Let us test Heron’s formula against some known cases: an isosceles triangle with equal sides a units and base 2b units.
Answer: s = a + b, so area = √((a + b)(b)(b)(a − b)) = b√(a2 − b2) sq. units, the same as 12 × base × height.
Step-by-step solution
Idea: Apply Heron’s formula, then check with the height from the apex, which meets the base at its midpoint.
- s = 12(a + a + 2b) = a + b. So s − a = b (twice) and s − 2b = a − b.½ mark
- Area = √((a + b)(a + b − a)(a + b − a)(a + b − 2b)) = √(b2(a + b)(a − b)) = b√(a2 − b2).1 mark
- Check: the height h from A meets BC at its midpoint D, so BD = b. Baudhāyana–Pythagoras: a2 − h2 = b2, so h = √(a2 − b2).1 mark
- Area = 12 × 2b × √(a2 − b2) = b√(a2 − b2) sq. units, the same as before.½ mark
Area = b√(a² − b²) sq. units by both methods.
Check: a = 5, b = 3 (sides 5, 5, 6): b√(a2 − b2) = 3 × 4 = 12; Heron with s = 8: √(8 × 3 × 3 × 2) = √144 = 12 ✓.
Answer to write in the exam
s = 12(a + a + 2b) = a + b
Area = √((a + b)(b)(b)(a − b)) = b√(a2 − b2)
Check: h2 = a2 − b2 ⇒ h = √(a2 − b2)
Area = 12 × 2b × √(a2 − b2)
∴ Area = b√(a2 − b2) sq. units
Common mistakes that cost marks
- Taking half the base as 2b. The base is 2b, so half of it is b.
- Writing √(a2 − b2) = a − b, which is false.
- Using s = 2a + 2b (the whole perimeter).
How this can come in the exam
MCQ (1 mark)
An isosceles triangle has equal sides 61 cm and base 22 cm. Its area is
- 330 cm2
- 671 cm2
- 1320 cm2
- 660 cm2
Show answer
(D) 660 cm2
b = 11: 11 × √(3721 − 121) = 11 × √3600 = 11 × 60 = 660 cm2.
Try one yourself
Find the area of an isosceles triangle with equal sides 17 cm and base 16 cm.
Show answer
b = 8: 8 × √(289 − 64) = 8 × 15 = 120 cm2.
More questions like this
- 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.
- Verify Brahmagupta’s formula for the case of an isosceles trapezium.