Let us test Heron’s formula against some known cases: a triangle with sides 3 units, 4 units and 5 units.
Answer: s = 6, area = √(6 × 3 × 2 × 1) = √36 = 6 sq. units. Since 32 + 42 = 52, it is right-angled, and 12 × 3 × 4 = 6 too.
Step-by-step solution
Idea: Heron’s formula first; then notice 32 + 42 = 52, so the triangle is right-angled and its legs are a base and a height.
- s = 12(3 + 4 + 5) = 6 units.½ mark
- Area = √(6 × (6 − 3) × (6 − 4) × (6 − 5)) = √(6 × 3 × 2 × 1) = √36 = 6 sq. units.1 mark
- 32 + 42 = 9 + 16 = 25 = 52, so by the converse of the Baudhāyana–Pythagoras theorem the triangle is right-angled (at the vertex between the sides 3 and 4).1 mark
- Area = 12 × 3 × 4 = 6 sq. units, the same as Heron’s formula.½ mark
Area = 6 sq. units by both methods.
Check: Both methods agree ✓.
Answer to write in the exam
s = 12(3 + 4 + 5) = 6
Area = √(6 × 3 × 2 × 1) = √36 = 6
32 + 42 = 52 ⇒ right-angled (converse of Baudhāyana–Pythagoras)
Area = 12 × 3 × 4 = 6
∴ Area = 6 sq. units
Common mistakes that cost marks
- Multiplying 6 × 3 × 2 × 1 and forgetting the square root (giving 36).
- Taking the hypotenuse 5 as base with height 4.
How this can come in the exam
MCQ (1 mark)
Using Heron’s formula, the area of a triangle with sides 5 cm, 12 cm, 13 cm is
- 30 cm2
- 60 cm2
- 65 cm2
- 78 cm2
Show answer
(A) 30 cm2
s = 15: √(15 × 10 × 3 × 2) = √900 = 30 cm2 (= 12 × 5 × 12).
Try one yourself
Find the area of a triangle with sides 6 cm, 8 cm and 10 cm by Heron’s formula, and check it another way.
Show answer
s = 12: √(12 × 6 × 4 × 2) = √576 = 24 cm2; right-angled, 12 × 6 × 8 = 24 ✓.
More questions like this
- In the same way we ask: can we find the area of a 4-gon if we only know the lengths of its sides? The figures below reveal the answer to this question. The problem (see the figure) is about a 4-gon whose sides are known to be 3, 3, 3, 3 (it is a ‘rhombus’). As you can see, the areas of the three figures are different. (We drew the figures using GeoGebra and found the areas using the ‘Area’ tool. Please try this exercise yourself, or by using four rods joined together at their ends.)
- Verify Brahmagupta’s formula for the case of a rectangle.
- The formula states that if the sides of the cyclic 4-gon have lengths a, b, c, d, and the semi-perimeter s is s = 12(a + b + c + d), then: Area of 4-gon = √((s − a)(s − b)(s − c)(s − d)). Please check out the formula against other special cases.
- Verify Brahmagupta’s formula for the case of an isosceles trapezium.
- Here is another example. Compare the following identities from algebra: (a + b)2 = a2 + b2 + 2ab, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca. Can you see that the first identity is a special case of the second one (put c = 0 in the second identity), and the second identity is a generalisation of the first one?