Theory
The area of a four-cornered figure on a circle
You know how to find the area of a triangle or a rectangle. But what about a general four-sided figure whose four corners all lie on a circle, a cyclic quadrilateral? Its area is not obvious. The Indian mathematician Brahmagupta gave an elegant formula for exactly this case, using only the four side lengths.
This topic covers Brahmagupta and his documented area formula, teaching the real mathematics with a clean worked example, and handling the history with the same care as the rest of this unit.
Watch out
Careful, factual framing
We name the author and text accurately, Brahmagupta and his treatise the Brahmasphutasiddhanta, and present only the DOCUMENTED formula and its mathematics. The verse number in the title is a pointer for locating the result; we do not reproduce or invent Sanskrit verse text or exact translations. For verbatim wording, consult the prescribed text. We present the result modestly, as genuine mathematics that stands on its own merit.
Theory
Brahmagupta and his formula
Brahmagupta (about 598 to 668 CE) was a major Indian mathematician and astronomer, author of the Brahmasphutasiddhanta. Among his many results is a formula for the area of a cyclic quadrilateral: a quadrilateral with sides a, b, c, d whose four vertices lie on a single circle.
First compute the semi-perimeter s, which is half the perimeter:
s = (a + b + c + d) / 2
Then the area is:
Area = the square root of (s - a)(s - b)(s - c)(s - d)
Four sides in, one area out. It is a strikingly clean rule for a shape that otherwise looks hard to measure.
Follow along
Worked example: sides 4, 5, 7, 10
- Add the sides 4 + 5 + 7 + 10 = 26 (the perimeter).
- Halve for the semi-perimeter s = 26 / 2 = 13.
- Subtract each side from s s - a = 9, s - b = 8, s - c = 6, s - d = 3.
- Multiply the four 9 x 8 x 6 x 3 = 1296.
- Take the square root Area = the square root of 1296 = 36.
Quiz
Using Brahmagupta's formula for a cyclic quadrilateral with sides 4, 5, 7, and 10, what is the area? (s = 13.)
- 26, the perimeter
- 36, because (13-4)(13-5)(13-7)(13-10) = 9 x 8 x 6 x 3 = 1296, and the square root of 1296 is 36
- 1296
- 13, the semi-perimeter
Show the answer
36, because (13-4)(13-5)(13-7)(13-10) = 9 x 8 x 6 x 3 = 1296, and the square root of 1296 is 36
With s = (4+5+7+10)/2 = 13, the four factors are (s-a)=9, (s-b)=8, (s-c)=6, (s-d)=3. Their product is 9 x 8 x 6 x 3 = 1296, and the area is the square root of 1296 = 36. Option A (26) is the perimeter, not the area. Option C (1296) is the product BEFORE taking the square root; you must take the root. Option D (13) is the semi-perimeter s, an intermediate value, not the area. So the answer is 36. The formula turns four side lengths into an area in a few clean steps: semi-perimeter, four subtractions, one product, one square root.
Think first
Where have you seen this shape of formula before?
Brahmagupta's formula looks a lot like a famous triangle-area formula. What is the connection? Then tap.
Show the answer
It is a GENERALISATION of Heron's formula for the area of a triangle. Heron's formula says a triangle with sides a, b, c has area equal to the square root of s(s-a)(s-b)(s-c), where s is its semi-perimeter. Brahmagupta's formula for a cyclic quadrilateral is the square root of (s-a)(s-b)(s-c)(s-d), with four sides. Watch what happens if you let one side of the quadrilateral shrink to zero, say d = 0: then (s - d) becomes (s - 0) = s, and the quadrilateral collapses into a triangle with sides a, b, c. Brahmagupta's formula turns into the square root of s(s-a)(s-b)(s-c), which is exactly Heron's formula. So the triangle case is just the special case where the fourth side vanishes. That connection is a lovely example of how a more general result contains a familiar one inside it, and it is a good way to remember both. One formula, and the triangle hiding within it.
Theory
Studying this well
Remember the documented formula: for a cyclic quadrilateral with sides a, b, c, d and semi-perimeter s = (a+b+c+d)/2, the area is the square root of (s-a)(s-b)(s-c)(s-d). Practise the worked steps (semi-perimeter, subtract, multiply, square root) until they are automatic, and note the link to Heron's formula. Treat the verse number as a pointer and cite the prescribed text. Next you will IMPLEMENT these Unit 4 results in code, which for this .NET subject you can do in C#.
Summary
Key takeaways
- Brahmagupta (about 598 to 668 CE) wrote the Brahmasphutasiddhanta and gave a formula for the area of a cyclic quadrilateral.
- A cyclic quadrilateral has four sides a, b, c, d with all four corners lying on a circle.
- Compute the semi-perimeter s = (a + b + c + d) / 2.
- Area = the square root of (s - a)(s - b)(s - c)(s - d).
- Worked example, sides 4, 5, 7, 10: s = 13; 9 x 8 x 6 x 3 = 1296; square root of 1296 = 36.
- The formula generalises Heron's formula for a triangle, which is the special case where the fourth side is 0.
- Memory hook: semi-perimeter, subtract each side, multiply the four, take the square root.