Heron’s Formula Class 9 Maths Worksheet with Answers Chapter 10

We are now in chapter 10 of the NCERT Class 9 book. This worksheet is structured into four parts – Basic, Standard, Advance, and HOTS. The difficulty of the questions increases in a ladder format. Finish the chapter thoroughly first, then solve in order, starting with easy questions and progressing to HOTS. Attempt this worksheet only after you have completed the chapter and revised the solved examples.

• Heron’s formula usage
• Semi-perimeter calculation
• Area from three sides
• Equilateral triangle area
• Word-based applications

Class 9 Maths Worksheet – Chapter 10: Heron’s Formula

Basic

1. Find the semi-perimeter (s) and area of a triangle with sides 13 cm, 14 cm, 15 cm using Heron’s formula.
2. A triangular park has sides 40 m, 32 m, and 24 m. Find its area using Heron’s formula.
3. Find the area of an equilateral triangle of side 10 cm using Heron’s formula.
4. Find the area of an isosceles triangle with equal sides 5 cm each and base 8 cm using Heron’s formula.
5. Two sides of a triangle are 8 cm and 11 cm and its perimeter is 32 cm. Find its area using Heron’s formula.
6. For a triangle with sides 7 cm, 9 cm, 10 cm:
   • Find s
   • Write (s−a), (s−b), (s−c)

Standard

1. A traffic signal board is an equilateral triangle with side a.
   • Write the area in terms of a using Heron’s formula.
   • If the perimeter is 180 cm, find its area.
2. A triangular park has sides 120 m, 80 m and 50 m.
   • Find the area to be grassed.
   • Find the cost of fencing at ₹20 per metre leaving a 3 m gate on one side.
3. Find the area of a triangle with two sides 18 cm and 10 cm and perimeter 42 cm.
4. The sides of a triangular plot are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.
5. A slide wall is triangular with sides 15 m, 11 m and 6 m. Find the area of the wall painted.
6. The sides of a triangle are in the ratio 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Advance

1. The triangular side wall of a flyover has sides 122 m, 22 m and 120 m.
   Advertisements earn ₹5000 per m² per year.
   If the wall is hired for 3 months, how much rent is paid?
2. Two sides of a triangle are 14 cm and 18 cm and its perimeter is 48 cm.
   Find the third side and then find the area.
3. Find the area of an isosceles triangle with sides 25 cm, 25 cm and 48 cm using Heron’s formula.
4. The area of a triangle with sides 13 cm, 14 cm and 15 cm is 84 cm².
   Find the height corresponding to base 14 cm.
5. Using Heron’s formula, find the area of a triangle with sides 9 cm, 40 cm, 41 cm.
   Also verify the area using the right triangle formula.
6. Check whether a triangle with sides 7 cm, 15 cm and 20 cm is possible.
   If yes, find its area using Heron’s formula.

HOTS

1. Using Heron’s formula, derive the area of an equilateral triangle of side a.
   Then find the area when a = 10 cm.
2. Two triangles have the same perimeter 30 cm:
   • Triangle A: 12 cm, 12 cm, 6 cm
   • Triangle B: 10 cm, 10 cm, 10 cm

   Without using a calculator, decide which has bigger area and show a short justification.

3. For sides 122 m, 120 m and 22 m:
   • Show it forms a right triangle (use squares).
   • Then find the area using 1/2 × base × height and compare with Heron’s result.
4. A triangular park has sides 40 m, 32 m and 24 m.
   Find the cost of fencing at ₹50 per metre leaving a 2 m gate.
5. The sides of a triangle are in the ratio 2 : 3 : 4 and the perimeter is 90 cm.
   Find its area.
6. For the triangle with sides 7 cm, 15 cm and 20 cm:
   Decide whether it is acute, right, or obtuse.
   (Use comparison of squares.)

Answer Key (Answers + Hints)

Basic – Answers

1. Ans: s = (13+14+15)/2 = 21 cm, area = √(21×8×7×6) = √7056 = 84 cm²
   Hint: Compute s, then (s−a)(s−b)(s−c).
2. Ans: s = 48 m, area = √(48×8×16×24) = √147456 = 384 m²
   Hint: Use s = (40+32+24)/2.
3. Ans: s = 15 cm, area = √(15×5×5×5) = √1875 = 25√3 cm²
   Hint: All sides equal ⇒ (s−a) repeats.
4. Ans: s = 9 cm, area = √(9×4×4×1) = √144 = 12 cm²
   Hint: s = (5+5+8)/2.
5. Ans: Third side = 32 − (8+11) = 13 cm, s = 16 cm,
   area = √(16×8×5×3) = √1920 = 8√30 cm²
   Hint: Find missing side first using perimeter.
6. Ans: s = (7+9+10)/2 = 13 cm
   (s−a)=6, (s−b)=4, (s−c)=3
   Hint: Just substitute into s and subtract.

Standard – Answers

1. Ans: For equilateral side a: s = 3a/2
   Area = √( (3a/2)(a/2)(a/2)(a/2) ) = (√3/4)a²
   If perimeter 180 ⇒ a = 60 ⇒ area = (√3/4)×60² = 900√3 cm²
   Hint: Convert perimeter to side, then apply formula.
2. Ans: s = (120+80+50)/2 = 125 m
   Area = √(125×5×45×75) = √2109375 = 375√15 m²
   Wire length = 250 − 3 = 247 m, cost = 247×20 = ₹4940
   Hint: Perimeter = 250 m, gate part not fenced.
3. Ans: Third side = 42 − (18+10) = 14 cm, s = 21 cm
   Area = √(21×3×11×7) = √4851 = 21√11 cm²
   Hint: Factor inside root if possible.
4. Ans: Ratio sum = 3+5+7 = 15, perimeter 300 ⇒ 1 part = 20
   Sides: 60 m, 100 m, 140 m; s = 150 m
   Area = √(150×90×50×10) = √6750000 = 1500√3 m²
   Hint: Convert ratio to actual sides first.
5. Ans: s = (15+11+6)/2 = 16 m
   Area = √(16×1×5×10) = √800 = 20√2 m²
   Hint: Multiply numbers to form perfect squares.
6. Ans: Ratio sum = 54, perimeter 540 ⇒ 1 part = 10
   Sides: 120 cm, 170 cm, 250 cm; s = 270 cm
   Area = √(270×150×100×20) = √81000000 = 9000 cm²
   Hint: Watch units: cm².

Advance – Answers

1. Ans: s = (122+22+120)/2 = 132
   Area = √(132×10×110×12) = √1742400 = 1320 m²
   Year earning = 1320×5000 = ₹6600000
   3 months = 1/4 year ⇒ rent = ₹1650000
   Hint: 3 months means multiply yearly by 1/4.
2. Ans: Third side = 48 − (14+18) = 16 cm, s = 24 cm
   Area = √(24×10×6×8) = √11520 = 48√5 cm²
   Hint: Always check triangle inequality (it holds here).
3. Ans: s = (25+25+48)/2 = 49 cm
   Area = √(49×24×24×1) = √28224 = 168 cm²
   Hint: One factor becomes 1, so it simplifies fast.
4. Ans: Height (to base 14) = 2×Area / base = 2×84/14 = 12 cm
   Hint: Use area = 1/2 × base × height.
5. Ans: s = (9+40+41)/2 = 45
   Area = √(45×36×5×4) = √32400 = 180 cm²
   Verification: 9² + 40² = 81+1600 = 1681 = 41² ⇒ right triangle
   Area = 1/2×9×40 = 180 cm²
   Hint: Largest side is hypotenuse if Pythagoras matches.
6. Ans: Triangle possible since 7+15 > 20
   s = 21, area = √(21×14×6×1) = √1764 = 42 cm²
   Hint: Check triangle inequality first.

HOTS – Answers

1. Ans: For equilateral side a:
   s = 3a/2 and (s−a)=a/2
   Area = √( (3a/2)(a/2)(a/2)(a/2) ) = (√3/4)a²
   For a=10 ⇒ area = (√3/4)×100 = 25√3 cm²
   Hint: Same factor repeats 3 times.
2. Ans:
   Triangle A area = √(15×3×3×9) = √1215 = 9√15 cm²
   Triangle B area = √(15×5×5×5) = √1875 = 25√3 cm²
   Since 25√3 > 9√15, Triangle B has bigger area.
   Hint: Compare squares: (25√3)² = 1875 and (9√15)² = 1215.
3. Ans: 120² + 22² = 14400 + 484 = 14884 = 122² ⇒ right triangle
   Area = 1/2×120×22 = 1320 m²
   Heron also gives 1320 m² (matches).
   Hint: Use the two smaller sides as base and height.
4. Ans: Perimeter = 40+32+24 = 96 m
   Wire needed = 96 − 2 = 94 m
   Cost = 94×50 = ₹4700
   Hint: Gate length is not fenced.
5. Ans: Ratio sum = 2+3+4 = 9, perimeter 90 ⇒ 1 part = 10
   Sides: 20, 30, 40; s = 45
   Area = √(45×25×15×5) = √84375 = 75√15 cm²
   Hint: Pull out perfect squares from inside √.
6. Ans: Largest side = 20
   20² = 400 and 7² + 15² = 49 + 225 = 274
   Since 400 > 274 ⇒ triangle is obtuse
   Hint: If c² > a²+b², angle opposite c is obtuse.

Worksheet for Other Class 9 Maths Chapters

• Number Systems Class 9 Maths Worksheet Chapter 1
• Polynomials Class 9 Maths Worksheet Chapter 2
• Linear Equations in Two Variables Class 9 Maths Worksheet Chapter 4
• Introduction to Euclid’s Geometry Class 9 Maths Worksheet Chapter 5
• Surface Areas and Volumes Class 9 Maths Worksheet Chapter 11
• Statistics Class 9 Maths Worksheet Chapter 12