Applications of Heron's Formula
Heron's Formula is not limited to finding the area of a single triangle. It extends to a wide range of practical problems involving quadrilaterals, irregular polygons, and real-life shapes.
The key technique is to divide complex shapes into triangles using diagonals, and then apply Heron's Formula to each triangle separately. The total area is the sum of the individual triangular areas.
In Class 9, these application problems appear in NCERT as word problems involving land plots, signboards, parks, and other real-world scenarios where side lengths are known but heights are not directly available.
What is Applications of Heron's Formula?
Area = √[s(s − a)(s − b)(s − c)]
where s = (a + b + c) / 2 is the semi-perimeter.
Application Strategy:
- For quadrilaterals: Divide into two triangles using a diagonal. Apply Heron's Formula to each.
- For polygons: Divide into multiple triangles from a single vertex or using diagonals.
- For word problems: Extract the side lengths from the problem, identify the shape, and apply the formula.
Applications of Heron's Formula Formula
Key Formulas Used:
1. Heron's Formula for each triangle:
- s = (a + b + c) / 2
- Area = √[s(s − a)(s − b)(s − c)]
2. Area of a quadrilateral:
Area = Area(▵1) + Area(▵2)
where the quadrilateral is divided by a diagonal into two triangles.
3. Area of equilateral triangle:
- Area = (√3/4) × a²
4. Cost or rate problems:
- Total cost = Area × Rate per unit area
Derivation and Proof
Method for Quadrilateral Area:
- Identify all four sides and at least one diagonal of the quadrilateral.
- The diagonal divides the quadrilateral into two triangles.
- For each triangle, calculate the semi-perimeter and apply Heron's Formula.
- Add the two areas to get the total area of the quadrilateral.
Method for Irregular Polygon Area:
- Choose a vertex and draw diagonals from it to all non-adjacent vertices.
- This divides an n-sided polygon into (n − 2) triangles.
- Apply Heron's Formula to each triangle.
- Sum all the areas.
Important: You need to know the lengths of the diagonals as well as the sides to apply this method.
Types and Properties
Types of Application Problems:
1. Quadrilateral Area Problems
- Given four sides and one diagonal.
- Divide into two triangles and use Heron's Formula on each.
2. Land and Field Problems
- Irregular plots of land with known boundary lengths.
- Find area for cost estimation (planting, fencing, paving).
3. Signboard and Painting Problems
- Triangular or quadrilateral signboards.
- Find area and multiply by cost per sq unit.
4. Park and Garden Problems
- Triangular parks, gardens, flower beds.
- Area needed for landscaping, grass planting, or tiling.
5. Verification Problems
- Verify the area obtained by Heron's Formula using the base × height method (for right triangles).
Solved Examples
Example 1: Example 1: Area of a quadrilateral field
Problem: A field ABCD has sides AB = 9 m, BC = 40 m, CD = 28 m, DA = 15 m, and diagonal AC = 41 m. Find the area.
Solution:
Triangle ABC (sides 9, 40, 41):
- s = (9 + 40 + 41)/2 = 45
- Area = √(45 × 36 × 5 × 4) = √32400 = 180 sq m
Triangle ACD (sides 41, 28, 15):
- s = (41 + 28 + 15)/2 = 42
- Area = √(42 × 1 × 14 × 27) = √15876 = 126 sq m
Total area = 180 + 126 = 306 sq m
Answer: The area of the field is 306 sq m.
Example 2: Example 2: Cost of planting grass
Problem: A triangular park has sides 120 m, 80 m, and 100 m. Find the cost of planting grass at Rs 8 per sq m.
Solution:
Given: a = 120, b = 80, c = 100
- s = (120 + 80 + 100)/2 = 150
- Area = √(150 × 30 × 70 × 50) = √15750000
- = √(15750000) = 50√6300 = 50 × 30√7 ≈ 3968.63 sq m
Cost = 3968.63 × 8 ≈ Rs 31,749
Answer: The cost of planting grass is approximately Rs 31,749.
Example 3: Example 3: Triangular signboard painting
Problem: A triangular signboard has sides 26 cm, 28 cm, and 30 cm. Find the cost of painting it on both sides at Rs 5 per sq cm.
Solution:
- s = (26 + 28 + 30)/2 = 42
- Area = √(42 × 16 × 14 × 12) = √112896 = 336 sq cm
- Both sides = 2 × 336 = 672 sq cm
- Cost = 672 × 5 = Rs 3,360
Answer: The cost of painting is Rs 3,360.
Example 4: Example 4: Rhombus area using Heron's Formula
Problem: A rhombus has side 13 cm and one diagonal 24 cm. Find its area.
Solution:
The diagonal divides the rhombus into two congruent triangles, each with sides 13, 13, 24.
- s = (13 + 13 + 24)/2 = 25
- Area of one triangle = √(25 × 12 × 12 × 1) = √3600 = 60 sq cm
- Area of rhombus = 2 × 60 = 120 sq cm
Answer: The area of the rhombus is 120 sq cm.
Example 5: Example 5: Trapezium area
Problem: A trapezium ABCD has AB = 7 cm, BC = 5 cm, CD = 13 cm, DA = 9 cm, and diagonal BD = 12 cm. Find the area.
Solution:
Triangle ABD (sides 7, 12, 9):
- s = (7 + 12 + 9)/2 = 14
- Area = √(14 × 7 × 2 × 5) = √980 = 14√5 ≈ 31.30 sq cm
Triangle BCD (sides 12, 5, 13):
- s = (12 + 5 + 13)/2 = 15
- Area = √(15 × 3 × 10 × 2) = √900 = 30 sq cm
Total area = 31.30 + 30 = 61.30 sq cm
Answer: The area of the trapezium is approximately 61.30 sq cm.
Example 6: Example 6: Irregular land plot
Problem: A plot of land is in the shape of a quadrilateral with sides 20 m, 30 m, 40 m, 50 m, and one diagonal of 32 m. Find the area.
Solution:
Triangle 1 (sides 20, 30, 32):
- s = (20 + 30 + 32)/2 = 41
- Area = √(41 × 21 × 11 × 9) = √85239 ≈ 292.0 sq m
Triangle 2 (sides 32, 40, 50):
- s = (32 + 40 + 50)/2 = 61
- Area = √(61 × 29 × 21 × 11) = √408639 ≈ 639.2 sq m
Total area ≈ 292.0 + 639.2 = 931.2 sq m
Answer: The area is approximately 931.2 sq m.
Example 7: Example 7: Equilateral triangle flowerbed
Problem: An equilateral triangular flowerbed has perimeter 36 m. Find the area using Heron's Formula and the cost of planting at Rs 15 per sq m.
Solution:
Given: Perimeter = 36 m, so side = 12 m.
- s = 36/2 = 18
- Area = √(18 × 6 × 6 × 6) = √3888 = 36√3 ≈ 62.35 sq m
- Cost = 62.35 × 15 ≈ Rs 935.25
Answer: Area ≈ 62.35 sq m; Cost ≈ Rs 935.
Example 8: Example 8: Fencing cost problem
Problem: A triangular plot has sides 50 m, 60 m, and 70 m. Find the cost of fencing at Rs 25 per metre and the cost of levelling at Rs 10 per sq m.
Solution:
Fencing cost:
- Perimeter = 50 + 60 + 70 = 180 m
- Fencing cost = 180 × 25 = Rs 4,500
Levelling cost:
- s = 180/2 = 90
- Area = √(90 × 40 × 30 × 20) = √2160000 = 1469.69 sq m (approx.)
- Levelling cost = 1469.69 × 10 ≈ Rs 14,697
Answer: Fencing = Rs 4,500; Levelling ≈ Rs 14,697.
Real-World Applications
Real-World Applications:
- Land Revenue and Surveying: Calculating the area of irregularly shaped fields for tax assessment and property registration.
- Construction Cost Estimation: Estimating material requirements for flooring, painting, and roofing triangular and quadrilateral surfaces.
- Agriculture: Determining the area of triangular and quadrilateral farm plots for seed, fertiliser, and irrigation planning.
- Park and Garden Design: Computing the area of flower beds, pathways, and lawns for landscaping budgets.
- Signage and Advertising: Calculating the area of triangular and irregularly shaped advertising boards for cost of painting or printing.
Key Points to Remember
- Heron's Formula can be applied to any triangle when three sides are known.
- For quadrilaterals, divide into two triangles using a diagonal and apply Heron's Formula to each.
- For polygons, divide into (n − 2) triangles from one vertex.
- You need side lengths and diagonal lengths to apply the method to quadrilaterals.
- Cost problems: Total cost = Area × Rate per unit area.
- Fencing problems: Total cost = Perimeter × Rate per unit length.
- Always check that the sides satisfy the triangle inequality before applying the formula.
- A rhombus can be split into two congruent triangles by either diagonal.
- These application problems are frequently asked in CBSE Class 9 examinations.
Practice Problems
- A quadrilateral ABCD has AB = 5 cm, BC = 12 cm, CD = 14 cm, DA = 9 cm, and diagonal AC = 13 cm. Find its area.
- A triangular garden has sides 100 m, 120 m, and 140 m. Find the cost of planting grass at Rs 6 per sq m.
- A rhombus has side 10 cm and one diagonal 16 cm. Find the area using Heron's Formula.
- A parallelogram has sides 10 cm and 14 cm with a diagonal of 16 cm. Find the area.
- A field in the shape of a trapezium has parallel sides 12 m and 18 m, and non-parallel sides 8 m and 10 m. If a diagonal is 15 m, find the area.
- The perimeter of a triangular plot is 300 m and its sides are in the ratio 3:5:7. Find the area and cost of fencing at Rs 20 per metre.
Frequently Asked Questions
Q1. How is Heron's Formula applied to quadrilaterals?
Divide the quadrilateral into two triangles using a diagonal. Apply Heron's Formula to each triangle and add the areas. You need all four sides and at least one diagonal.
Q2. Can Heron's Formula be used for irregular shapes?
Yes. Any polygon can be divided into triangles. As long as all side lengths and necessary diagonals are known, Heron's Formula can be applied to each triangle.
Q3. What data is needed to use Heron's Formula for a quadrilateral?
You need the lengths of all four sides and at least one diagonal. The diagonal divides the quadrilateral into two triangles, each with three known sides.
Q4. How do you find the cost of painting a triangular surface?
Find the area using Heron's Formula. Multiply by the rate per square unit. If painted on both sides, multiply the area by 2 first.
Q5. Is this topic important for CBSE Class 9 exams?
Yes. Application problems of Heron's Formula are a standard question type in CBSE Class 9 exams, typically carrying 3-5 marks.
Q6. Can the area of a parallelogram be found using Heron's Formula?
Yes. A diagonal divides a parallelogram into two congruent triangles. Find the area of one triangle using Heron's Formula and multiply by 2.
Q7. What if the diagonal length is not given for a quadrilateral?
Without a diagonal, you cannot split the quadrilateral into triangles with all sides known. Additional information (such as an angle or the other diagonal) is needed.
Q8. What is the difference between fencing and levelling cost problems?
Fencing cost depends on the perimeter (boundary length). Levelling or planting cost depends on the area (surface). Use perimeter for fencing and Heron's Formula for area-based costs.
