Heron's Formula
CBSE Class 9 · Mathematics · Notes, formulas and practice questions
Revise Heron's formula for Class 9: find the area of any triangle from its three sides, see its special cases for equilateral and isosceles triangles, and apply it to quadrilaterals by splitting them along a diagonal.
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Most area questions you previously solved needed a base and an altitude. When you know all three sides of a triangle but not its height, the direct formula ½ × base × height fails. Heron's formula comes to the rescue: it needs only the three sides and a quantity called the semi-perimeter. The chapter introduces this formula, shows why semi-perimeter is convenient, and then uses it in triangles with special shapes. It also demonstrates that many real-life plots are not right triangles or rectangles, so Heron's formula is a powerful tool.
The key steps are always the same: first find s = (a + b + c)/2, then substitute into A = √(s(s − a)(s − b)(s − c)). For a scalene triangle this is the whole procedure. For an equilateral triangle of side a, substituting the equal sides makes the algebra collapse to A = √3 a² / 4. For an isosceles triangle you may either substitute the three lengths directly or use the altitude and the base; the Heron method is fully consistent with that earlier calculation.
The most important application in this chapter is to quadrilaterals. A plot or field in the shape of a quadrilateral is divided into two triangles by one of its diagonals. If the diagonal is not directly given, it is often easy to find because two adjacent sides meet at a right angle and Pythagoras' theorem applies. Once the diagonal is known, each triangle may have all three sides known, so Heron's formula is applied separately to them, and the two areas are added. This gives the total area of the quadrilateral.
Students commonly mix up units, forget to take the square root after multiplying, or use the formula even when the sides do not form a triangle. Always remember that 2 + 3 < 6 gives no triangle, so Heron's formula simply cannot be used. In calculations, it is wise to simplify radical products step by step rather than trying to compute large numbers in one go. Carefully keep the square units in the answer: sides in metres give area in square metres, sides in centimetres give area in square centimetres.
Key terms
- Semi-perimeter
- Half the perimeter of a triangle, denoted by the letter s. For a triangle with side lengths a, b and c, s = (a + b + c)/2. All expressions under Heron's formula use s because it makes the formula symmetric and easier to handle.
- Heron's formula
- A formula that gives the area of a triangle from the lengths of its three sides. If the sides are a, b, c and s is the semi-perimeter, then area = √(s(s − a)(s − b)(s − c)). It is valid for every triangle whose sides satisfy the triangle inequality.
- Equilateral triangle
- A triangle in which all three sides are equal. Let each side be a. Then s = 3a/2, and using Heron's formula the area simplifies to A = √3 a² / 4. You may use this shortcut instead of substituting step by step.
- Isosceles triangle
- A triangle with two sides of equal length. In this chapter, if the equal sides are l and the base is b, you can substitute a = l, b = l, c = b directly into Heron's formula. The answer agrees with the area found via altitude.
- Triangle inequality
- The fundamental condition for three lengths to form a triangle: the sum of any two sides must be greater than the third side. If this fails, no triangle exists, and Heron's formula is not applicable because the expression under the square root could be negative.
- Diagonal of a quadrilateral
- The segment joining two opposite vertices. To find the area of a quadrilateral using Heron's formula, draw one diagonal so that the quadrilateral becomes two triangles. If the diagonal is unknown, compute it from a known right angle using Pythagoras' theorem, or use a given measurement.
- Square units
- Area is always measured in square units. If side lengths are in centimetres, the area is in cm²; if in metres, the area is in m². Convert all side lengths to the same unit before applying Heron's formula.
- Radical expression under Heron's formula
- The product s(s − a)(s − b)(s − c) appears inside the square root sign. For an actual triangle this product is positive, and its square root is the area. For a degenerate (collinear) triangle the product is zero; for impossible sides it is negative.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Semi-perimeter of a triangle | s = (a + b + c) / 2 | a, b and c are the lengths of the three sides. s is half the perimeter and is always positive. |
| Area of a triangle using Heron's formula | A = √(s(s − a)(s − b)(s − c)) | s is the semi-perimeter from above; a, b and c are the three side lengths. The sides must satisfy the triangle inequality, and all lengths must be in the same unit. A is the area in square units of that unit. |
| Area of an equilateral triangle (derived from Heron's formula) | A = (√3 a²) / 4 | a is the side of the equilateral triangle. This result comes from putting a = b = c = a in Heron's formula and is valid only for equilateral triangles. |
Practice questions with answers
1. Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.
Semi-perimeter s = (13 + 14 + 15)/2 = 21 cm. Using Heron's formula, area = √(21 × (21 − 13) × (21 − 14) × (21 − 15)) = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
2. Find the area of an equilateral triangle of side 6 cm.
Here a = 6 cm, so s = 18/2 = 9 cm. Heron's formula gives area = √(9 × 3 × 3 × 3) = √243 = 9√3 cm². The shortcut A = √3 a² / 4 also gives A = √3 × 36 / 4 = 9√3 cm².
3. Find the area of an isosceles triangle whose equal sides are 10 cm each and whose base is 12 cm.
Let the sides be 10, 10 and 12. Then s = (10 + 10 + 12)/2 = 16 cm, so area = √(16 × 6 × 6 × 4) = √2304 = 48 cm². As a check, the altitude to the base is √(10² − 6²) = 8 cm, and area = ½ × 12 × 8 = 48 cm².
4. A triangle has sides 10 cm, 24 cm and 26 cm. Verify that it is right-angled and find its area by Heron's formula.
Since 10² + 24² = 100 + 576 = 676 = 26², the sides form a right triangle with hypotenuse 26 cm. Using Heron's formula, s = 30 and area = √(30 × 20 × 6 × 4) = √14400 = 120 cm². This agrees with the usual area ½ × 10 × 24 = 120 cm².
5. For a triangle with sides 5 cm, 7 cm and 8 cm, write down the semi-perimeter and the value of s(s − a)(s − b)(s − c).
Semi-perimeter s = (5 + 7 + 8)/2 = 10 cm. Taking a = 5, b = 7 and c = 8 gives s − a = 5, s − b = 3 and s − c = 2, so s(s − a)(s − b)(s − c) = 10 × 5 × 3 × 2 = 300.
6. A quadrilateral park ABCD has AB = 5 m, BC = 12 m, CD = 15 m, DA = 14 m, and ∠ABC = 90°. Find the area of the park.
In right triangle ABC, diagonal AC = √(5² + 12²) = √169 = 13 m, and area(ABC) = ½ × 5 × 12 = 30 m². In triangle ACD, sides are 13 m, 14 m and 15 m; semi-perimeter s = 21 m, so area(ACD) = √(21 × 8 × 7 × 6) = 84 m². Total area of the park is 30 + 84 = 114 m².
7. The sides of a triangular lawn are 20 m, 34 m and 42 m. Find its area and the cost of levelling it at ₹5 per square metre.
Semi-perimeter s = (20 + 34 + 42)/2 = 48 m. Area = √(48 × 28 × 14 × 6) = √112896 = 336 m². Levelling cost = 336 × ₹5 = ₹1680.
8. Why can Heron's formula not be used for sides 2 cm, 3 cm and 6 cm?
The given sides do not form a triangle because the sum of the two smaller sides, 2 + 3 = 5, is less than 6. If one naively substitutes them, s = 5.5, giving the factor s − 6 = −0.5, so the product inside the square root is negative. Hence there is no real area.
9. The semi-perimeter of a triangle is 14 cm, and two sides are 8 cm and 9 cm. Find the third side and the area.
Let the third side be x. From s = 14, (8 + 9 + x)/2 = 14, so 17 + x = 28 and x = 11 cm. The sides are 8, 9 and 11. Since s = 14, the factors are s − 8 = 6, s − 9 = 5 and s − 11 = 3, giving area = √(14 × 6 × 5 × 3) = √1260 = 6√35 cm².
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