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.
Answer: Square (side a): s = 2a, √(a · a · a · a) = a2 ✓. Kite with two right angles, sides 3, 4, 4, 3: s = 7, √(4 × 3 × 3 × 4) = 12 = 2 × 12 × 3 × 4 ✓. Triangle (d = 0): it becomes Heron’s formula ✓.
Step-by-step solution
Idea: Choose shapes that are cyclic and whose area we already know another way, then compare.
- Square of side a: cyclic. s = 12(4a) = 2a, each s − side = a. Area = √(a4) = a2 ✓.1 mark
- Kite ABCD with AB = AD = 3, CB = CD = 4 and right angles at B and D: ∠B + ∠D = 180°, so it is cyclic. Brahmagupta: s = 12(3 + 4 + 4 + 3) = 7, area = √((7 − 3)(7 − 4)(7 − 4)(7 − 3)) = √144 = 12. Directly: AC splits it into two right triangles of area 12 × 3 × 4 = 6, total 12 ✓.1½ marks
- Triangle (fourth side 0): a triangle is always cyclic, and putting d = 0 gives √(s(s − a)(s − b)(s − c)), which is Heron’s formula ✓.½ mark
The formula checks out: square a², right-angled kite 3, 4, 4, 3 gives 12 (correct), and with d = 0 it becomes Heron’s formula.
Answer to write in the exam
Square: s = 2a; √(a · a · a · a) = a2 ✓
Kite 3, 4, 4, 3 (∠B = ∠D = 90°, cyclic): s = 7; √(4 × 3 × 3 × 4) = 12
Direct: 2 × 12 × 3 × 4 = 12 ✓
d = 0: √(s(s − a)(s − b)(s − c)) = Heron ✓
∴ The formula gives the correct area in each special case.
Common mistakes that cost marks
- Testing the formula on a non-cyclic 4-gon such as a leaning rhombus. It gives the wrong area there, because the formula needs the 4-gon to be cyclic.
- Adding a fifth factor s as in Heron’s formula: Brahmagupta’s formula has only the four factors (s − a)(s − b)(s − c)(s − d).
How this can come in the exam
MCQ (1 mark)
A cyclic 4-gon has sides 5, 5, 5, 5. By Brahmagupta’s formula its area is
- 10
- 20
- 25
- 625
Show answer
(C) 25
s = 10; √(5 × 5 × 5 × 5) = 25. (A cyclic rhombus must be a square.)
Try one yourself
A kite has sides 5, 12, 12, 5 with right angles between the 5 and 12 sides. Find its area with Brahmagupta’s formula.
Show answer
s = 17; √(12 × 5 × 5 × 12) = 60 (= 2 × 12 × 5 × 12 ✓).
More questions like this
- 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?
- Try to work out why this method works. You will find that it is a geometrical translation of the formula (a + b2)2 − (a − b2)2 = ab.
- What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?
- Find the area of triangle ADE in the figure.