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Half base times height is easy when you know the height. Often you do not: a surveyor measuring a triangular plot can walk its three sides but cannot easily measure a height across it. Heron's formula solves this by working from the sides alone. This guide explains the formula, works four examples you can check, and shows the link to Brahmagupta's formula that the new syllabus adds.
Where is Heron's formula in the 2026-27 syllabus?
In NCERT's new textbook Ganita Manjari, Heron's formula is section 6.8.1 of Chapter 6, Measuring Space: Perimeter and Area. The same chapter covers the perimeter and area of circles, arcs and sectors, and Brahmagupta's formula for the area of a cyclic quadrilateral. The CBSE Class IX curriculum for 2026-27 lists Heron's formula, Brahmagupta's formula and "Heron's formula as a special case of Brahmagupta's formula" under Area and Perimeter. The Mensuration unit carries 14 of the 80 theory marks.
Heron of Alexandria was a geometer and inventor. The MacTutor biography at the University of St Andrews dates him to about AD 10 to 75, using a lunar eclipse in AD 62 that he mentions in one of his works, and notes that Heron proves the formula in Book I of his work on measurement, the Metrica.
What is the formula?
For a triangle with sides a, b and c:
Find the semi-perimeter, half the perimeter: s = (a + b + c)/2.
Area = square root of [s(s − a)(s − b)(s − c)].
The NCERT chapter checks the formula on an equilateral triangle, an isosceles triangle and the 3-4-5 right triangle, and says proofs will come in Grade 10.
Worked example 1: the 13-14-15 triangle
Sides 13 cm, 14 cm and 15 cm.
s = (13 + 14 + 15)/2 = 21.
s − a = 8, s − b = 7, s − c = 6.
Area = square root of (21 × 8 × 7 × 6) = square root of 7056 = 84 cm².
A tip for the arithmetic: factorise before multiplying. 21 × 8 × 7 × 6 = (3 × 7) × (2 × 2 × 2) × 7 × (2 × 3) = 2⁴ × 3² × 7², so the square root is 2² × 3 × 7 = 84.
Worked example 2: an answer that is not a whole number
Sides 5 cm, 6 cm and 7 cm.
s = 9, and s − a = 4, s − b = 3, s − c = 2.
Area = square root of (9 × 4 × 3 × 2) = square root of 216 = 6√6, about 14.70 cm².
Leave the answer as 6√6 unless the question asks for a decimal, and round only at the end.
Worked example 3: check against half base times height
An isosceles triangle has sides 10 cm, 10 cm and 12 cm.
Heron: s = 16, so area = square root of (16 × 6 × 6 × 4) = square root of 2304 = 48 cm².
Check: the height to the 12 cm base splits it into 6 and 6, so height = square root of (100 − 36) = 8. Half base times height = (1/2) × 12 × 8 = 48 cm².
The two methods agree, which is a good way to test your arithmetic on any triangle where you can find a height.
Worked example 4: a four-sided field
Quadrilateral ABCD has AB = 3 m, BC = 4 m, CD = 12 m, DA = 13 m and diagonal AC = 5 m. Find its area.
Split along the diagonal into triangles ABC (3, 4, 5) and ACD (5, 12, 13).
Triangle ABC: s = 6, area = square root of (6 × 3 × 2 × 1) = 6 m².
Triangle ACD: s = 15, area = square root of (15 × 10 × 3 × 2) = square root of 900 = 30 m².
Total = 36 m².
How does Brahmagupta's formula extend it?
For a cyclic quadrilateral (one whose vertices lie on a circle) with sides a, b, c and d and s = (a + b + c + d)/2, Brahmagupta's formula gives area = square root of [(s − a)(s − b)(s − c)(s − d)]. Set d = 0, so the quadrilateral collapses to a triangle, and s − d becomes s: you get Heron's formula. The chapter uses this to explain what a "special case" and a "generalisation" mean. Quick check with a square of side a: s = 2a, so the area is the square root of a⁴, which is a².
What mistakes cost marks?
Using the perimeter instead of the semi-perimeter. Always halve.
Rounding early. Keep exact values until the last step.
Missing units. Area is in square units (cm², m²).
Forgetting to check the sides make a triangle.
A practice routine for Heron's formula
Mixed practice trains you to choose a method. Rohrer and Taylor (2007) found that shuffling mathematics problems improved learning compared with practising one type at a time. So mix Heron questions with half-base-times-height ones:
Do six area problems: three where a height is easy to find and three where only the sides are given. Choose the method before starting each.
For two of them, solve both ways and compare.
Two days later, do one split-quadrilateral problem and one Brahmagupta check without notes.
If the arithmetic under the square root keeps going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. Heron's formula feeds into areas related to circles in Class 10 and sits alongside surface areas and volumes in Class 9.
Try worked example 1 without looking, factorising before you multiply, and check that you get 84.
