60-second revision
- Heron: area = sqrt(s(s − a)(s − b)(s − c)), s = half the perimeter.
- Equilateral triangle area = sqrt(3)/4 × a²; rhombus = half the product of diagonals.
- Cone slant height l = sqrt(r² + h²); curved surface = pi r l.
- Hemisphere total surface = 3 pi r²; sphere = 4 pi r².
- Melting and recasting keeps the volume the same.
- Scale lengths by k: areas scale by k², volumes by k³.
- 1 cubic metre = 1,000 litres; 1 hectare = 10,000 sq m.
2D formulas
| Shape | Area | Perimeter / other |
|---|---|---|
| Rectangle | l × b | 2(l + b); diagonal sqrt(l² + b²) |
| Square | a² (or diagonal²/2) | 4a; diagonal a sqrt(2) |
| Triangle | 1/2 × base × height | Sum of sides |
| Triangle (three sides) | sqrt(s(s − a)(s − b)(s − c)) | s = (a + b + c)/2 |
| Equilateral triangle | sqrt(3)/4 × a² | Height sqrt(3)a/2 |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | 1/2 × d1 × d2 | 4a, with a² = (d1/2)² + (d2/2)² |
| Trapezium | 1/2 × (sum of parallel sides) × height | Sum of sides |
| Circle | pi r² | 2 pi r |
| Semicircle | pi r²/2 | pi r + 2r |
| Sector of angle A | (A/360) × pi r² | Arc = (A/360) × 2 pi r |
Worked example: a circle of area 154 sq cm has r² = 154 × 7/22 = 49, so r = 7 cm and circumference = 44 cm.
Worked example (Heron): sides 9, 10, 17 give s = 18 and area = sqrt(18 × 9 × 8 × 1) = sqrt(1296) = 36 sq units.
Worked example: a rhombus with diagonals 16 cm and 12 cm has area 1/2 × 16 × 12 = 96 sq cm; its half-diagonals 8 and 6 give a side of 10 cm, so the perimeter is 40 cm.
Worked example: a sector of radius 14 cm and angle 90° has area 1/4 × 22/7 × 196 = 154 sq cm and arc length 1/4 × 88 = 22 cm, so its perimeter is 22 + 28 = 50 cm.
3D formulas
| Solid | Volume | Curved / lateral surface | Total surface |
|---|---|---|---|
| Cuboid | l × b × h | 2h(l + b) | 2(lb + bh + hl) |
| Cube | a³ | 4a² | 6a² |
| Cylinder | pi r² h | 2 pi r h | 2 pi r (r + h) |
| Cone | 1/3 pi r² h | pi r l | pi r (l + r) |
| Sphere | 4/3 pi r³ | 4 pi r² | 4 pi r² |
| Hemisphere | 2/3 pi r³ | 2 pi r² | 3 pi r² |
| Frustum (radii R, r) | pi h (R² + r² + Rr)/3 | pi l (R + r) | Curved + pi R² + pi r² |
Cuboid diagonal = sqrt(l² + b² + h²); cube diagonal = a sqrt(3). A cone's slant height is l = sqrt(r² + h²).
Worked example: a cone with r = 5 cm and h = 12 cm has l = 13 cm, curved surface 65 pi sq cm and volume 1/3 × pi × 25 × 12 = 100 pi cu cm.
Melting and recasting
When a solid is melted and remade, volume is conserved. Write volume before = volume after, cancel pi, and solve.
Worked example: 27 small cubes of side 2 cm are melted into one big cube. Total volume = 27 × 8 = 216, so the big cube has side 6 cm. Worked example: a sphere of radius 4 cm can be recast into 4³ = 64 spheres of radius 1 cm.
Paths and borders
- Path of width w outside an l × b rectangle: area = 2w(l + b + 2w).
- Path of width w inside: area = 2w(l + b − 2w).
- Two crossroads of width w through the middle: area = w(l + b − w).
Worked example: a 25 m × 15 m lawn with a 2 m path outside: 2 × 2 × (25 + 15 + 4) = 176 sq m. Check: 29 × 19 − 375 = 551 − 375 = 176.
Scaling and percentage change
- All lengths × k: areas × k², volumes × k³.
- Each side up x%: area up (2x + x²/100)%. A 10% rise in the side gives a 21% rise in area.
- All three dimensions up 10%: volume × 1.331, i.e. up 33.1%.
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