Chapter 15 of 19

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SSC CGL Quantitative Aptitude★ Tier 1: 2–3 Q · Tier 2: 3–4 Q — half 2D, half 3D

Mensuration — 2D & 3D

2–3 questions in Tier 1 and 3–4 in Tier 2 — areas, volumes and surface areas of the standard solids, plus SSC's favourite composites: melting and recasting, wire-drawing, rolling a sheet into a cylinder, and the cone-hemisphere-cylinder family on equal bases.

⏱ 24 min readMedium difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

SSC CGL PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Recall the 2D bank (Heron, rhombus via diagonals, trapezium, sector) and the 3D bank (cuboid to frustum) without hesitation
2Apply the scaling reflex: lengths k ⇒ areas k² ⇒ volumes k³
3Solve melting/recasting by volume conservation, including sphere-to-wire and n-spheres counts
4Unroll curved surfaces: cylinder ⇒ rectangle 2πr × h; cone ⇒ sector of radius l
5Use the 1:2:3 cone-hemisphere-cylinder family and Archimedes' sphere-cylinder ratios
6Handle flow questions: pipe volume rate = cross-section × speed

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Why this chapter matters in SSC CGL

Mensuration contributes 2–3 Tier-1 and 3–4 Tier-2 questions and is the most formula-convertible topic after algebra: the entire question set is the standard 2D/3D bank plus two ideas — melting conserves volume, surfaces unroll into rectangles/sectors. SSC's composites (cone-in-cylinder, sphere-to-wire, 1:2:3 family) repeat so reliably that prepared aspirants treat mensuration as substitution, not problem-solving.

Before you start — revise these

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Geometry — Triangles, Circles & Quadrilaterals

Pythagoras, triples and the rhombus-diagonal relation feed every slant-height computation.

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Percentage

Percentage-change questions on areas/volumes use the successive-change formula with k²/k³.

Mensuration — SSC CGL Quantitative Aptitude

Mensuration is a formula bank plus two ideas: volume survives melting (recasting problems) and surfaces are unrolled rectangles (curved-surface problems). SSC composes the standard solids into predictable hybrids — a cone scooped from a cylinder, a sphere melted into wire. Know the bank cold and each question is substitution.

1. What SSC actually asks

Tier 1: 2–3 Q · Tier 2: 3–4 Q, split roughly half 2D / half 3D:

2D: triangles, circles (sectors), quadrilaterals, polygons; area↔perimeter interplay; paths around fields.
3D: cube/cuboid, cylinder, cone (+frustum), sphere/hemisphere, prism/pyramid; melting & recasting; equal-base comparisons.

2. The 2D bank

Figure	Area	Perimeter / extras
Triangle	$\frac{1}{2} bh$ ; Heron: $s (s - a) (s - b) (s - c)$	Equilateral: $\frac{3}{4} a^{2}$ , h = $\frac{3}{2} a$
Rectangle	$l b$	diagonal $l^{2} + b^{2}$
Square	$a^{2} = \frac{d ^{2}}{2}$	diagonal $a 2$
Parallelogram	$bh$
Rhombus	$\frac{1}{2} d_{1} d_{2}$	side $^{2} = (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$
Trapezium	$\frac{1}{2} (a + b) h$
Circle	$π r^{2}$	$C = 2 π r$
Sector (θ°)	$\frac{θ}{360} π r^{2}$	arc $= \frac{θ}{360} 2 π r$ ; sector area $= \frac{1}{2} \times arc \times r$

Ratio reflex: all lengths scale k ⇒ areas scale $k^{2}$ . Radius +40% → area +96% (factor $1. 4^{2} = 1.96$ ).

Paths: outer path around an $l \times b$ field, width w: area $= 2 w (l + b + 2 w)$ ; inner path: $2 w (l + b - 2 w)$ .

Circle↔square classics: same perimeter → circle has the larger area; wire bent from square (side a) into circle → $r = \frac{2 a}{π}$ .

3. The 3D bank

Solid	Volume	Curved/Lateral SA	Total SA
Cuboid	$l bh$	$2 h (l + b)$	$2 (l b + bh + h l)$ ; diagonal $l^{2} + b^{2} + h^{2}$
Cube	$a^{3}$	$4 a^{2}$	$6 a^{2}$ ; diagonal $a 3$
Cylinder	$π r^{2} h$	$2 π r h$	$2 π r (r + h)$
Cone	$\frac{1}{3} π r^{2} h$	$π r l$ , $l = r^{2} + h^{2}$	$π r (l + r)$
Sphere	$\frac{4}{3} π r^{3}$	—	$4 π r^{2}$
Hemisphere	$\frac{2}{3} π r^{3}$	$2 π r^{2}$	$3 π r^{2}$
Frustum	$\frac{π h}{3} (R^{2} + R r + r^{2})$	$π (R + r) l$ , $l = h^{2} + (R - r)^{2}$	+ $π (R^{2} + r^{2})$
Prism	base area × h	perimeter × h	+ 2 × base
Pyramid	$\frac{1}{3}$ base area × h	$\frac{1}{2}$ × perimeter × slant	+ base

The 1 : 2 : 3 family: cone, hemisphere and cylinder on the same base and same height ( $h = r$ ) have volumes in ratio $\frac{1}{3} : \frac{2}{3} : 1$ = 1 : 2 : 3.

Sphere in cylinder (Archimedes): sphere inscribed in a cylinder (h = 2r): volume ratio sphere : cylinder = 2 : 3; curved surfaces equal ( $4 π r^{2}$ each).

4. The two big ideas

Melting conserves volume. Recasting questions equate volumes:

Sphere (r = 3) melted into wire (r = 0.1): $\frac{4}{3} π 27 = π (0.1)^{2} L$ → $L = \frac{36}{0.01} \times \frac{1}{3}$ … compute: $36 π = 0.01 π L$ → L = 3600 cm = 36 m.
N small spheres from one big: $n = (R / r)^{3}$ . Radius halves → 8 spheres. (Surface area total grows by factor $n^{1/3}$ .)

Surfaces unroll. A cylinder's curved surface is a rectangle $2 π r \times h$ — so a rectangular sheet rolled along its length $L$ makes a cylinder with $2 π r = L$ . A cone's surface unrolls into a sector of radius $l$ (slant) and arc $2 π r$ .

Rain/tank flow: volume through a pipe per second = cross-section area × flow speed. River 3 m deep, 40 m wide, 2 km/h → water into sea per minute = $3 \times 40 \times \frac{2000}{60}$ m³ = 4,000 m³.

5. Solved PYQ-style examples

Q1. The perimeter of a rhombus is 52 cm and one diagonal is 24 cm. Its area? Solution. Side 13; half-diagonals 12 and $169 - 144 = 5$ → d₂ = 10 → area $= \frac{1}{2} \times 24 \times 10 =$ 120 cm².

Q2. A wire bent as a circle of radius 42 cm is re-bent into a square. The side? Solution. Length $= 2 π (42) = 264$ cm → side $=$ 66 cm.

Q3. Three metal cubes of edges 3, 4, 5 cm are melted into one cube. Its edge? Solution. $27 + 64 + 125 = 216 = 6^{3}$ → 6 cm. (The one recasting question everyone must know.)

Q4. A conical tent has radius 7 m and height 24 m. Canvas required (CSA)? Solution. $l = 49 + 576 = 25$ → CSA $= π r l = \frac{22}{7} \times 7 \times 25 =$ 550 m².

Q5. The radii of two spheres are in ratio 2 : 3. Ratio of surface areas? Of volumes? Solution. SA $\propto r^{2}$ → 4 : 9; V $\propto r^{3}$ → 8 : 27.

Q6 (Tier 2). A hemispherical bowl of internal radius 9 cm is full of liquid, to be filled into cylindrical bottles of radius 1.5 cm and height 4 cm. How many bottles? Solution. $\frac{\frac{2}{3} π \cdot 729}{π \cdot 2.25 \cdot 4} = \frac{486}{9} =$ 54 bottles.

Q7. The volume of a cube is numerically equal to its total surface area. Its edge? Solution. $a^{3} = 6 a^{2}$ → a = 6 units.

6. Exam protocol

Identify the solid(s), write the formula(s) before inserting numbers.
Melting/recasting → equate volumes; count questions → divide volumes.
Ratio questions → k² for areas, k³ for volumes; never re-derive.
Keep $π = \frac{22}{7}$ when 7 divides a radius; $π \approx 3.14$ otherwise — SSC engineers radii around 7s.
Unit check: mixing cm and m is the top mensuration error under the clock — convert before, not after.
45-second cap Tier 1; frustum/composite questions in Tier 2 get 90 seconds.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Heron / equilateral

√(s(s−a)(s−b)(s−c)); equilateral: (√3/4)a²

s = semi-perimeter.

Rhombus

area = ½ d₁d₂; side² = (d₁/2)² + (d₂/2)²

Perimeter → side → second diagonal via Pythagoras.

Sector

area = (θ/360)πr² = ½ · arc · r; arc = (θ/360)2πr

Cylinder

V = πr²h · CSA = 2πrh · TSA = 2πr(r+h)

CSA unrolls to a 2πr × h rectangle.

Cone

V = ⅓πr²h · CSA = πrl · l = √(r²+h²)

7-24-25: r=7, h=24 → l=25.

Sphere / hemisphere

V = 4/3πr³, SA = 4πr² · hemi: V = ⅔πr³, CSA = 2πr², TSA = 3πr²

Hemisphere TSA includes the flat disc.

Frustum

V = (πh/3)(R² + Rr + r²); CSA = π(R+r)l, l = √(h² + (R−r)²)

Tier-2 favourite (bucket questions).

1:2:3 family

cone : hemisphere : cylinder = 1 : 2 : 3 (same r, h = r)

Sphere : circumscribing cylinder = 2 : 3 (volumes AND curved surfaces equal 4πr²).

Recasting

melted volume = recast volume; n spheres: n = (R/r)³

3-4-5 cubes → edge-6 cube (27+64+125 = 216).

Scaling reflex

lengths ×k ⇒ areas ×k² ⇒ volumes ×k³

Radius +40% → area +96%.

Diagonals

cube a√3 · cuboid √(l²+b²+h²) · square a√2

Paths

outer: 2w(l+b+2w) · inner: 2w(l+b−2w)

w = path width around an l×b field.

SSC CGL test series

Ready to be tested on this?

Mensuration — 2D & 3D shows up in every SSC CGL paper — sit a timed one and find out where you actually stand.

2–3Qs in Tier 1

Highpriority

Full SSC CGL mocks, real clock
Quantitative Aptitude-only sectionals & chapter tests
All-India rank + topic-level analysis

Free forever · solver-verified questions

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Traps SSC CGL sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Using diameter as radius (or mixing cm with m)

✓ Circle every given dimension and label it r or d on the sketch; convert all units BEFORE substituting. Unit mixing is the single biggest mensuration mark-killer.

WATCH OUT

✗ Forgetting the ⅓ in cone and pyramid volumes

✓ Pointed solids take one-third of their prism/cylinder counterpart. Sanity anchor: the 1:2:3 family — cone is exactly ⅓ of the cylinder on the same base and height.

WATCH OUT

✗ Using h instead of slant height l in curved surface areas

✓ CSA of a cone is πrl, never πrh. Compute l = √(r²+h²) first — SSC picks r, h from Pythagorean triples so l is always clean.

WATCH OUT

✗ Hemisphere TSA as 2πr²

✓ 2πr² is only the curved part; the TOTAL adds the flat disc: 3πr². Read whether the bowl is open (CSA) or solid (TSA).

WATCH OUT

✗ Scaling volumes by k² (or areas by k)

✓ One reflex: k, k², k³ for lengths, areas, volumes. Radii 2:3 → surfaces 4:9 → volumes 8:27.

WATCH OUT

✗ In recasting, equating surface areas instead of volumes

✓ Melting conserves VOLUME only — surface area actually increases when one solid becomes many. Equate volumes, always.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Mensuration — 2D & 3D?

11 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

11 questions~8 min worth ~6 marks in SSC CGL exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•2D: Heron; rhombus ½d₁d₂ (side via half-diagonals); trapezium ½(a+b)h; sector (θ/360)πr².
•3D volumes: cuboid lbh · cylinder πr²h · cone ⅓πr²h · sphere 4/3πr³ · hemi ⅔πr³.
•Surfaces: cylinder CSA 2πrh · cone πrl (l = √(r²+h²)) · sphere 4πr² · hemi TSA 3πr².
•Frustum: V = (πh/3)(R²+Rr+r²); l = √(h²+(R−r)²).
•1:2:3 (cone:hemi:cylinder, h = r); sphere:cylinder = 2:3.
•Melting conserves volume; n = (R/r)³ spheres; 3-4-5 cubes → 6.
•Cylinder unrolls to 2πr × h; cone unrolls to a sector of radius l.
•k / k² / k³ for lengths / areas / volumes.
•Diagonals: square a√2, cube a√3, cuboid √(l²+b²+h²).
•π = 22/7 when 7 divides a dimension; 3.14 otherwise.

SSC CGL question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: Tier 1: 4–6 marks (2–3 Q × 2) · Tier 2: 9–12 marks (3–4 Q × 3)

Question style	Marks each	Typical count	What it tests
2D area/perimeter	2–3	1–2	Rhombus/sector/path formulas with triples
3D volume/surface	2–3	1–2	Formula substitution, CSA vs TSA reading
Composite (recast/frustum)	3	0–1	Volume conservation, frustum slant

Prep strategy

Write the two banks (2D + 3D tables) daily for a week — this topic IS its formula sheet.
Drill the three composite patterns: recasting, 1:2:3 family, unrolled surfaces.
Every practice miss: classify as formula-gap vs unit-slip vs wrong-solid — the fix differs.

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Formula first, numbers second — write it before substituting.
Sketch and label r/d/h; convert units before computing.
Melting → volumes; counting → volume ÷ volume; never surfaces.
Expect a Pythagorean triple whenever slant height is needed.
Ratio questions: k²/k³ reflex, no derivation.
45 s Tier 1; composites in Tier 2 get 90 s with the frustum formula pre-written.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Construction estimates

Cement for a cylindrical pillar, paint for a wall, water in a tank — literal daily mensuration in PWD-type postings.

Packaging & logistics

Carton volumes and material minimisation are cuboid/cylinder optimisation in practice.

Water management

The pipe-flow formula (area × speed) is how reservoir inflows and irrigation are actually computed.

Manufacturing

Melting-recasting is real foundry math — volume conservation with machining losses.

Where else this topic is tested

Prepare once, score in every exam that asks it.

SSC CHSL / CPOVery high — same bank, same composites

CDSHigh — mensuration is a CDS maths staple

RRB NTPC / Group DHigh — lighter 2D-heavy versions

State PSC aptitudeHigh — recycled SSC questions

📝 My notes

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

All the standard solids appear, but three composites dominate: melting/recasting (volume conservation), the cone-hemisphere-cylinder family on equal bases, and cone/cylinder questions engineered around Pythagorean triples (7-24-25 especially). The frustum/bucket is Tier 2's favourite 'hard' question.

Memorise CSA and add the flat faces mentally: cylinder TSA = CSA + two discs; hemisphere TSA = CSA + one disc; cone TSA = CSA + base. This halves the memory load and prevents the 2πr²-vs-3πr² hemisphere slip.

SSC plants a 7 (or 21, 42, 3.5) in the dimensions whenever 22/7 is intended — the π cancels. If your arithmetic has a stray 22/7 that won't cancel, re-check which value the question expects.

Cube-root sanity: n small spheres from a big one must be a perfect cube ((R/r)³); melted cubes give edge³ sums (27+64+125 = 216). If your count isn't a clean integer, the volume equation has a slip.

Occasionally in Tier 2: volume = base × height (prism) and ⅓ base × height (pyramid), with hexagonal bases needing the equilateral-triangle area. Low frequency — learn the two general formulas and move on.