1/2) * 24 * 13

     = 156 square meters

Calculate the total area of the quadrilateral:

The total area of the quadrilateral is the sum of the areas of the two triangles:

Total area  

= Area of triangle ABE + Area of triangle CDE

 = 96 + 156

= 252 square meters

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

Ans : 

Formula for the area of a rhombus:

• Area of a rhombus = (1/2) * (product of diagonals)

Given:

• Diagonal 1 (d₁) = 7.5 cm
• Diagonal 2 (d₂) = 12 cm

Calculation:

• Area = (1/2) * d₁ * d₂ = (1/2) * 7.5 cm * 12 cm 
• = 45 cm²

6. 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.

Ans : 

Area of the Rhombus

Given: Side = 5 cm, Altitude = 4.8 cm

Area of a rhombus = Base * Height

Area = 5 cm * 4.8 cm 

= 24 cm²

Finding the Other Diagonal

Given: One diagonal (d1) 

= 8 cm, Area = 24 cm²

Formula for the area of a rhombus: 

Area = (1/2) * d1 * d2

where d1 and d2 are the diagonals.

Substituting the given values:

24 = (1/2) * 8 * d2

d2 = (24 * 2) / 8 = 6 cm

So, the area of the rhombus is 24 cm² and the length of the other diagonal is 6 cm.

7. 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.

Ans : 

Step 1: Find the area of one tile:

• Area of a rhombus = (1/2) * (product of diagonals)
  • Area of one tile = (1/2) * 45 cm * 30 cm = 675 cm²

Step 2: Find the total area of all tiles:

• Total area = Area of one tile * Number of tiles = 675 cm² * 3000 = 2025000 cm²

Step 3: Convert area to square meters:

• 1 m² = 10000 cm²
• Total area = 2025000 cm² / 10000 = 202.5 m²

Step 4: Calculate the total cost:

• Cost of polishing = Area * Rate per m² = 202.5 m² * Rs. 4/m² 
• = Rs. 810

8. 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.

Ans : 

Area of a trapezium = (1/2) * (sum of parallel sides) * (perpendicular distance between them)

10500 = (1/2) * (x + 2x) * 100

10500 = (1/2) * 3x * 100

10500 = 150x

x = 10500 / 150

x = 70 meters

So, the length of the side along the road is 70 meters, and the length of the side along the river is 2 * 70 

= 140 meters.

9. The 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.

Ans : 

Area of the octagonal surface = area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF

Area of trapezium ABCH = Area of trapezium GDEF

= 1/2 (a + b) × h

= 

1/2 (11 + 5) × 4

= 1/2 × 16 × 4

= 32 m2

Area of rectangle HCDG = l × b = 11 m × 5 m = 55 m2

Area of the octagonal surface = 32 m2 + 55 m2 + 32 m2 = 119 m2

Hence, the required area = 119 m2.

10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans : 

Jyoti’s Diagram:

• Jyoti divided the pentagon into two trapeziums.
• Each trapezium has parallel sides of 15m and 30m, and a height of 15m.
• Area of a trapezium = 1/2 * (sum of parallel sides) * height
• Area of one trapezium = 1/2 * (15 + 30) * 15 = 337.5 m²

Kavita’s Diagram:

• Kavita divided the pentagon into a square and a right-angled triangle.
• The square has a side of 15m.
• The triangle has a base of 15m and a height of 15m.
• Area of the square = side * side = 15m * 15m = 225 m²
• Area of the triangle = 1/2 * base * height = 1/2 * 15m * 15m = 112.5 m²
• Total area of the park = Area of square + Area of triangle = 225 m² + 112.5 m² = 337.5 m²

Therefore, the area of the pentagonal park is 337.5 square meters, regardless of the method used.

11. Diagram of the 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 the same.

Ans :

Exercise 9.2

1. There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?

Ans : 

(a) Volume of the cuboid = l × b × h 

= 60 × 40 × 50

 = 120000 cm3

(b) Volume of cube = (Side)3 = (50)3 = 50 × 50 × 50 = 125000 cm3

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

Ans : 

A suitcase is a cuboid.

Total Surface Area (TSA) of a cuboid = 2(lb + bh + hl)

TSA = 2(80 * 48 + 48 * 24 + 24 * 80) cm² = 2(3840 + 1152 + 1920) cm² = 13824 cm²

Calculate the area of tarpaulin required for one suitcase:

• Since the tarpaulin will cover all six faces of the suitcase, the area of tarpaulin required for one suitcase is equal to the total surface area of the suitcase.
• Area of tarpaulin for one suitcase = 13824 cm²

Calculate the total area of tarpaulin required for 100 suitcases:

• Total area = Area per suitcase * Number of suitcases = 13824 cm² * 100 = 1382400 cm²

Calculate the length of tarpaulin required:

• We know the width of the tarpaulin is 96 cm.
• Length of tarpaulin = Total area / Width = 1382400 cm² / 96 cm = 14400 cm = 144 m (converting cm to m)

Therefore, 144 meters of tarpaulin cloth is required to cover 100 such suitcases.

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

Ans : 

• Formula for surface area of a cube:
  • Surface area = 6a²
• Given: Surface area = 600 cm²
• Substituting the values in the formula:
  • 600 = 6a²
  • a² = 600 / 6
  • a² = 100
  • a = √100
  • a = 10 cm

Therefore, the side of the cube is 10 cm.

4. 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?

Ans : 

Given:

• Dimensions of the cabinet: 1 m (length) x 2 m (width) x 1.5 m (height)
• Rukhsar painted all except the bottom of the cabinet.

To find:

• The surface area Rukhsar painted.

Solution:

Calculate the total surface area of the cabinet:

Total Surface Area (TSA) = 2(lb + bh + hl)

TSA = 2(1 x 2 + 2 x 1.5 + 1.5 x 1)

    = 2(2 + 3 + 1.5)

    = 2 x 6.5

    = 13 m²

Calculate the surface area of the bottom of the cabinet:
The bottom of the cabinet is a rectangle with dimensions 1 m x 2 m.
Area of bottom = length x width = 1 x 2 = 2 m²

Calculate the surface area Rukhsar painted:
To find the surface area Rukhsar painted, we subtract the area of the bottom from the total surface area.
Painted area =

 Total surface area – Area of bottom

= 13 m² – 2 m²

= 11 m²

5. 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 the area is painted. How many cans of paint will she need to paint the room?

Ans : 

• Since we’re painting the walls and ceiling, we need to find the lateral surface area and the area of the ceiling.
• Lateral Surface Area (LSA) = 2 * height * (length + breadth) = 2 * 7 * (15 + 10) = 2 * 7 * 25 = 350 m²
• Area of ceiling = length * breadth = 15 * 10 = 150 m²
• Total area to be painted = LSA + Area of ceiling = 350 + 150 = 500 m²

Calculate the number of paint cans required:

• Each can covers 100 m².
• Number of cans = Total area to be painted / Area covered by one can = 500 m² / 100 m²/can = 5 cans

6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Ans : 

Similarities:

• Both figures have the same height: The height of both the cylinder and the cube is 7 cm.
• Both figures have the same width: The width of both the cylinder and the cube is 7 cm.

Differences:

• Shape: The first figure is a cylinder, while the second figure is a cube.
• Base: The cylinder has a circular base, while the cube has a square base.

Lateral Surface Area:

• Cylinder:
  In this case, r = 7 cm and h = 7 cm. 

 2 * π * 7 * 7 = 308 cm².  

• Cube:

The lateral surface area of a cube is given by the formula 4a², where a is the  side of the cube. In this case, a = 7 cm. 

4 * 7² = 196 cm².  

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

Ans : 

• TSA = 2πr(h + r) where:
• r = radius of the base
• h = height of the cylinder
• π = 22/7 (approximately)

Given values:

• Radius (r) = 7 m
• Height (h) = 3 m

Calculation:

• TSA = 2 * (22/7) * 7 (3 + 7) = 44 * 10 = 440 m²

8. 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.

Ans : 

h = 128 cm

l = 128 cm

b = 33 cm

Perimeter of the sheet = 2(l + b) = 2(128 + 33) = 2 × 161 = 322 cm

Hence, the required perimeter = 322 cm.

9. 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.

Ans : 

10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Ans : 

Solution:

1. Calculate the radius of the cylinder:
   Radius (r) = Diameter / 2 

= 14 cm / 2 

= 7 cm

2. Calculate the height of the label:
   Since the label is placed 2 cm from the top and bottom, the height of the label is:
   Height of label = Total height – 2 * 2 cm = 20 cm – 4 cm = 16 cm
3. Calculate the area of the label:
   The area of the label is the same as the lateral surface area of the cylinder with radius 7 cm and height 16 cm.
   Lateral Surface Area (LSA) of a cylinder = 2πrh
   Area of label = 2 * (22/7) * 7 cm * 16 cm = 704 cm²

Therefore, the area of the label is 704 cm².

Exercise 9.3

1. Given a cylindrical tank, in which situation will you find the 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.

Ans : 

(a) To find how much it can hold: We need to find the volume of the cylindrical tank. Volume measures the capacity or how much a container can hold.

(b) Number of cement bags required to plaster it: We need to find the surface area of the cylindrical tank. Surface area measures the total area of the outer surface of an object, which is what needs to be plastered.

(c) To find the number of smaller tanks that can be filled with water from it: We need to find the volume of the cylindrical tank to determine how much water it can hold, and then compare it to the volume of the smaller tanks to find out how many times it can be filled.

2. 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 by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Ans : 

• Volume = πr²h

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

Ans : 

Solution

• Formula for volume of a cuboid:
  • Volume = Base Area * Height
• Given:
  • Base area = 180 cm²
  • Volume = 900 cm³
• Substituting the values in the formula:
  • 900 cm³ = 180 cm² * Height
• Finding the height:
  • Height = 900 cm³ / 180 cm² = 5 cm

Therefore, the height of the cuboid is 5 cm.

4. 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?

Ans : 

Solution:

1. Find the number of cubes along each side of the cuboid:
   • Along the length: 60 cm / 6 cm/cube = 10 cubes
   • Along the breadth: 54 cm / 6 cm/cube = 9 cubes
   • Along the height: 30 cm / 6 cm/cube = 5 cubes
2. Find the total number of cubes:
   • Total number of cubes = number of cubes along length * number of cubes along breadth * number of cubes along height
   • Total number of cubes = 10 cubes * 9 cubes * 5 cubes = 450 cubes

Therefore, 450 small cubes with a side length of 6 cm can be placed in the given cuboid.

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

Ans : 

Given:

• Volume of the cylinder = 1.54 m³
• Diameter of the base = 140 cm = 1.4 m (converting to meters)

Formula for the volume of a cylinder:

• Volume = πr²h Where:
  • r is the radius of the base
  • h is the height of the cylinder
  • π is a constant, approximately equal to 22/7  

Calculations:

• Radius (r) = Diameter / 2 = 1.4 m / 2 = 0.7 m  

Substituting the values in the formula:

• 1.54 = (22/7) * (0.7)² * h  
• 1.54 = 1.54 * h
• h = 1.54 / 1.54
• h = 1 m

6. 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.

Ans :

Solution:

1. Identify the shape: The milk tank is in the shape of a cylinder.
2. Given information:
   • Radius (r) of the cylinder = 1.5 m
   • Length (height) (h) of the cylinder = 7 m
3. Formula for the volume of a cylinder:
   • Volume (V) = πr²h
4. Calculation:
   • Substituting the given values into the formula, we get: V = (22/7) * (1.5)² * 7 V = 49.5 m³
5. Conversion to liters:
   • 1 m³ = 1000 liters
   • So, 49.5 m³ = 49.5 * 1000 = 49500 liters

Therefore, the quantity of milk that can be stored in the tank is 49500 liters.

7. If each edge of a cube is doubled,

(i) how many times will it be surface area increase?

(ii) how many times will its volume increase?

Ans : 

(i) Change in Surface Area

• Original surface area = 6a² square units
• If each edge is doubled, the new edge becomes ‘2a’ units.
• New surface area = 6(2a)² = 6 * 4a² = 24a² square units
• The new surface area is 4 times the original surface area.

Therefore, the surface area increases by 4 times when the edge of a cube is doubled.

(ii) Change in Volume

• Original volume = a³ cubic units
• New volume = (2a)³ = 8a³ cubic units

Therefore, the volume increases by 8 times when the edge of a cube is doubled.

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

Ans : 

1. Convert units:
   • Volume of reservoir = 108 m³ = 108 * 1000 liters = 108000 liters (since 1 m³ = 1000 liters)
   • Water pouring rate = 60 liters/minute
2. Calculate the time to fill the reservoir:
   • Time = Total volume / Filling rate = 108000 liters / 60 liters/minute = 1800 minutes
3. Convert minutes to hours:
   • 1 hour = 60 minutes
   • Time in hours = 1800 minutes / 60 minutes/hour 
   • = 30 hours