Cuboids, cylinders, cones, spheres — every 3D shape has its formula. Learn them with the shortcuts that make CAT problems solvable in 90 seconds.
Everything in the real world is 3-dimensional. A water tank, a cricket ball, an ice cream cone, a dice — these are all 3D objects.
3D Mensuration deals with:
CAT tests these concepts both directly (find volume of a solid) and indirectly (melting/recasting, painting costs, water overflow).
3D Mensuration appears in 2–3 CAT questions every year. Common question types include:
| Shape | Volume | Total Surface Area | Lateral Surface Area |
|---|---|---|---|
| Cube (side a) | a³ | 6a² | 4a² |
| Cuboid (l,b,h) | lbh | 2(lb+bh+hl) | 2(l+b)h |
| Cylinder (r,h) | πr²h | 2πr(r+h) | 2πrh |
| Cone (r,l,h) | ⅓πr²h | πr(r+l) | πrl |
| Sphere (r) | (4/3)πr³ | 4πr² | — |
| Hemisphere (r) | (2/3)πr³ | 3πr² | 2πr² |
Where l = slant height of cone = √(r² + h²).
Cube diagonals:
A cube of side 4 cm. Volume = 64 cm³. TSA = 6 × 16 = 96 cm².
Space diagonal = 4√3 cm.
Cuboid: A room 5×4×3 m. Volume = 60 m³. TSA = 2(20+12+15) = 2×47 = 94 m².
Volume = πr²h (same as circle area × height).
Curved/Lateral SA = 2πrh (rectangle rolled into cylinder: 2πr × h).
Total SA = 2πr(r+h).
Cylinder: r = 7 cm, h = 10 cm.
Volume = π × 49 × 10 = 490π ≈ 1540 cm³.
Curved SA = 2π × 7 × 10 = 140π ≈ 440 cm².
Total SA = 2π × 7 × (7+10) = 14π × 17 = 748 cm² (approx 2354 mm² wrong, recalc: 14×22/7×17 = 44×17 = 748 cm²).
Slant height l = √(r² + h²).
Volume = ⅓πr²h (exactly 1/3 of a cylinder with same base and height).
Curved SA = πrl.
Total SA = πr(r+l) = πr² + πrl.
Cone: r = 6, h = 8.
l = √(36+64) = √100 = 10.
Volume = ⅓ × π × 36 × 8 = 96π ≈ 301.7 cm³.
Curved SA = π × 6 × 10 = 60π ≈ 188.5 cm².
Key relation: Cone volume = ⅓ × Cylinder volume (same base, same height).
Sphere: V = (4/3)πr³, SA = 4πr².
Mnemonic: 4πr² looks like a circle with 4 faces.
Hemisphere: V = (2/3)πr³ (half sphere).
Curved SA = 2πr² (half sphere surface).
Flat face = πr².
Total SA = 3πr² (curved + flat).
Sphere radius 3. V = (4/3)π(27) = 36π. SA = 4π(9) = 36π.
(Interesting: for r=3, volume = surface area numerically!)
When a solid is melted and recast, volume is conserved.
A metallic sphere of radius 6 is melted and recast into small spheres of radius 1. How many?
Volume of big sphere = (4/3)π(216) = 288π.
Volume of each small sphere = (4/3)π(1) = (4/3)π.
Number = 288π / (4π/3) = 288 × 3/4 = 216 small spheres.
A cone of r=5 and h=12 melted into a cylinder of r=5. Find height of cylinder.
Volume of cone = ⅓ × π × 25 × 12 = 100π.
Volume of cylinder = π × 25 × h = 25πh.
100π = 25πh → h = 4 cm.
Hollow cylinder (inner radius r, outer radius R, height h):
Volume = π(R²-r²)h.
Outer curved SA = 2πRh. Inner curved SA = 2πrh.
Two rings (top+bottom) area = 2π(R²-r²).
Total SA = 2π(R+r)(R-r) + 2π(R+r)h = 2π(R+r)(R-r+h).
Hollow cylinder: outer r=10, inner r=8, h=14.
Volume = π(100-64)(14) = π × 36 × 14 = 504π cm³.
When a cone is cut parallel to its base, the remaining bottom portion is a frustum.
| Frustum property | Formula |
|---|---|
| Slant height | l = √(h² + (R-r)²) |
| Volume | (πh/3)(R² + r² + Rr) |
| Curved SA | π(R+r)l |
| Total SA | π(R+r)l + πR² + πr² |
Frustum: R=10, r=6, h=8.
l = √(64+16) = √80 = 4√5.
Volume = (π×8/3)(100+36+60) = (8π/3)(196) = 1568π/3 ≈ 1641 cm³.
| Term | Meaning |
|---|---|
| Volume | Cubic space occupied by 3D solid |
| Surface Area | Total area of all faces |
| Lateral/Curved SA | Area of sides only (excluding top/bottom) |
| Slant Height | Diagonal distance along cone's surface |
| Frustum | Cone with apex cut by a plane parallel to base |
| Recasting | Melting a solid into same material but different shape |
❌ Using height instead of slant height in cone surface area.
❌ Forgetting the flat base when computing hemisphere Total SA.
❌ Confusing volume formula of cone (1/3 × cylinder) with hemisphere (2/3 × sphere).
3D shapes have volume and surface area. Key relationships: cone = 1/3 cylinder; hemisphere = 2/3 sphere for volume. Surface area distinguishes curved (lateral) from total. Melting and recasting conserves volume. Frustum formulas apply when a cone is truncated. Always draw a cross-section for hollow or complex shapes. Unit consistency is critical.
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