Triangle & Quadrilateral Area
The right area formula depends entirely on what you are given. Base and height ⇒ ½bh. Three sides ⇒ Heron’s formula with s = (a+b+c)/2, Area = √[s(s−a)(s−b)(s−c)]. Two sides and the angle between them ⇒ ½ab·sinC, which is the fastest route whenever a 30°, 45°, 60° or 90° angle appears. For the equilateral triangle memorise Area = (√3/4)a² and height = (√3/2)a outright. Quadrilaterals reduce to known pieces: a parallelogram is base × height, a rhombus is ½·d₁·d₂ (half the product of diagonals), a trapezium is ½(a+b)·h where a and b are the parallel sides, and any irregular quadrilateral can be split along a diagonal into two triangles. A CAT favourite is the 3-4-5 / 5-12-13 / 8-15-17 right triangle hiding inside a larger figure — spot the Pythagorean triple and the height appears for free.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Triangle & quadrilateral areas
| Triangle (base–height) | Area = ½ × b × h |
|---|---|
| Triangle (Heron, s = (a+b+c)/2) | Area = √[s(s−a)(s−b)(s−c)] |
| Triangle (two sides + angle) | Area = ½ × a × b × sinC |
| Equilateral triangle (side a) | Area = (√3 / 4) × a² ; height = (√3/2)a |
| Parallelogram / Rhombus | b × h ; rhombus = ½ × d₁ × d₂ |
| Trapezium (parallel sides a, b) | Area = ½ × (a + b) × h |
Circle, sector & segment
| Circle area & circumference | Area = πr² ; Circumference = 2πr |
|---|---|
| Sector area (angle θ°) | (θ/360) × πr² |
| Arc length (angle θ°) | (θ/360) × 2πr |
| Minor segment area | Sector − triangle = (θ/360)πr² − ½r²·sinθ |
| Ring (annulus) area | π(R² − r²) |