Let us test Heron’s formula against some known cases: an equilateral triangle with side a units.
Answer: s = 3a2, so area = √(3a2 × a2 × a2 × a2) = √34a2 sq. units, the same as 12 × base × height.
Step-by-step solution
Idea: Heron’s formula: area = √(s(s − a)(s − b)(s − c)) with s = half the perimeter. Check it against 12 × base × height, with the height from the Baudhāyana–Pythagoras theorem.
- All three sides are a, so s = 12(a + a + a) = 32a units, and s − a = 12a for each side.½ mark
- Heron: area = √(32a × 12a × 12a × 12a) = √(316a4) = √34a2 sq. units.1 mark
- Check: drop the height h from A to the midpoint D of BC. Then a2 − h2 = a24, so h2 = 3a24 and h = √32a.1 mark
- Area = 12 × a × √32a = √34a2 sq. units. Same formula!½ mark
Area of an equilateral triangle of side a = (√3/4)a² sq. units, by Heron’s formula and by ½ × base × height.
Check: For a = 2: Heron gives √(3 × 1 × 1 × 1) = √3, and √34 × 4 = √3 ✓.
Answer to write in the exam
s = 12(a + a + a) = 3a2
Area = √(s(s − a)3) = √(3a2 × a2 × a2 × a2) = √34a2
Check: h2 = a2 − a24 = 3a24 ⇒ h = √32a
Area = 12 × a × √32a = √34a2
∴ Both give √34a2 sq. units.
Common mistakes that cost marks
- Using the perimeter 3a instead of the semi-perimeter 3a2 for s.
- Forgetting the square root in Heron’s formula.
- Taking the height as a (the slant side).
How this can come in the exam
MCQ (1 mark)
The area of an equilateral triangle of side 6 cm is
- 9√3 cm2
- 18 cm2
- 36√3 cm2
- 6√3 cm2
Show answer
(A) 9√3 cm2
√34 × 36 = 9√3 cm2 (Heron: √(9 × 3 × 3 × 3) = 9√3).
Try one yourself
Use Heron’s formula to find the area of an equilateral triangle of side 4 cm.
Show answer
s = 6; √(6 × 2 × 2 × 2) = √48 = 4√3 cm2 ≈ 6.93 cm2.
More questions like this
- Let us test Heron’s formula against some known cases: an isosceles triangle with equal sides a units and base 2b units.
- Let us test Heron’s formula against some known cases: a triangle with sides 3 units, 4 units and 5 units.
- 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.