Current Affairs updated daily
KalviGavaniKalviGavani
Class 9 · Tamil Nadu State Board

Samacheer Kalvi 9th Maths Mensuration Exercise Solutions | KalviGavani

TN Class 9 Maths Mensuration exercise solutions — Heron's formula, surface area and volume of solids, step-by-step worked answers.

Updated 13 June 202622 min read

How to Use This Page

1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.

Key Concepts (Quick Recap)

Heron's Formula

  • Area of a triangle = √[s(s − a)(s − b)(s − c)]
  • Semi-perimeter s = (a + b + c) / 2, where a, b, c are the sides
  • Works for any triangle — right, acute, or obtuse
  • For an equilateral triangle of side a: Area = (√3/4) × a² (derived from Heron's formula)
  • For a quadrilateral, split it along a diagonal into two triangles and find each area separately

Cube (side = a)

  • Lateral Surface Area (LSA) = 4a²
  • Total Surface Area (TSA) = 6a²
  • Volume = a³
  • Diagonal = a√3

Cuboid (length l, breadth b, height h)

  • LSA = 2(l + b)h
  • TSA = 2(lb + bh + hl)
  • Volume = l × b × h
  • Diagonal = √(l² + b² + h²)

Cylinder (radius r, height h)

  • Curved Surface Area (CSA) = 2πrh
  • TSA = 2πr(r + h) = 2πrh + 2πr²
  • Volume = πr²h

Cone (radius r, height h, slant height l)

  • Slant height l = √(r² + h²)
  • CSA = πrl
  • TSA = πrl + πr² = πr(l + r)
  • Volume = (1/3)πr²h

Sphere (radius r)

  • TSA = 4πr²
  • Volume = (4/3)πr³

Hemisphere (radius r)

  • CSA = 2πr²
  • TSA = 3πr² (CSA + base circle)
  • Volume = (2/3)πr³

Key Conversions

  • 1 m³ = 1000 litres = 1000 dm³
  • 1 cm³ = 1 mL
  • 1 litre = 1000 cm³
  • 1 m = 100 cm

I. Choose the Best Answer (MCQs)

Q1. The semi-perimeter of a triangle with sides 5 cm, 12 cm, and 13 cm is: (a) 15 cm (b) 30 cm (c) 20 cm (d) 10 cm

s = (a + b + c) / 2 = (5 + 12 + 13) / 2 = 30 / 2 = 15 cm Answer: (a) 15 cm

Q2. The area of a triangle with sides 3 cm, 4 cm, and 5 cm is: (a) 6 cm² (b) 12 cm² (c) 10 cm² (d) 7.5 cm²

s = (3 + 4 + 5) / 2 = 6 Area = √[6(6−3)(6−4)(6−5)] = √[6 × 3 × 2 × 1] = √36 = 6 cm² Answer: (a) 6 cm²

Q3. The total surface area of a cube of side 4 cm is: (a) 64 cm² (b) 96 cm² (c) 16 cm² (d) 48 cm²

TSA = 6a² = 6 × 4² = 6 × 16 = 96 cm² Answer: (b) 96 cm²

Q4. The volume of a cuboid with dimensions 5 cm × 3 cm × 2 cm is: (a) 10 cm³ (b) 30 cm³ (c) 15 cm³ (d) 20 cm³

Volume = l × b × h = 5 × 3 × 2 = 30 cm³ Answer: (b) 30 cm³

Q5. The curved surface area of a cylinder of radius 7 cm and height 10 cm is (take π = 22/7): (a) 440 cm² (b) 220 cm² (c) 308 cm² (d) 616 cm²

CSA = 2πrh = 2 × (22/7) × 7 × 10 = 2 × 22 × 10 = 440 cm² Answer: (a) 440 cm²

Q6. The volume of a cone with radius 3 cm and height 4 cm is: (a) 12π cm³ (b) 36π cm³ (c) 16π cm³ (d) 24π cm³

Volume = (1/3)πr²h = (1/3) × π × 3² × 4 = (1/3) × π × 9 × 4 = 12π cm³ Answer: (a) 12π cm³

Q7. The total surface area of a sphere of radius 3.5 cm is (take π = 22/7): (a) 154 cm² (b) 77 cm² (c) 44 cm² (d) 308 cm²

TSA = 4πr² = 4 × (22/7) × (3.5)² = 4 × (22/7) × 12.25 = 4 × 22 × 1.75 = 154 cm² Answer: (a) 154 cm²

Q8. The slant height of a cone with radius 6 cm and height 8 cm is: (a) 14 cm (b) 10 cm (c) 12 cm (d) 2 cm

l = √(r² + h²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm Answer: (b) 10 cm

Q9. If the area of an equilateral triangle is 9√3 cm², its side is: (a) 6 cm (b) 3 cm (c) 9 cm (d) 12 cm

Area of equilateral triangle = (√3/4) × a² 9√3 = (√3/4) × a² 9 = a² / 4 a² = 36 a = 6 cm Answer: (a) 6 cm

Q10. The volume of a hemisphere of radius 3 cm is: (a) 18π cm³ (b) 36π cm³ (c) 12π cm³ (d) 54π cm³

Volume = (2/3)πr³ = (2/3) × π × 3³ = (2/3) × π × 27 = 18π cm³ Answer: (a) 18π cm³

Q11. A right triangle with sides 6 cm, 8 cm, and 10 cm is rotated about its hypotenuse. The resulting solid has: (a) two cones (b) one cone (c) one cylinder (d) two hemispheres

When rotated about the hypotenuse, the solid formed consists of two cones sharing the same base (the altitude drawn to the hypotenuse creates two right triangles that generate two cones). Answer: (a) two cones

Q12. The lateral surface area of a cuboid with l = 5, b = 4, h = 3 is: (a) 60 (b) 54 (c) 94 (d) 27

LSA = 2(l + b)h = 2(5 + 4) × 3 = 2 × 9 × 3 = 54 Answer: (b) 54

II. Fill in the Blanks

1. The semi-perimeter of a triangle with sides 13 cm, 14 cm, and 15 cm is ______.

s = (13 + 14 + 15) / 2 = 42 / 2 = 21 cm Answer: 21 cm

2. The area of an equilateral triangle with side 4 cm is ______ cm².

Area = (√3/4) × a² = (√3/4) × 16 = 4√3 cm² Answer: 4√3 cm²

3. The total surface area of a cube of side 5 cm is ______ cm².

TSA = 6 × 5² = 6 × 25 = 150 cm² Answer: 150 cm²

4. The volume of a cylinder with radius 7 cm and height 10 cm is ______ cm³ (take π = 22/7).

V = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 70 = 1540 cm³ Answer: 1540 cm³

5. The curved surface area of a cone with slant height 10 cm and radius 7 cm is ______ cm² (take π = 22/7).

CSA = πrl = (22/7) × 7 × 10 = 220 cm² Answer: 220 cm²

6. The volume of a sphere of radius 3 cm is ______ cm³.

V = (4/3)πr³ = (4/3) × π × 27 = 36π cm³ Answer: 36π cm³

7. The total surface area of a hemisphere of radius r is ______.

TSA = 3πr² Answer: 3πr²

8. 1 litre = ______ cm³.

1 litre = 1000 cm³ Answer: 1000

9. The diagonal of a cuboid with dimensions 3 cm × 4 cm × 5 cm is ______ cm.

Diagonal = √(l² + b² + h²) = √(9 + 16 + 25) = √50 = 5√2 cm Answer: 5√2 cm

10. If the side of a cube is doubled, its volume becomes ______ times the original.

Original volume = a³ New volume = (2a)³ = 8a³ Answer: 8

11. The volume of a cone with radius r and height h is one-third the volume of a ______ with the same base and height.

A cone has exactly 1/3 the volume of a cylinder with the same base radius and height. Answer: cylinder

12. The curved surface area of a hemisphere of radius 7 cm is ______ cm² (take π = 22/7).

CSA = 2πr² = 2 × (22/7) × 49 = 2 × 22 × 7 = 308 cm² Answer: 308 cm²

III. True or False

1. Heron's formula can be used to find the area of any triangle.
AnswerTrue
Heron's formula works for all triangles — right, acute, or obtuse — as long as the three side lengths are known.
2. The curved surface area of a cylinder equals 2πr².
AnswerFalse
Correction: CSA of a cylinder = 2πrh (not 2πr²). 2πr² is the area of one circular base. The formula 2πr(r + h) gives TSA.
3. The volume of a cone is one-third the volume of a cylinder with the same base and height.
AnswerTrue
V_cone = (1/3)πr²h and V_cylinder = πr²h, so V_cone = (1/3)V_cylinder.
4. The total surface area of a hemisphere equals 2πr².
AnswerFalse
Correction: CSA of a hemisphere = 2πr², but TSA = 3πr² (which includes the circular base πr²).
5. Heron's formula requires knowing the height of the triangle.
AnswerFalse
Correction: Heron's formula uses only the three sides. No height or angle measurement is needed. That is the key advantage.
6. The lateral surface area of a cuboid equals the sum of areas of its four vertical faces.
AnswerTrue
LSA = 2(l + b)h, which is exactly the sum of the four side faces: 2lh (front and back) + 2bh (left and right).
7. If the radius of a sphere is halved, its surface area becomes one-fourth.
AnswerTrue
TSA = 4πr². If r becomes r/2, new TSA = 4π(r/2)² = 4πr²/4 = (1/4)(4πr²). So the surface area becomes one-fourth.
8. The volume of a cube of side 1 cm is equal to 1 mL.
AnswerTrue
Volume = 1³ = 1 cm³ = 1 mL. These are equivalent units.

IV. Match the Following

Set 1:

Column AColumn B
1. CSA of cylinder(a) 2πrh
2. TSA of cube(b) 6a²
3. Volume of cone(c) (1/3)πr²h
4. TSA of sphere(d) 4πr²
5. Volume of cuboid(e) l × b × h
1 → (a) CSA of cylinder = 2πrh 2 → (b) TSA of cube = 6a² 3 → (c) Volume of cone = (1/3)πr²h 4 → (d) TSA of sphere = 4πr² 5 → (e) Volume of cuboid = l × b × h

Set 2:

Column AColumn B
1. CSA of hemisphere(a) 2πr²
2. TSA of cylinder(b) 2πr(r + h)
3. Volume of sphere(c) (4/3)πr³
4. CSA of cone(d) πrl
5. Diagonal of cuboid(e) √(l² + b² + h²)
1 → (a) CSA of hemisphere = 2πr² 2 → (b) TSA of cylinder = 2πr(r + h) 3 → (c) Volume of sphere = (4/3)πr³ 4 → (d) CSA of cone = πrl 5 → (e) Diagonal of cuboid = √(l² + b² + h²)

V. Short Answer Questions

Q1. Find the area of a triangle with sides 7 cm, 24 cm, and 25 cm.
s = (7 + 24 + 25) / 2 = 56 / 2 = 28 Area = √[s(s − a)(s − b)(s − c)] = √[28(28 − 7)(28 − 24)(28 − 25)] = √[28 × 21 × 4 × 3] = √[7056] = 84 cm² Verification: 7² + 24² = 49 + 576 = 625 = 25² — right triangle. Area = (1/2) × 7 × 24 = 84 cm² ✓ Answer: 84 cm²
Q2. Find the area of an equilateral triangle with side 6 cm.
Area = (√3/4) × a² = (√3/4) × 36 = 9√3 cm² Answer: 9√3 cm² (≈ 15.59 cm²)
Q3. Find the total surface area of a cuboid with dimensions 10 cm × 8 cm × 6 cm.
TSA = 2(lb + bh + hl) = 2(10 × 8 + 8 × 6 + 6 × 10) = 2(80 + 48 + 60) = 2 × 188 = 376 cm² Answer: 376 cm²
Q4. Find the volume of a cube of side 5 cm.
Volume = a³ = 5³ = 125 cm³ Answer: 125 cm³
Q5. Find the curved surface area of a cylinder with radius 14 cm and height 20 cm (take π = 22/7).
CSA = 2πrh = 2 × (22/7) × 14 × 20 = 2 × 22 × 2 × 20 = 2 × 880 = 1760 cm² Answer: 1760 cm²
Q6. Find the slant height of a cone with radius 8 cm and height 15 cm.
l = √(r² + h²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm Answer: 17 cm
Q7. Find the volume of a sphere of radius 7 cm (take π = 22/7).
Volume = (4/3)πr³ = (4/3) × (22/7) × 7³ = (4/3) × (22/7) × 343 = (4/3) × 22 × 49 = (4/3) × 1078 = 4312/3 ≈ 1437.33 cm³ Answer: 4312/3 cm³ (or approximately 1437.33 cm³)
Q8. A cuboidal tank has dimensions 6 m × 4 m × 3 m. Find its capacity in litres.
Volume = 6 × 4 × 3 = 72 m³ = 72 × 1000 litres (since 1 m³ = 1000 litres) = 72000 litres Answer: 72000 litres
Q9. Find the area of a triangle whose sides are 13 cm, 14 cm, and 15 cm.
s = (13 + 14 + 15) / 2 = 42 / 2 = 21 Area = √[21(21 − 13)(21 − 14)(21 − 15)] = √[21 × 8 × 7 × 6] = √[7056] = 84 cm² Answer: 84 cm²
Q10. Find the total surface area of a hemisphere of radius 7 cm (take π = 22/7).
TSA of hemisphere = 3πr² = 3 × (22/7) × 7² = 3 × (22/7) × 49 = 3 × 22 × 7 = 462 cm² Answer: 462 cm²

VI. Detailed Answer Questions

Q1. A triangular park has sides 30 m, 26 m, and 28 m. Find its area using Heron's formula. Also find the cost of levelling it at Rs 15 per m².
Step 1: Find semi-perimeter s = (30 + 26 + 28) / 2 = 84 / 2 = 42 m Step 2: Apply Heron's formula Area = √[s(s − a)(s − b)(s − c)] = √[42(42 − 30)(42 − 26)(42 − 28)] = √[42 × 12 × 16 × 14] Step 3: Simplify under the square root 42 × 12 = 504 16 × 14 = 224 504 × 224 = 112896 √112896 = 336 Area = 336 m² Step 4: Find cost Cost = Area × Rate = 336 × 15 = Rs 5040 Answer: Area = 336 m², Cost = Rs 5040
Q2. The sides of a quadrilateral are 10 m, 15 m, 12 m, and 13 m, and the diagonal connecting the first and third sides is 18 m. Find the area of the quadrilateral.
The diagonal divides the quadrilateral into two triangles. Triangle 1: sides 10 m, 15 m, 18 m s₁ = (10 + 15 + 18) / 2 = 43 / 2 = 21.5 Area₁ = √[21.5(21.5 − 10)(21.5 − 15)(21.5 − 18)] = √[21.5 × 11.5 × 6.5 × 3.5] = √[5635.375] ≈ 75.07 m² Triangle 2: sides 12 m, 13 m, 18 m s₂ = (12 + 13 + 18) / 2 = 43 / 2 = 21.5 Area₂ = √[21.5(21.5 − 12)(21.5 − 13)(21.5 − 18)] = √[21.5 × 9.5 × 8.5 × 3.5] = √[6070.1875] ≈ 77.91 m² Total area = Area₁ + Area₂ ≈ 75.07 + 77.91 ≈ 150.98 m² Answer: Approximately 151 m²
Q3. A cylindrical tank has radius 1.4 m and height 3 m. Find (i) its curved surface area, (ii) its total surface area, and (iii) its capacity in litres. (Take π = 22/7)
Given: r = 1.4 m, h = 3 m (i) CSA = 2πrh = 2 × (22/7) × 1.4 × 3 = 2 × (22/7) × 4.2 = 2 × 22 × 0.6 = 26.4 m² (ii) TSA = 2πr(r + h) = 2 × (22/7) × 1.4 × (1.4 + 3) = 2 × (22/7) × 1.4 × 4.4 = 2 × 22 × 0.2 × 4.4 = 38.72 m² (iii) Volume = πr²h = (22/7) × (1.4)² × 3 = (22/7) × 1.96 × 3 = 22 × 0.28 × 3 = 18.48 m³ = 18.48 × 1000 = 18480 litres Answer: (i) 26.4 m², (ii) 38.72 m², (iii) 18480 litres
Q4. A cone has slant height 25 cm and curved surface area 550 cm². Find (i) the radius of the base and (ii) the total surface area. (Take π = 22/7)
Given: l = 25 cm, CSA = 550 cm² (i) CSA = πrl 550 = (22/7) × r × 25 550 = (550/7) × r r = 550 × 7 / 550 = 7 cm (ii) TSA = πr(l + r) = (22/7) × 7 × (25 + 7) = 22 × 32 = 704 cm² Answer: (i) Radius = 7 cm, (ii) TSA = 704 cm²
Q5. A spherical ball of radius 3 cm is melted and recast into a cylinder of radius 1 cm. Find the height of the cylinder.
Volume of sphere = Volume of cylinder (mass is conserved) Volume of sphere = (4/3)πr³ = (4/3)π(3)³ = (4/3)π(27) = 36π cm³ Volume of cylinder = πR²h = π(1)²h = πh cm³ Equating: 36π = πh h = 36 cm Answer: Height of cylinder = 36 cm
Q6. The radius of a sphere is 5 cm. Find its surface area and volume. (Take π = 22/7)
Surface Area = 4πr² = 4 × (22/7) × 5² = 4 × (22/7) × 25 = 2200/7 ≈ 314.29 cm² Volume = (4/3)πr³ = (4/3) × (22/7) × 125 = (4/3) × (2750/7) = 11000/21 ≈ 523.81 cm³ Answer: Surface area ≈ 314.29 cm², Volume ≈ 523.81 cm³
Q7. A cuboidal box is 5 cm long, 4 cm wide, and 3 cm high. Find (i) its volume, (ii) its total surface area, and (iii) its diagonal.
Given: l = 5 cm, b = 4 cm, h = 3 cm (i) Volume = l × b × h = 5 × 4 × 3 = 60 cm³ (ii) TSA = 2(lb + bh + hl) = 2(5 × 4 + 4 × 3 + 3 × 5) = 2(20 + 12 + 15) = 2 × 47 = 94 cm² (iii) Diagonal = √(l² + b² + h²) = √(25 + 16 + 9) = √50 = 5√2 cm Answer: (i) 60 cm³, (ii) 94 cm², (iii) 5√2 cm
Q8. The area of a triangular field is 336 m² and its sides are in the ratio 3:5:7. Find the perimeter of the field.
Let the sides be 3k, 5k, 7k for some positive value k. Semi-perimeter s = (3k + 5k + 7k) / 2 = 15k / 2 Using Heron's formula: Area = √[s(s − a)(s − b)(s − c)] 336 = √[(15k/2)(15k/2 − 3k)(15k/2 − 5k)(15k/2 − 7k)] 336 = √[(15k/2)(9k/2)(5k/2)(k/2)] 336 = √[(15 × 9 × 5 × 1) × k⁴ / 16] 336 = √[675k⁴/16] 336 = k²√675 / 4 336 × 4 = k²√675 1344 = k² × 15√3 k² = 1344 / (15√3) = 1344√3 / 45 This is getting complex. Let us try a direct approach. Since the sides are in ratio 3:5:7, let sides be 3x, 5x, 7x. s = (3x + 5x + 7x)/2 = 7.5x Area = √[7.5x × (7.5x − 3x) × (7.5x − 5x) × (7.5x − 7x)] = √[7.5x × 4.5x × 2.5x × 0.5x] = √[7.5 × 4.5 × 2.5 × 0.5 × x⁴] = x²√[42.1875] = x²√[675/16] = x²√675 / 4 336 = x² × 15√3 / 4 x² = 1344 / (15√3) = 1344√3 / 45 Check: 3:5:7 sides with s = (3k+5k+7k)/2 = 7.5k 336² = 7.5 × 4.5 × 2.5 × 0.5 × k⁴ 112896 = 42.1875 k⁴ k⁴ = 112896 / 42.1875 ≈ 2676 k ≈ 7.16... Perimeter = 3k + 5k + 7k = 15k Note: This problem with ratio 3:5:7 leads to irrational side lengths. The perimeter is 15k where k satisfies the equation above. The exact perimeter is 15k = 15 × √(1344/(15√3)). For a clean integer answer, a common exam variant uses sides in ratio 13:14:15. Answer: Perimeter = 15k where k is determined from the Heron's formula equation.
Q9. A well of diameter 3 m is dug 14 m deep. The earth taken out is spread around the well to form a circular embankment of width 4 m. Find the height of the embankment.
Volume of earth dug out = Volume of the well (cylinder) r = 3/2 = 1.5 m, h = 14 m Volume of earth = πr²h = π × (1.5)² × 14 = π × 2.25 × 14 = 31.5π m³ The embankment is a hollow cylinder: Inner radius = r₁ = 1.5 m (same as well radius) Outer radius = r₂ = 1.5 + 4 = 5.5 m Let height of embankment = H Volume of embankment = π(r₂² − r₁²)H = π(5.5² − 1.5²)H = π(30.25 − 2.25)H = 28πH Equating volumes: 31.5π = 28πH H = 31.5/28 = 1.125 m Answer: Height of embankment = 1.125 m
Q10. How many litres of water can a cylindrical tank hold if its radius is 1.4 m and height is 2.5 m? (Take π = 22/7)
Volume = πr²h = (22/7) × (1.4)² × 2.5 = (22/7) × 1.96 × 2.5 = (22/7) × 4.9 = 22 × 0.7 = 15.4 m³ Capacity = 15.4 × 1000 = 15400 litres Answer: 15400 litres

Common Mistakes

1. Forgetting to compute semi-perimeter before using Heron's formula: Always calculate s = (a + b + c)/2 first, then substitute into √[s(s−a)(s−b)(s−c)]. Students often use the perimeter instead of the semi-perimeter.

2. Confusing CSA and TSA: CSA (curved/lateral surface area) includes only the sides. TSA adds the base(s). For a cylinder, CSA = 2πrh but TSA = 2πrh + 2πr² = 2πr(r + h). Always check which one the question asks for.

3. Wrong formula for cone volume: The volume of a cone is (1/3)πr²h, NOT πr²h (that is the cylinder). The factor of 1/3 is easy to forget and costs marks every time.

4. Using height instead of slant height in cone CSA: CSA of cone = πrl where l is the slant height, NOT the vertical height h. If only h and r are given, first find l = √(r² + h²).

5. Forgetting unit conversions: 1 m = 100 cm means 1 m² = 10000 cm² and 1 m³ = 1000000 cm³. Also, 1 m³ = 1000 litres. Mixing up these conversions is a common source of error.

6. Hemisphere TSA vs CSA: TSA of a hemisphere is 3πr² (curved surface 2πr² PLUS the flat circular base πr²). Students often give 2πr² as the TSA, which is only the CSA.

7. Incorrect ratio reasoning for scaled solids: If the side of a cube is doubled, volume becomes 2³ = 8 times, not 2 times. Surface area becomes 2² = 4 times, not 2 times. Use the correct power: linear (1), area (2), volume (3).

8. Not reading "diameter" as "radius/2": When the question gives the diameter, remember to halve it before substituting into formulas. r = d/2. This is the single most common arithmetic mistake.

Exam Checklist

  • [ ] I can calculate semi-perimeter and apply Heron's formula correctly
  • [ ] I can find the area of an equilateral triangle using (√3/4)a²
  • [ ] I can split a quadrilateral along a diagonal and find its area using two Heron's formula applications
  • [ ] I can compute LSA, TSA, and Volume for a cube given its side
  • [ ] I can compute LSA, TSA, and Volume for a cuboid given l, b, h
  • [ ] I can compute CSA, TSA, and Volume for a cylinder given r and h
  • [ ] I can find the slant height of a cone from r and h, then compute CSA and TSA
  • [ ] I can compute CSA, TSA, and Volume for a cone
  • [ ] I can compute TSA and Volume for a sphere
  • [ ] I can compute CSA, TSA, and Volume for a hemisphere
  • [ ] I can convert between m³, cm³, and litres correctly
  • [ ] I can solve "melt and recast" problems (volume is conserved)
  • [ ] I always check whether the question gives radius or diameter before substituting
  • [ ] I can set up and solve word problems involving cost per unit area

Internal Links

  • [Class 9 Maths Home](/class/9/maths) — All chapters and resources
  • [Class 9 Maths Mensuration Chapter Guide](/class/9/maths/mensuration) — Concept explanations and examples
  • [Class 9 Maths Real Numbers Solutions](/class/9/maths/solutions/real-numbers-exercise-solutions) — Previous chapter
  • [Class 9 Maths Set Language Solutions](/class/9/maths/solutions/set-language-exercise-solutions) — Related chapter
  • [TN State Board Class 9 Home](/class/9?board=tamil-nadu-state-board) — All subjects
  • [KalviGavani Home](/) — Classes 6–12 resources

Frequently Asked Questions

Area of a triangle = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2 is the semi-perimeter and a, b, c are the sides.
← Browse all Class 9 Mathematics Exercise Solutions

Other question types — same chapter

Chapter Notes

Tamil Nadu Class 9 Maths Mensuration — Samacheer Kalvi

Exercise Solutions

Samacheer Kalvi 9th Maths Algebra Exercise Solutions | KalviGavani

Exercise Solutions

Samacheer Kalvi 9th Maths Coordinate Geometry Exercise Solutions | KalviGavani