Heron's Formula Chapter 10 Class 9 Maths Revision Notes

Heron’s Formula is an important topic in Chapter 10 of Class 9 Maths. In this chapter, students learn how to find the area of a triangle using only the lengths of its three sides, without knowing the height. This method is very useful in geometry and also helps in solving real-life problems. These Heron's Formula Chapter 10 class 9 maths notes are designed to give clear understanding with simple steps and examples.

In heron's formula class 9, we use the concept of semi-perimeter to calculate the area. It is a part of class 9 maths heron's formula syllabus and is often asked in exams. Students can also practice from heron's formula class 9 solutions to improve problem-solving skills. All important concepts, including heron's formula class 9 all formulas, are explained in an easy way so that learners can understand without confusion.

These CBSE notes are also helpful for quick revision, and many students search for heron's formula class 9 maths notes pdf download for offline study. By learning this chapter properly, students can build a strong base in mensuration and triangle geometry. Some questions may look tricky at first, but with practice it becomes easy and intresting.

Introduction to Mensuration

Mensuration is the branch of mathematics that deals with the measurement of geometric figures their lengths, perimeters, areas, and volumes. It is one of the oldest and most practically applied branches of mathematics, forming the foundation for fields like architecture, engineering, land surveying, and design.

In Class 9 CBSE Mathematics, the chapter on Heron's Formula specifically focuses on calculating the area of triangles including those where the height is not directly given using only the lengths of the three sides. This is made possible through a powerful, elegant formula attributed to the ancient Greek mathematician Heron of Alexandria (also called Hero), who lived around 10–70 AD.

Understanding Heron's Formula equips students with a versatile tool: rather than requiring a triangle's height (which can be difficult to measure in real-world scenarios), you only need the three side lengths.

What is Perimeter? 

The perimeter of any plane figure is the total length of its boundary. For a triangle or polygon, this is simply the sum of the lengths of all its sides.

Perimeter of a Triangle = Sum of all three sides

If a triangle has sides of length a, b, and c, then:

Perimeter = a + b + c

Why Perimeter Matters

Perimeter is used to determine the total length of fencing needed around a field, the length of a picture frame, the boundary of a sports court, and much more.

Units of Perimeter

The unit of perimeter is the same as the unit of length. Here is a complete reference:

What is Area?

The area of a plane figure is the measure of the surface enclosed within its boundary. For a triangle or polygon, it is the total region covered by its sides.

Units of Area

Important: 1 hectare = 10,000 m² this conversion is frequently tested in exams.

Heron's Formula – The Core Concept The Formula

For a triangle with sides a, b, and c:

Step 1 – Find the Semi-Perimeter (s):

s = (a + b + c) / 2

Step 2 – Apply Heron's Formula:

Area = √[s(s − a)(s − b)(s − c)]

Where:

• s = semi-perimeter of the triangle
• a, b, c = lengths of the three sides

Why is it called a "Semi-Perimeter"?

Because s is exactly half the perimeter. The semi-perimeter acts as a clever averaging mechanism that makes the formula dimensionally consistent and computationally elegant.

Historical Note

Heron of Alexandria proved this formula in his work Metrica around 60 AD. The beauty of the formula lies in the fact that it requires no height measurement only the three side lengths. This makes it indispensable in surveying, construction, and navigation, where heights of triangular land plots are rarely known directly.

Types of Triangles and Their Formulas

Right-Angled Triangle

Let b = base, h = perpendicular height, d = hypotenuse.

Isosceles Right-Angled Triangle

Let a = the two equal sides.

Equilateral Triangle

All three sides equal a. All angles = 60°.

Derivation of Height:
Using the Pythagorean theorem on the half-triangle formed by the altitude:
(a/2)² + h² = a²
→ h² = a² − a²/4 = 3a²/4
→ h = (√3/2) × a

Step-by-Step Method to Apply Heron's Formula

Follow these four steps every time:

Step 1: Identify the three side lengths a, b, and c.

Step 2: Calculate the semi-perimeter:
s = (a + b + c) / 2

Step 3: Compute each factor:
(s − a), (s − b), (s − c)

Step 4: Apply the formula:
Area = √[s × (s − a) × (s − b) × (s − c)]

Pro Tip: Always check your answer using the unit. If sides are in cm, your area must be in cm². If you get a unit mismatch, re-check your working.

Heron's Formula Chapter 10 Class 9 Maths Solved Examples 

Example 1 – Basic Heron's Formula Application

Q: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.

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

Area = √[21 × (21−13) × (21−14) × (21−15)]
= √[21 × 8 × 7 × 6]
= √[7056]
= 84 cm² 

Example 2 – Triangle with Base and Altitude Relationship

Q: The area of a triangle is 30 cm². Find the base if the altitude exceeds the base by 7 cm.

Solution:
Let base = x cm, altitude = (x + 7) cm

Area = ½ × base × height
30 = ½ × x × (x + 7)
60 = x² + 7x
x² + 7x − 60 = 0
x² + 12x − 5x − 60 = 0
x(x + 12) − 5(x + 12) = 0
(x − 5)(x + 12) = 0

x = 5 or x = −12 → Since x > 0, x = 5

Base = 5 cm, Altitude = 12 cm 

Example 3 – Cost-Based Area Problem

Q: The cost of turfing a triangular field at Rs. 45 per 100 m² is Rs. 900. Find the height if double the base equals 5 times the height.

Solution:
Let height = h metres
2 × base = 5h → base = 5h/2

Area = ½ × (5h/2) × h = 5h²/4 ...(i)

Area from cost = Total cost ÷ Rate per m²
= 900 ÷ (45/100) = 900 × 100/45 = 2000 m² ...(ii)

From (i) and (ii):
5h²/4 = 2000
5h² = 8000
h² = 1600
h = 40 m 

Example 4 – Interior Point of Equilateral Triangle

Q: From an interior point of an equilateral triangle, perpendiculars to the three sides are 8 cm, 10 cm, and 11 cm. Find the area.

Solution:
Let each side = x cm. Interior point O drops perpendiculars OD = 11 cm, OE = 8 cm, OF = 10 cm.

Area of △ABC = Area(△OBC) + Area(△OCA) + Area(△OAB)
= ½·x·11 + ½·x·8 + ½·x·10
= x/2 × (11 + 8 + 10) = 29x/2 ...(i)

Also, area of equilateral triangle = (√3/4)x² ...(ii)

Setting (i) = (ii):
(√3/4)x² = 29x/2
x = (4 × 29)/(2√3) = 58/√3 cm

Area = (29/2) × (58/√3) = 841/√3 ≈ 841/1.732 ≈ 486.1 cm² 

Example 5 – Right-Angled Triangle: Perimeter

Q: The difference between the two legs of a right-angled triangle is 14 cm. Its area is 120 cm². Find the perimeter.

Solution:
Let legs be x and (x − 14):
½ × x × (x − 14) = 120
x(x − 14) = 240
x² − 14x − 240 = 0
x² − 24x + 10x − 240 = 0
(x − 24)(x + 10) = 0
x = 24 (neglecting x = −10)

Legs: 24 cm and 10 cm
Hypotenuse = √(24² + 10²) = √(576 + 100) = √676 = 26 cm
Perimeter = 24 + 10 + 26 = 60 cm 

Example 6 – Percentage Increase in Area on Doubling Sides

Q: Find the percentage increase in area of a triangle if each side is doubled.

Solution:
Let original sides = a, b, c; semi-perimeter = s
New sides = 2a, 2b, 2c; new semi-perimeter s' = 2s

Original Area (Δ) = √[s(s−a)(s−b)(s−c)]

New Area (Δ') = √[2s(2s−2a)(2s−2b)(2s−2c)]
= √[16 × s(s−a)(s−b)(s−c)]
= 4Δ

Increase = 4Δ − Δ = 3Δ
% Increase = (3Δ/Δ) × 100 = 300%

Example 7 – Umbrella Problem (Application)

Q: An umbrella is made by stitching 10 triangular pieces (5 of each colour), each measuring 20 cm, 50 cm, 50 cm. Find the cloth required per colour.

Solution:
s = (20 + 50 + 50)/2 = 60 cm

Area of one piece = √[60 × 40 × 10 × 10]
= √24000 = 200√6 cm²

Cloth per colour (5 pieces) = 5 × 200√6 = 1000√6 cm²

Example 8 – Equilateral Triangle Area

Q: Find the area and perimeter of an equilateral triangle with side 12 cm.

Solution:
Area = (√3/4) × 12² = (√3/4) × 144 = 36√3 ≈ 62.35 cm²
Perimeter = 3 × 12 = 36 cm 

Example 9 – Scalene Triangle

Q: Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm.

Solution:
s = (5 + 12 + 13)/2 = 15 cm

Area = √[15 × 10 × 3 × 2] = √900 = 30 cm²

Note: 5² + 12² = 25 + 144 = 169 = 13², so this is actually a right-angled triangle. Verification: ½ × 5 × 12 = 30 cm²

Example 10 – Isosceles Triangle

Q: Find the area of an isosceles triangle with equal sides 10 cm and base 12 cm.

Solution:
s = (10 + 10 + 12)/2 = 16 cm

Area = √[16 × 6 × 6 × 4] = √[2304] = 48 cm²

Example 11 – Land Survey Problem

Q: A triangular park has sides 120 m, 80 m, and 50 m. Find its area.

Solution:
s = (120 + 80 + 50)/2 = 125 m

Area = √[125 × 5 × 45 × 75]
= √[125 × 5 × 45 × 75]
= √[2,109,375]
≈ 1452.6 m² 

Example 12 – Triangle with Given Semi-Perimeter

Q: The semi-perimeter of a triangle is 18 cm and its sides are in the ratio 3:4:5. Find the area.

Solution:
Let sides = 3k, 4k, 5k
s = (3k + 4k + 5k)/2 = 6k = 18 → k = 3

Sides: 9 cm, 12 cm, 15 cm
Area = √[18 × 9 × 6 × 3] = √2916 = 54 cm² 

Example 13 – Finding a Side Given Area

Q: The area of an equilateral triangle is 9√3 cm². Find its side.

Solution:
(√3/4) × a² = 9√3
a² = 36
a = 6 cm 

Example 14 – Quadrilateral Divided into Triangles

Q: A quadrilateral ABCD has diagonal AC = 10 cm. Perpendiculars from B and D to AC are 5 cm and 3 cm respectively. Find the area of ABCD.

Solution:
Area of △ABC = ½ × 10 × 5 = 25 cm²
Area of △ACD = ½ × 10 × 3 = 15 cm²
Area of ABCD = 40 cm² 

Example 15 – Isosceles Right-Angled Triangle

Q: The hypotenuse of an isosceles right-angled triangle is 10√2 cm. Find its area.

Solution:
Hypotenuse = a√2 = 10√2 → a = 10 cm
Area = ½ × a² = ½ × 100 = 50 cm² 

Example 16 – Cost of Painting a Triangular Wall

Q: A triangular wall has sides 7 m, 8 m, and 9 m. Cost of painting is Rs. 50 per m². Find the total cost.

Solution:
s = (7 + 8 + 9)/2 = 12 m

Area = √[12 × 5 × 4 × 3] = √720 = 6√20 = 12√5 ≈ 26.83 m²

Cost = 26.83 × 50 ≈ Rs. 1341.6

Example 17 – Percentage Change with Side Increase

Q: Each side of an equilateral triangle is increased by 10%. Find the percentage increase in area.

Solution:
Original Area = (√3/4)a²
New side = 1.1a
New Area = (√3/4)(1.1a)² = (√3/4) × 1.21a²

Increase = (1.21 − 1)/1 × 100 = 21%

Example 18 – Combined Area Problem

Q: A rhombus-shaped field has diagonals 12 m and 16 m. Find the cost of fencing at Rs. 30/m.

Solution:
Side of rhombus = √[(12/2)² + (16/2)²] = √[36 + 64] = √100 = 10 m
Perimeter = 4 × 10 = 40 m
Cost = 40 × 30 = Rs. 1200 

Example 19 – Trapezium Using Triangles

Q: A trapezium PQRS has PQ ∥ RS. PQ = 18 cm, RS = 10 cm, and the diagonal PR = 14 cm. QR = 12 cm. Find the area.

Solution:
Draw QT ⊥ PR (height of △PQR).
Use Heron's formula on △PQR with sides 18, 12, 14:
s = (18 + 12 + 14)/2 = 22
Area(△PQR) = √[22 × 4 × 10 × 8] = √7040 ≈ 83.9 cm²
Height from Q = 2 × 83.9/18 ≈ 9.32 cm
Area of trapezium = ½ × (18 + 10) × 9.32 ≈ 130.5 cm²

Example 20 – Real-World Application: Triangular Plot

Q: A triangular plot of land has sides 50 m, 60 m, and 70 m. A farmer wants to grow grass on it. If 1 kg of seed covers 10 m², how many kg of seed is needed?

Solution:
s = (50 + 60 + 70)/2 = 90 m

Area = √[90 × 40 × 30 × 20]
= √[2,160,000]
= 1469.7 m² (approx.)

Seeds needed = 1469.7 / 10 ≈ 147 kg

Heron's Formula Chapter 10 Class 9 Maths Formulas Sheet

s = (a + b + c) / 2 for all Heron's Formula applications.

Heron's Formula Chapter 10 Class 9 Maths Concepts to Remember for Exams

• The semi-perimeter (s) is always half the perimeter not the full perimeter.
• Heron's Formula works for all triangles scalene, isosceles, equilateral, right-angled without needing the height.
• When each side of a triangle is doubled, its area becomes 4 times the original (300% increase).
• When each side is scaled by factor k, the area scales by k².
• Always verify that the given sides can form a triangle: the sum of any two sides must be greater than the third side.
• If the answer under the square root is a perfect square, the area will be a rational number.
• For equilateral triangles on a test, the formula (√3/4)a² is faster than Heron's use it when you can.
• In cost problems, always convert the rate to cost per m² before applying to the area.