Mensuration (3D)

medium importance~2 Q in Tier 120 formulas⚡ 10 shortcuts5 subtopics

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Cube and cuboid

For a cube of edge $a$ and a cuboid $l\times b\times h$ :

Quantity	Cuboid	Cube
Volume	$lbh$	$a^3$
Lateral (4 walls)	$2(l+b)h$	$4a^2$
Total surface	$2(lb+bh+lh)$	$6a^2$
Space diagonal	$\sqrt{l^2+b^2+h^2}$	$a\sqrt3$

The space diagonal is the longest rod that fits in the room. Wall questions use the lateral area only.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 14 practice Q

Detailed notes

The four numbers of a box

For a cuboid of dimensions $l \times b \times h$ (and a cube, where $l = b = h = a$ ):

Quantity	Cuboid	Cube
Volume	$V = lbh$	$a^3$
Total surface area	$2(lb + bh + lh)$	$6a^2$
Lateral (4 walls)	$2h(l+b)$	$4a^2$
Diagonal	$\sqrt{l^2+b^2+h^2}$	$a\sqrt{3}$
Sum of edges	$4(l+b+h)$	$12a$

These six pairs answer nearly every cube/cuboid question directly. The diagonal formula is the 2-D Pythagoras applied twice — it is the hypotenuse of a right triangle whose legs are a face diagonal and the third edge.

Units and capacity

Volume in $\text{cm}^3$ converts to millilitres 1:1; $1\ \text{m}^3 = 10^6\ \text{cm}^3 = 1000$ litres. Tank and pit questions: compute $V$ in $\text{m}^3$ , multiply by 1000 for litres. Reverse: a "capacity of 50000 litres" tank is $50\ \text{m}^3$ . Area-based costs (painting, whitewashing) use the surface area asked — TOTAL for all faces, LATERAL for the four walls of a room (floor and ceiling excluded).

Composite box work

Cuboid cut into cubes: number of cubes $= \frac{lbh}{a^3}$ when $a$ divides all dimensions.
Painting vs carpeting: carpeting is a floor area ( $l \times b$ ); painting four walls is lateral area; painting a closed box is total area. Read which one is priced.
Sum-of-edges questions: a frame or the total length of edges uses $4(l+b+h)$ — a different quantity from both area and volume.

Reading the ask, a worked mini-case

"A room 10 m x 6 m x 4 m is to be whitewashed, including the ceiling at Rs 5 per sq m": whitewash covers four walls plus the ceiling (not the floor) $= 2h(l+b) + lb = 128 + 60 = 188$ sq m; cost $= 188 \times 5 = 940$ . The same room carpeted is $lb = 60$ sq m; the same room painted fully outside is the TSA. One set of dimensions, four different answers — the exam is testing which surface you price.

The cube number family

Cubes $1^3$ to $10^3$ (1, 8, 27, 64, 125, 216, 343, 512, 729, 1000) and their $6a^2$ (6, 24, 54, 96, 150, 216, 294, 384, 486, 600) cover most reverse questions at sight: TSA 294 → side 7 → diagonal $7\sqrt3$ in two steps. The 3-D diagonal triplets reuse the 2-D ones: edges $(3, 4, 12)$ give diagonal $13$ ( $9 + 16 + 144 = 169$ ), and edges $(2, 3, 6)$ give diagonal $7$ ( $4 + 9 + 36 = 49$ ).

Quick revision

$V = lbh = a^3$ ; TSA $= 2(lb+bh+lh) = 6a^2$ ; LSA $= 2h(l+b) = 4a^2$ .
Diagonal $\sqrt{l^2+b^2+h^2} = a\sqrt3$ ; edges $4(l+b+h) = 12a$ .
$1\ \text{m}^3 = 1000$ L; $1\ \text{cm}^3 = 1$ mL.
Four walls of a room $= 2h(l+b)$ ; total incl. ceiling $= 2h(l+b) + lb$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Volume and surface area of cuboid/cubevery common3 practice Q

How to spot it:

Dimensions given; volume or surface area (total/lateral) asked directly.

V = lbh;\quad \text{TSA} = 2(lb+bh+lh);\quad \text{LSA} = 2h(l+b)

Read whether TOTAL (all 6 faces), LATERAL (4 walls) or VOLUME is asked.
Substitute; for a cube use the $a$ -forms ( $a^3$ , $6a^2$ , $4a^2$ ).
Keep factor form until the last step to avoid arithmetic slips.

Why: each quantity prices a different real thing — capacity, full paint, wall paint — so the ask determines the formula.

Example: Find the total surface area of a cuboid 12 cm x 10 cm x 8 cm.

TSA $= 2(12{\cdot}10 + 10{\cdot}8 + 12{\cdot}8) = 2(120+80+96) = 592$ sq cm.

Type 2: Cube specifics: diagonal, lateral area, edgescommon3 practice Q

How to spot it:

A cube's face diagonal/body diagonal, lateral surface, or the total edge length is asked.

d = a\sqrt{3},\quad \text{LSA} = 4a^2,\quad \text{edges} = 12a

Body diagonal: side times $\sqrt3$ ; face diagonal: side times $\sqrt2$ — do not mix them.
Lateral area (4 faces) vs total (6 faces): check the wording.
Edge-total questions (frames, wire along edges): $12a$ .

Why: the body diagonal spans a face diagonal and the third edge ( $a^2 + 2a^2 = 3a^2$ ), giving the clean $a\sqrt3$ .

Example: The length of the diagonal of a cube of side 6 cm is:

$d = a\sqrt3 = 6\sqrt3$ cm.

Type 3: Tank capacity in litres / volume costvery common2 practice Q

How to spot it:

A tank, pit or room's capacity asked in litres — or digging/painting cost from volume/area at a rate.

V(\text{m}^3) \times 1000 = \text{litres}

Compute the volume in the stated units; convert BEFORE comparing to rates.
$1\ \text{m}^3 = 1000$ L; $1\ \text{cm}^3 = 1$ mL.
Cost = (area or volume) x rate — decide which from the wording.

Why: rates are quoted per litre/sq m/cu m, so the conversion is where marks are won or lost.

Example: A tank is 1.5 m long, 1 m wide and 0.8 m deep. Its capacity in litres is:

$V = 1.5 \times 1 \times 0.8 = 1.2\ \text{m}^3 = 1200$ litres.

Type 4: Reverse: from volume/area back to a sidecommon2 practice Q

How to spot it:

Volume (or surface area) of a cube given — the side, diagonal or edge-sum asked.

a = \sqrt[3]{V},\quad a = \sqrt{\text{TSA}/6}

Cube: take the cube root of the volume for the side.
From surface area: divide by 6, take the square root.
Then answer the actual ask (diagonal $a\sqrt3$ , edges $12a$ , face area $a^2$ ).

Why: every cube quantity is a clean power of $a$ , so any one of them rebuilds the whole box.

Example: The volume of a cube is 343 cu cm. Its total surface area is:

$a = \sqrt[3]{343} = 7$ ; TSA $= 6 \times 49 = 294$ sq cm.

Formulas

Volume

V=lbh,\qquad V_{\text{cube}}=a^3

Lateral surface

LSA=2(l+b)h,\qquad LSA_{\text{cube}}=4a^2

Total surface

TSA=2(lb+bh+lh),\qquad TSA_{\text{cube}}=6a^2

Space diagonal

d=\sqrt{l^2+b^2+h^2},\qquad d_{\text{cube}}=a\sqrt3

Shortcut tricks

⚡ Longest rod = space diagonal

A rod can poke corner-to-corner through the room: $\sqrt{l^2+b^2+h^2}$ . Triplets like (3,4,12,13) exist: $3^2+4^2+12^2=169=13^2$ .

Example: Find the length of the longest rod that can be placed in a room of dimensions $12\text{ m}\times9\text{ m}\times8\text{ m}$ .

$\sqrt{12^2+9^2+8^2}=\sqrt{144+81+64}=\sqrt{289}=17$ m.

⚡ Painting/flooring picks the right area

Four walls → $2(l+b)h$ (ignore floor and roof). Full painting including roof and floor → TSA. Cost = area × rate.

Example: A room is 8 m long, 6 m wide and 4 m high. Find the cost of painting its four walls at ₹10 per sq m.

Four walls $=2(8+6)\times4=112$ sq m; cost $=112\times10=$ ₹1,120.

Where students lose marks

Using TSA where the question asks only for the four walls (or the reverse).
Cube diagonal $a\sqrt3$ confused with the face diagonal $a\sqrt2$ .
Height left out of the wall formula: walls need $h$ ; floor area does not.

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.