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Arithmetic - Mensuration

July 11, 2026
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Arithmetic - Chapter 10: Mensuration

Mensuration

Mensuration
Figure 1: Mensuration deals with the measurement of area, perimeter, and volume of geometric figures.

1. 2D Figures

TRIANGLE

  • Perimeter: a + b + c = 2s (s = semi-perimeter)
  • Area: ½ × base × height = √[s(s-a)(s-b)(s-c)]
  • Right-angled triangle: p = √(b² + h²)
  • Equilateral triangle: Area = (√3/4) × a², Height = (√3/2) × a
QUADRILATERALS

  • Rectangle: Perimeter = 2(a+b), Area = a×b
  • Square: Perimeter = 4a, Area = a², Diagonal = a√2
  • Parallelogram: Area = base × height
  • Rhombus: Area = ½ × d₁ × d₂ (d₁, d₂ = diagonals)
  • Trapezium: Area = ½ × (sum of parallel sides) × height
CIRCLE

  • Circumference: 2πr = πd
  • Area: πr²
  • Sector Area: (θ/360) × πr²
  • Length of arc: (θ/360) × 2πr

2. 3D Figures

CUBOID & CUBE

  • Cuboid: Surface Area = 2(lb + bh + hl), Volume = l×b×h
  • Cube: Surface Area = 6a², Volume = a³, Diagonal = a√3
CYLINDER

  • Curved Surface Area: 2πrh
  • Total Surface Area: 2πr(r + h)
  • Volume: πr²h
CONE

  • Slant Height (l): √(r² + h²)
  • Curved Surface Area: πrl
  • Total Surface Area: πr(r + l)
  • Volume: ⅓πr²h
SPHERE & HEMISPHERE

  • Sphere: Surface Area = 4πr², Volume = 4/3πr³
  • Hemisphere: Curved Surface Area = 2πr², Total Surface Area = 3πr², Volume = 2/3πr³
FRUSTUM OF A CONE

  • Slant Height (l): √[h² + (r₁ - r₂)²]
  • Curved Surface Area: πl(r₁ + r₂)
  • Total Surface Area: πl(r₁ + r₂) + πr₁² + πr₂²
  • Volume: ⅓πh(r₁² + r₂² + r₁r₂)
IMPORTANT RELATIONS

  • If radius of a circle is diminished by r%, area diminishes by (2r - r²/100)%
  • If the side of a square is increased by x%, area increases by (2x + x²/100)%

3. Exercise (Questions 1-10)

Q1. The length and breadth of a rectangle are in the ratio 9:5. If its area is 720 m², find its perimeter.

  • (a) 112 metre
  • (b) 115 metre
  • (c) 110 metre
  • (d) 118 metre
Answer: (a) 9x × 5x = 720 ⇒ x=4. Length=36, Breadth=20. Perimeter=2(36+20)=112 m

Q2. How many squares are there in a 5 inch by 5 inch square grid, if the grid is made up one inch by one inch squares?

  • (a) 50
  • (b) 150
  • (c) 55
  • (d) 25
Answer: (d) Total squares = 5²/1² = 25

Q3. A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle?

  • (a) 88 cm²
  • (b) 154 cm²
  • (c) 1250 cm²
  • (d) 616 cm²
Answer: (d) Perimeter = 2(18+26)=88 cm. 2πr=88 ⇒ r=14 cm. Area = π×14² = 616 cm²

Q4. When the circumference and area of a circle are numerically equal, then the diameter is numerically equal to:

  • (a) area
  • (b) circumference
  • (c) 4
  • (d) 2π
Answer: (c) 2πr = πr² ⇒ r=2 ⇒ diameter=4

Q5. A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze?

  • (a) 154 cm²
  • (b) 308 m²
  • (c) 150 m²
  • (d) None of these
Answer: (a) Area = (1/4)π×14² = 154 m²

Q6. From a square piece of paper having each side equal to 10 cm, the largest possible circle is cut out. The ratio of the area of the circle to the area of the original square is nearly:

  • (a) 4
  • (b) 3/5
  • (c) 5/6
  • (d) 6/7
Answer: (a) Area of circle = π×5² = 78.5. Area of square = 100. Ratio ≈ 0.785 ≈ 4/5

Q7. If the area of a circle decreases by 36%, then the radius of a circle decreases by:

  • (a) 20%
  • (b) 18%
  • (c) 36%
  • (d) 64%
Answer: (a) 100 - 10√(100-36) = 100 - 80 = 20%

Q8. A typist uses a paper 12" by 5" length wise and leaves a margin of 1" at top and bottom and a margin of ½" on either side. What fractional part of the paper is available for typing?

  • (a) 2/3
  • (b) 1/2
  • (c) 1/3
  • (d) 5/7
Answer: (a) Typing area = (12-2)×(5-1) = 10×4 = 40. Total area = 60. Fraction = 40/60 = 2/3

Q9. A rectangular lawn 70m×30m has two roads each 5 metres wide, running in the middle, one parallel to length and other parallel to breadth. Find the cost of gravelling the road at ₹4 per square metre.

  • (a) ₹2,000
  • (b) ₹1,800
  • (c) ₹1,900
  • (d) ₹1,700
Answer: (c) Area = 70×5 + 30×5 - 5×5 = 350+150-25 = 475. Cost = 475×4 = ₹1900

Q10. A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed, the height of the heap being 12 cm. The radius of the heap at the base is:

  • (a) 63 cm
  • (b) 53 cm
  • (c) 56 cm
  • (d) 66 cm
Answer: (a) π×21²×36 = ⅓π×r²×12 ⇒ r² = 21²×9 ⇒ r = 63 cm

4. Exercise (Questions 11-20)

Q11. The area of a square field is 576 km². How long will it take for a horse to run around at the speed of 12 km/h?

  • (a) 12 h
  • (b) 10 h
  • (c) 8 h
  • (d) 6 h
Answer: (c) Side = 24 km. Perimeter = 96 km. Time = 96/12 = 8 hours

Q12. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space?

  • (a) 46
  • (b) 81
  • (c) 126
  • (d) 252
Answer: (c) l=9, l+2b=37 ⇒ b=14. Area = 9×14 = 126 sq ft

Q13. A farmer wishes to start a 100 square metres rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side. The dimension of the garden is:

  • (a) 15m×6.67m
  • (b) 20m×5m
  • (c) 30m×3.33m
  • (d) 40m×2.5m
Answer: (b) 2b + l = 30, l×b = 100. Solving: b=5, l=20

Q14. A rectangular tank measuring 5m×4.5m×2.1m is dug in the centre of the field measuring 13.5m×2.5m. The earth dug out is spread evenly over the remaining portion. How much is the level of the field raised?

  • (a) 4.0 m
  • (b) 4.1 m
  • (c) 4.2 m
  • (d) 4.3 m
Answer: (c) Volume = 5×4.5×2.1 = 47.25 m³. Area remaining = 13.5×2.5 - 5×4.5 = 11.25 m². Height = 47.25/11.25 = 4.2 m

Q15. Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is:

  • (a) 24 cm²
  • (b) 42 cm²
  • (c) 154 cm²
  • (d) 196 cm²
Answer: (b) Area = 14² - 4×(1/4)π×7² = 196 - 154 = 42 cm²

Q16. The ratio between length and breadth of a rectangular park is 3:2. If a man cycling along the boundary at 12 km/h completes one round in 8 minutes, then the area of the park is:

  • (a) 15360
  • (b) 153600
  • (c) 30720
  • (d) 307200
Answer: (b) Perimeter = 12000/60×8 = 1600 m. 2(3x+2x)=1600 ⇒ x=160. Area = 480×320 = 153600

Q17. The water in a rectangular reservoir having base 80m×60m is 6.5 metres deep. In what time can the water be emptied by a pipe whose cross section is a square of side 20 cm, if water runs at 15 km per hour?

  • (a) 52 hrs
  • (b) 26 hrs
  • (c) 65 hrs
  • (d) 42 hrs
Answer: (a) Volume = 80×60×6.5 = 31200 m³. Pipe volume per hour = 0.2×0.2×15000 = 600 m³. Time = 31200/600 = 52 hrs

Q18. A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be:

  • (a) 3520 cm²
  • (b) 6400 cm²
  • (c) 7744 cm²
  • (d) 8800 cm²
Answer: (c) Length = 2π×56 = 352 cm. Side of square = 88 cm. Area = 88² = 7744 cm²

Q19. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, then find the edge of the third smaller cube.

  • (a) 10 cm
  • (b) 14 cm
  • (c) 12 cm
  • (d) 16 cm
Answer: (a) 12³ = 6³ + 8³ + x³ ⇒ 1728 = 216 + 512 + x³ ⇒ x³ = 1000 ⇒ x = 10 cm

Q20. A copper sphere of radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is:

  • (a) 9 m
  • (b) 12 m
  • (c) 18 m
  • (d) 36 m
Answer: (d) 4/3π×3³ = π×0.1²×h ⇒ h = 3600 cm = 36 m

5. Exercise (Questions 21-30)

Q21. The length, breadth and height of a cuboid are in the ratio 1:2:3. The dimensions are increased by 100%, 200% and 200% respectively. The increase in volume will be:

  • (a) 5 times
  • (b) 6 times
  • (c) 12 times
  • (d) 17 times
Answer: (d) Original volume = 6x³. New dimensions = 2x, 6x, 9x. New volume = 108x³. Increase = 102x³/6x³ = 17 times

Q22. The surface area of a cube is 150 m². The length of its diagonal is:

  • (a) 5√3 m
  • (b) 5 m
  • (c) 10/√3 m
  • (d) 15 m
Answer: (a) 6a²=150 ⇒ a=5. Diagonal = 5√3 m

Q23. The length of the longest rod that can be placed in a room which is 12m long, 9m broad and 8m high is:

  • (a) 27 m
  • (b) 19 m
  • (c) 17 m
  • (d) 13 m
Answer: (c) Diagonal = √(12²+9²+8²) = √289 = 17 m

Q24. If the volume of a sphere is divided by its surface area, the result is 27 cm. The radius of the sphere is:

  • (a) 9 cm
  • (b) 27 cm
  • (c) 81 cm
  • (d) 243 cm
Answer: (c) (4/3πr³)/(4πr²) = r/3 = 27 ⇒ r = 81 cm

Q25. The volume of water measured on a rectangular field 500m×300m is 3000 m³. Find the depth of water.

  • (a) 2 cm
  • (b) 3 cm
  • (c) 4 cm
  • (d) 5 cm
Answer: (a) 3000 = 500×300×h ⇒ h = 0.02 m = 2 cm

Q26. A cylindrical rod of radius 6 cm and length 28 cm is melted and recast into small spherical balls of radius 3/4 cm. How many balls can be made?

  • (a) 1792
  • (b) 1872
  • (c) 1972
  • (d) 2072
Answer: (a) Volume of rod = π×6²×28 = 1008π. Volume of ball = 4/3π×(3/4)³ = 9π/16. Balls = 1008π × 16/9π = 1792

Q27. A cylindrical vessel of radius 60 cm is filled with water up to some height. If 30 spherical balls of radius 1 cm are dropped, the water level rises by:

  • (a) 3 m
  • (b) 3 cm
  • (c) 3 mm
  • (d) 30 cm
Answer: (b) Volume of 30 balls = 30×4/3π×1³ = 40π. π×60²×h = 40π ⇒ h = 40/3600 = 1/90 m = 1.11 cm

Q28. A cylindrical tank of height 400 cm can hold 246.4 litres of water. The diameter of its base is:

  • (a) 28 cm
  • (b) 14 cm
  • (c) 56 cm
  • (d) 42 cm
Answer: (a) Volume = 246400 cm³. πr²×400 = 246400 ⇒ r² = 196 ⇒ r=14. Diameter=28 cm

Q29. A reservoir is supplied from a pipe 6 cm in diameter. How many pipes of 3 cm diameter would discharge the same quantity, supposing velocity is same?

  • (a) 4
  • (b) 5
  • (c) 6
  • (d) 7
Answer: (a) Area ratio = π×3² : π×1.5² = 9:2.25 = 4:1. So 4 pipes.

Q30. If the radius of a sphere is increased by 2 cm, then its surface area increases by 352 cm². The radius of the sphere before the increase was:

  • (a) 3 cm
  • (b) 4 cm
  • (c) 5 cm
  • (d) 6 cm
Answer: (d) 4π[(r+2)² - r²] = 352 ⇒ (r+2)² - r² = 28 ⇒ 4r+4=28 ⇒ r=6 cm

Complete Answer Key

1. (a)2. (d)3. (d)4. (c)5. (a)
6. (a)7. (a)8. (a)9. (c)10. (a)
11. (c)12. (c)13. (b)14. (c)15. (b)
16. (b)17. (a)18. (c)19. (a)20. (d)
21. (d)22. (a)23. (c)24. (c)25. (a)
26. (a)27. (b)28. (a)29. (a)30. (d)
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