Last Updated: May 2026
IPMAT 2027 — Mensuration overview
For IPMAT 2027 Mensuration, expect 2-4 questions in the QA-MCQ and QA-SA sections combined. The IPMAT favours 3D solids (cylinder, cone, sphere, hemisphere, frustum, cuboid) more than 2D shapes. Mastering the 12 core formulas and the cube-root simplification trick is enough to score 90%+ on this topic.
Master Formula Sheet
2D Shapes
| Shape | Area | Perimeter / Circumference |
|---|---|---|
| Square (side a) | a² | 4a |
| Rectangle (l × b) | l × b | 2(l + b) |
| Triangle (base b, height h) | ½bh | a + b + c |
| Equilateral triangle (side a) | (√3/4)a² | 3a |
| Circle (radius r) | πr² | 2πr |
| Sector (radius r, angle θ) | (θ/360)πr² | 2r + (θ/360)2πr |
| Parallelogram (b × h) | b × h | 2(a + b) |
| Rhombus (diagonals d₁, d₂) | ½d₁d₂ | 4a |
| Trapezium (parallel a, b, h) | ½(a+b)h | sum of all sides |
3D Solids
| Solid | Volume | Surface Area (Total) | Curved/Lateral SA |
|---|---|---|---|
| Cube (side a) | a³ | 6a² | 4a² |
| Cuboid (l × b × h) | l·b·h | 2(lb+bh+hl) | 2h(l+b) |
| Cylinder (r, h) | πr²h | 2πr(r+h) | 2πrh |
| Cone (r, h) | (1/3)πr²h | πr(r+l) | πrl (l = √(r²+h²)) |
| Sphere (r) | (4/3)πr³ | 4πr² | 4πr² |
| Hemisphere (r) | (2/3)πr³ | 3πr² | 2πr² |
| Frustum (R, r, h) | (1/3)πh(R²+r²+Rr) | πl(R+r)+π(R²+r²) | πl(R+r); l = √((R−r)²+h²) |
| Prism | (area of base) × h | 2(base) + (perim. of base × h) | perim. of base × h |
| Pyramid (square base a, slant l) | (1/3)a²h | a² + 2al | 2al |
Six Shortcut Tricks
- Diagonal of cube: a√3. Diagonal of cuboid: √(l²+b²+h²).
- Volume cube ratio scaling: If side ratio a:b, volume ratio is a³:b³.
- Sphere area to volume: SA/V = 3/r. Useful for “smallest sphere with V/A ratio = X” type problems.
- Cylinder + Hemisphere combo: often appears as fuel-tank or capsule problems. Volume = πr²h + (2/3)πr³.
- Cone height when slant given: h = √(l² − r²); then volume.
- Surface area painted ratio: If a cube is painted and cut into smaller cubes of side a/n, fraction with 0/1/2/3 painted faces follow a fixed pattern: corners (8) have 3 faces, edges have 2, faces have 1, interior have 0.
Worked Examples
Q. A cylinder has radius 7 cm and height 10 cm. Find its CSA and volume.
A. CSA = 2πrh = 2 × 22/7 × 7 × 10 = 440 cm². Volume = πr²h = 22/7 × 49 × 10 = 1540 cm³.
Q. A solid cone of radius 10 cm and height 24 cm is melted and recast into a sphere. Find sphere’s radius.
A. Volume conserved. (1/3)π × 100 × 24 = (4/3)πr³ → r³ = 100 × 6 = 600 → r ≈ 8.43 cm.
40 Practice Problems
- Area of a square with side 8 = (A) 8 (B) 16 (C) 32 (D) 64
- Perimeter of rectangle l=10, b=4 = (A) 14 (B) 28 (C) 40 (D) 50
- Area of equilateral Δ side 6 = (A) 9√3 (B) 18 (C) 36 (D) 12
- Area of circle r=7 = (A) 154 (B) 49π (C) 22 (D) 308
- Volume of cube side 5 = (A) 25 (B) 125 (C) 75 (D) 250
- Surface area of cube side 4 = (A) 16 (B) 32 (C) 64 (D) 96
- Diagonal of cube side 6 = (A) 6 (B) 6√2 (C) 6√3 (D) 12
- Volume of cylinder r=7, h=10 = (A) 1540 (B) 770 (C) 220 (D) 154
- CSA of cylinder r=7, h=10 = (A) 154 (B) 220 (C) 440 (D) 880
- Volume of cone r=3, h=4 = (A) 12π (B) 36π (C) 9π (D) 24π
- Slant height of cone r=3, h=4 = (A) 5 (B) 7 (C) 4 (D) √7
- Volume of sphere r=3 = (A) 36π (B) 4π (C) 12π (D) 27π
- SA of sphere r=7 = (A) 154 (B) 196π (C) 4π (D) 49π
- Volume of hemisphere r=6 = (A) 144π (B) 72π (C) 36π (D) 288π
- SA (total) of hemisphere r=2 = (A) 4π (B) 8π (C) 12π (D) 16π
- If side of cube doubles, volume becomes — (A) 2× (B) 4× (C) 8× (D) 16×
- If radius of sphere doubles, volume becomes — (A) 2× (B) 4× (C) 8× (D) 16×
- If radius doubles, SA of sphere becomes — (A) 2× (B) 4× (C) 8× (D) 16×
- Volume of cuboid 4×5×6 = (A) 60 (B) 120 (C) 240 (D) 30
- Diagonal of cuboid 3, 4, 12 = (A) 13 (B) 12 (C) 5 (D) √169
- Curved SA of cone with r=5, l=13 = (A) 65π (B) 130π (C) 30π (D) 25π
- Total SA of cone (r=5, l=13) = (A) 90π (B) 65π (C) 25π (D) 75π
- Volume of frustum (R=5, r=3, h=12) — (A) 196π (B) 49π (C) 312π (D) 124π
- If base of triangle doubles and height halves, area — (A) doubles (B) halves (C) same (D) quadruples
- Area of trapezium (parallel sides 6, 10; height 4) = (A) 16 (B) 32 (C) 40 (D) 24
- Area of rhombus (diagonals 8, 6) = (A) 14 (B) 24 (C) 32 (D) 48
- Sector area (r=4, θ=90°) = (A) π (B) 4π (C) 8π (D) 16π
- Volume of cube circumscribed around sphere r=3 = (A) 27 (B) 216 (C) 8 (D) 36
- Length of wire to make a cube edge 2 cm — (A) 16 (B) 24 (C) 12 (D) 36
- If a cube of side 6 is cut into 27 smaller equal cubes, each side = (A) 1 (B) 2 (C) 3 (D) 4
- How many faces of unit cubes (cube cut from side-3 cube into 27) have 0 painted face — (A) 0 (B) 1 (C) 6 (D) 8
- How many unit cubes have exactly 3 painted faces in side-3 cube → (A) 6 (B) 8 (C) 12 (D) 0
- How many edges in a cube — (A) 6 (B) 8 (C) 12 (D) 14
- How many vertices — (A) 6 (B) 8 (C) 12 (D) 14
- Diagonal of square side 5 = (A) 5 (B) 5√2 (C) 5√3 (D) 10
- Surface area of cylinder closed (r=2, h=5) = (A) 28π (B) 30π (C) 24π (D) 14π
- Volume of pyramid base 5×5, height 6 = (A) 50 (B) 30 (C) 25 (D) 75
- If 4 spheres r=2 melted into 1 sphere, new r — (A) 2∛4 (B) 2 (C) 4 (D) 8
- Ratio of surface areas of 2 cubes with volumes 8:27 — (A) 4:9 (B) 2:3 (C) 8:27 (D) 1:1
- The shape with maximum SA-to-V ratio for a given V — (A) sphere (B) cylinder (C) cube (D) tetrahedron
Answer Key
1-D, 2-B, 3-A, 4-A, 5-B, 6-D, 7-C, 8-A, 9-C, 10-A, 11-A, 12-A, 13-B, 14-A, 15-C, 16-C, 17-C, 18-B, 19-B, 20-A, 21-A, 22-A, 23-A, 24-C, 25-B, 26-B, 27-B, 28-B, 29-B, 30-B, 31-B, 32-B, 33-C, 34-B, 35-B, 36-A, 37-A, 38-A, 39-A, 40-D (tetrahedron has highest SA/V among Platonic solids; sphere has the LOWEST SA/V — common reverse confusion. Sphere minimises SA for given V.)
FAQ
How many Mensuration questions in IPMAT 2027?
Typically 2–4 questions across QA-MCQ and QA-SA. Mensuration is recurring across both Indore and Rohtak IPMAT papers.
Which 3D solid is most asked in IPMAT?
Cylinder and cone — usually combined in melt-recast or composite-solid problems. Sphere appears in volume conservation problems frequently.
Are 2D shape problems important?
Yes but usually as part of larger problems. Pure 2D area problems are 1 in 4 mensuration questions; the rest are 3D.
What’s the painted-cube trick for IPMAT?
For an n×n×n cube cut into unit cubes: 8 cubes have 3 painted faces (corners), 12(n−2) have 2 (edges), 6(n−2)² have 1 (faces), (n−2)³ have 0. Apply directly to count.