Verify Brahmagupta’s formula for the case of an isosceles trapezium.
Step-by-step solution
Idea: All isosceles trapezia are cyclic, so the formula should apply. Simplify it and compare with 12(sum of parallel sides) × height, where the height comes from the Baudhāyana–Pythagoras theorem.
- Perimeter = 2a + 2b + 2c, so s = a + b + c. Area by the formula = √((s − 2a)(s − 2b)(s − c)(s − c)).1 mark
- s − 2a = c + b − a, s − 2b = c + a − b, s − c = a + b. So area = (a + b)√((c + b − a)(c + a − b)).1 mark
- (c + (b − a))(c − (b − a)) = c2 − (b − a)2, so area = (a + b)√(c2 − (b − a)2).1 mark
- Height: the perpendicular from B to DC cuts off a right triangle with hypotenuse c and base (2b − 2a) ÷ 2 = b − a, so h = √(c2 − (b − a)2). The formula gives (a + b)h = 12(2a + 2b) × h, the usual trapezium area ✓.1 mark
Check: Parallel sides 6 and 12 (a = 3, b = 6), slant sides 5: h = √(25 − 9) = 4, area = 9 × 4 = 36. Brahmagupta: s = 14, √(8 × 2 × 9 × 9) = √1296 = 36 ✓.
Answer to write in the exam
s = a + b + c
Area = √((s − 2a)(s − 2b)(s − c)2) = (a + b)√((c + b − a)(c + a − b))
= (a + b)√(c2 − (b − a)2)
h = √(c2 − (b − a)2) (Baudhāyana–Pythagoras)
∴ Area = (a + b)h = 12(2a + 2b)h ✓
Common mistakes that cost marks
- Taking the overhang as 2b − 2a instead of half of it, b − a.
- Applying the formula to a trapezium that is not isosceles; such a trapezium is not cyclic.
- Losing the factor (a + b) when taking the square root of (a + b)2.
How this can come in the exam
An isosceles trapezium has parallel sides 10 cm and 22 cm and slant sides 10 cm. Find its area.
Show answer
Overhang = (22 − 10) ÷ 2 = 6; h = √(100 − 36) = 8 cm (1 mark). Area = 12(10 + 22) × 8 = 128 cm2 (1 mark). (Brahmagupta: s = 26, √(16 × 4 × 16 × 16) = 128 ✓.)Try one yourself
Use Brahmagupta’s formula for an isosceles trapezium with parallel sides 4 and 10 and slant sides 5.
Show answer
s = 12: √(8 × 2 × 7 × 7) = √784 = 28. Check: h = √(25 − 9) = 4, 12(14)(4) = 28 ✓.
More questions like this
- 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.
- The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.