🧊 Math Formulas
31 formulas in 10 sections. For each one you get the formula, what every symbol means, where it is used, an explanation in plain words and a worked example.
Sections: Cube · Cuboid · Cylinder · Cone · Sphere · Hemisphere · Prism · Pyramid · Frustum · Capacity
V = a³
| a | length of an edge |
|---|
Where it is used: To find the space inside a cube — boxes, dice, storage cubes, and volume problems.
In simple words: A cube has equal sides. Multiply one edge by itself three times to find how much it can hold.
Example: a = 4 cm → V = 4³ = 64 cm³
TSA = 6 × a²
| a | length of an edge |
|---|
Where it is used: To find the total outer area of a cube — how much paint or paper covers all 6 faces.
In simple words: A cube has 6 equal square faces. Find one face's area and multiply by 6.
Example: a = 4 → TSA = 6 × 16 = 96 cm²
d = a × √3
| a | edge length |
|---|---|
| d | longest inside diagonal |
Where it is used: To find the longest straight line inside a cube (corner to opposite corner).
In simple words: The longest line through a cube is its edge times about 1.732.
Example: a = 4 → d = 4√3 ≈ 6.93 cm
LSA = 4 × a²
| a | edge length |
|---|
Where it is used: To find the area of the four side faces of a cube (not top and bottom).
In simple words: The four walls of a cube — four square faces.
Example: a=5 → 4×25 = 100 cm²
V = (d ÷ √3)³
| d | the long diagonal |
|---|
Where it is used: To find a cube's volume when only its longest diagonal is known.
In simple words: Get the edge from the diagonal first, then cube it.
Example: d=√3 → edge 1 → V = 1 cm³
V = l × b × h
| l | length |
|---|---|
| b | breadth |
| h | height |
Where it is used: To find the space inside a box shape — rooms, bricks, tanks, cartons.
In simple words: Multiply length, width and height to find how much a box-shaped thing can hold.
Example: l=5, b=4, h=3 → V = 5×4×3 = 60 cm³
TSA = 2 × (lb + bh + hl)
| l | length |
|---|---|
| b | breadth |
| h | height |
Where it is used: To find the total outer area of a box — wrapping paper, paint for a room's walls and ceiling.
In simple words: A box has 6 rectangular faces in 3 matching pairs. Add the three different faces and double it.
Example: l=5, b=4, h=3 → TSA = 2(20+12+15) = 94 cm²
d = √(l² + b² + h²)
| l | length |
|---|---|
| b | breadth |
| h | height |
Where it is used: To find the longest straight line inside a box — packing long items into cartons.
In simple words: Square all three dimensions, add them, and take the square root to get the longest inside line.
Example: l=3, b=4, h=12 → d = √(9+16+144) = 13
LSA = 2h(l + b)
| l | length |
|---|---|
| b | breadth |
| h | height |
Where it is used: To find the area of the four walls of a room (excluding floor and ceiling) — painting.
In simple words: The area of the four walls of a box-shaped room.
Example: l=5,b=4,h=3 → 2×3×9 = 54 m²
Area = 2 × height × (length + breadth)
| length, breadth, height | room dimensions |
|---|
Where it is used: To find how much paint or wallpaper the walls of a room need.
In simple words: Same as lateral surface area — just the four walls.
Example: 5×4 room, 3 high → 2×3×9 = 54 m²
V = π × r² × h
| r | radius of the circular base |
|---|---|
| h | height |
Where it is used: To find the space inside a cylinder — pipes, tanks, tins, glasses, gas cylinders.
In simple words: Find the round base's area (πr²) and multiply by the height, like stacking many circles.
Example: r=7, h=10 → V = 22/7 × 49 × 10 = 1540 cm³
CSA = 2 × π × r × h
| r | radius |
|---|---|
| h | height |
Where it is used: To find the curved (side) area of a cylinder — the label on a tin, paint on a pipe's outside.
In simple words: Unroll the curved side and it becomes a rectangle: its area is the base's circumference times the height.
Example: r=7, h=10 → CSA = 2 × 22/7 × 7 × 10 = 440 cm²
TSA = 2πr(r + h)
| r | radius |
|---|---|
| h | height |
Where it is used: To find the whole outer area of a closed cylinder — the curved side plus the two circle ends.
In simple words: Add the curved side and the two round lids to get the full outside area.
Example: r=7, h=10 → 2×22/7×7×17 = 748 cm²
V = π × h × (R² − r²)
| R | outer radius |
|---|---|
| r | inner radius |
| h | length |
Where it is used: To find the material volume of a pipe or tube.
In simple words: Subtract the hollow inside from the full outside to get the pipe's material.
Example: R=5,r=3,h=10 → 22/7×10×16 ≈ 502.9 cm³
V = ⅓ × π × r² × h
| r | radius of the base |
|---|---|
| h | height |
Where it is used: To find the space inside a cone — ice-cream cones, funnels, conical tents and heaps.
In simple words: A cone holds exactly one-third of the cylinder with the same base and height.
Example: r=3, h=7 → V = ⅓ × 22/7 × 9 × 7 = 66 cm³
CSA = π × r × l
| r | radius |
|---|---|
| l | slant height |
Where it is used: To find the sloping surface area of a cone — the paper of a party hat or the cloth of a conical tent.
In simple words: Multiply pi by the radius and by the slanting side length to get the curved surface.
Example: r=7, l=25 → CSA = 22/7 × 7 × 25 = 550 cm²
l = √(r² + h²)
| l | slant height |
|---|---|
| r | radius |
| h | height |
Where it is used: To find the sloping side of a cone from its radius and height, before finding curved surface area.
In simple words: The slant side, the height and the radius form a right triangle, so use Pythagoras to find the slant.
Example: r=3, h=4 → l = √(9+16) = 5
TSA = πr(l + r)
| r | radius |
|---|---|
| l | slant height |
Where it is used: To find the whole outer area of a solid cone — the sloping side plus the circular base.
In simple words: Add the slanting surface and the round base to get the full outside area.
Example: r=7, l=25 → 22/7×7×32 = 704 cm²
h = 3V ÷ (π r²)
| V | volume |
|---|---|
| r | radius |
Where it is used: To find a cone's height when its volume and radius are known.
In simple words: Rearranges the cone-volume formula to give the height.
Example: V=66, r=3 → h = 198/(22/7×9) = 7
V = 4/3 × π × r³
| r | radius |
|---|
Where it is used: To find the space inside a ball-shaped object — balls, globes, bubbles, planets.
In simple words: For a perfectly round ball, cube the radius, multiply by pi, then by four-thirds.
Example: r=3 → V = 4/3 × 22/7 × 27 ≈ 113.1 cm³
SA = 4 × π × r²
| r | radius |
|---|
Where it is used: To find the outer area of a ball — leather for a football, paint for a dome.
In simple words: The whole surface of a ball is four times the area of the flat circle across its middle.
Example: r=7 → SA = 4 × 22/7 × 49 = 616 cm²
V = (1/6) × π × d³
| d | diameter |
|---|
Where it is used: To find a sphere's volume directly from its diameter.
In simple words: If you know the full width instead of the radius, use this form.
Example: d=6 → 1/6×22/7×216 ≈ 113.1 cm³
V = 2/3 × π × r³
| r | radius |
|---|
Where it is used: To find the space inside a half-ball — bowls, domes, half-orange shapes.
In simple words: A hemisphere is half a ball, so it holds half the volume of the full sphere.
Example: r=3 → V = 2/3 × 22/7 × 27 ≈ 56.6 cm³
CSA = 2πr²
| r | radius |
|---|
Where it is used: To find the curved (dome) surface area of a half-ball — bowls, domes.
In simple words: The curved part of half a ball is twice the area of the flat circle across it.
Example: r=7 → 2×22/7×49 = 308 cm²
TSA = 3πr²
| r | radius |
|---|
Where it is used: To find the whole outer area of a solid half-ball — the dome plus its flat circular face.
In simple words: Add the curved dome and the flat circle to get three times the circle's area.
Example: r=7 → 3×22/7×49 = 462 cm²
V = Base area × Height
| Base area | area of the cross-section |
|---|---|
| Height | length of the prism |
Where it is used: To find the space inside any prism — triangular, hexagonal, or other uniform cross-section.
In simple words: Find the area of the shape at the end and multiply by how long the prism is.
Example: triangular base 6 cm², length 10 → V = 60 cm³
LSA = Perimeter of base × Height
| Perimeter | of the base shape |
|---|---|
| Height | length of the prism |
Where it is used: To find the side surface area of a prism.
In simple words: Unroll the sides — their area is the base's perimeter times the length.
Example: base perimeter 12, length 10 → 120 cm²
V = ⅓ × Base area × Height
| Base area | area of the base |
|---|---|
| Height | vertical height |
Where it is used: To find the space inside any pyramid — square, triangular, etc.
In simple words: A pyramid holds one-third of the prism with the same base and height.
Example: base 36 cm², height 10 → V = 120 cm³
V = ⅓ × π × h × (R² + r² + Rr)
| R | larger radius |
|---|---|
| r | smaller radius |
| h | height |
Where it is used: To find the space inside a cone with the tip cut off — buckets, glasses, lampshades.
In simple words: A bucket shape: this gives its capacity from the two circle radii and the height.
Example: R=6,r=3,h=7 → ⅓×22/7×7×(36+9+18) = 462 cm³
CSA = π × l × (R + r)
| l | slant height |
|---|---|
| R, r | the two radii |
Where it is used: To find the sloping surface area of a bucket-shaped frustum.
In simple words: The slanting side area of a cut-off cone.
Example: R=6,r=3,l=5 → 22/7×5×9 ≈ 141.4 cm²
1000 cm³ = 1 litre, 1 m³ = 1000 litres
| cm³, m³ | volume units |
|---|
Where it is used: To convert a container's volume into how much liquid it holds.
In simple words: A box of 1000 cubic centimetres holds exactly one litre.
Example: 2000 cm³ tank holds 2 litres