Verify Brahmagupta’s formula for the case of a rectangle.
Answer: Sides a, b, a, b: s = a + b, so area = √(b × a × b × a) = √(ab · ab) = ab, which is correct.
Step-by-step solution
Idea: Brahmagupta’s formula for a cyclic 4-gon: area = √((s − a)(s − b)(s − c)(s − d)), s = half the perimeter. Every rectangle is cyclic (its diagonals are equal and bisect each other, so all four corners are at the same distance from their meeting point).
- All rectangles are cyclic, so the formula should apply. Sides in order: a, b, a, b.½ mark
- s = 12(2a + 2b) = a + b.½ mark
- Area = √((a + b − a)(a + b − b)(a + b − a)(a + b − b)) = √(b · a · b · a) = √(ab · ab).1½ marks
- = ab, which is the correct area of a rectangle.½ mark
Brahmagupta’s formula gives √(ab · ab) = ab for an a × b rectangle, which is correct.
Check: 4 × 3 rectangle: s = 7; √(3 × 4 × 3 × 4) = 12 = 4 × 3 ✓.
Answer to write in the exam
Rectangle is cyclic; sides a, b, a, b
s = 12(2a + 2b) = a + b
Area = √((s − a)(s − b)(s − a)(s − b)) = √(b · a · b · a)
∴ Area = ab ✓
Common mistakes that cost marks
- Using the full perimeter for s.
- Forgetting that the formula only applies to cyclic 4-gons (a rectangle is one; a general parallelogram is not).
How this can come in the exam
MCQ (1 mark)
For a square of side 5 cm, Brahmagupta’s formula gives the area as
- √5 cm2
- 10 cm2
- 25 cm2
- 625 cm2
Show answer
(C) 25 cm2
s = 10; √(5 × 5 × 5 × 5) = 25 cm2.
Try one yourself
Use Brahmagupta’s formula to find the area of a rectangle 9 cm by 4 cm.
Show answer
s = 13; √(4 × 9 × 4 × 9) = √1296 = 36 cm2.
More questions like this
- 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?
- 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?