3D Mensuration: Methods & Shortcuts — SSC CGL Maths: Methods & Shortcuts Notes (English Medium)

A flat figure (square, circle) has only length and breadth. A solid (box, ball, pipe) has a third measure too — height or depth. For every solid the exam asks only three things:

Volume — how much space is inside (how much water, earth or metal it holds). Unit: cu cm, cu m.
Lateral or curved surface (LSA/CSA) — only the side walls, leaving out the top and bottom (the four walls of a room, the label of a tin).
Total surface (TSA) — every face you can touch from outside (painting the whole box). Unit: sq cm, sq m.

Convert	Rule	Why
length	$1 m = 100 cm$	one power
area	$1 m^{2} = 1 0^{4} cm^{2}$	$100 \times 100$
volume	$1 m^{3} = 1 0^{6} cm^{3}$	$100 \times 100 \times 100$
capacity	1 cu m = 1000 litres; 1 litre = 1000 cu cm	definition

RememberFirst job in every question: bring every measure to one unit. The most common mistake is mixing cm and m. Second job: read whether space is asked (volume) or covering/painting (area).

Take $π = \frac{22}{7}$ when the radius is a multiple of 7; otherwise leave the answer in terms of $π$ (like $240 π$ ). In this chapter every method has solved examples followed by questions to do yourself — open the solution only after you have tried.

1

Box-shaped solids

Cuboid · cube · room · prism

Solids with flat faces: cuboid, cube, rooms and boxes, and a prism on any base.

Cuboid and cube

Volume · surface · diagonal

Why the cuboid/cube formulas work, reverse questions, the diagonal, the face-product trick and ratio questions.

A cuboid is an ordinary box: 6 rectangular faces, opposite faces equal. Say the box is 5 cm long, 3 cm wide and 2 cm high. Lay 1-cm cubes on the floor — one layer holds $5 \times 3 = 15$ cubes; 2 such layers fit, so 30 cubes. Write this counting as a formula and you get volume $= l \times b \times h$ .

The faces come in three pairs: top-bottom ( $l b$ twice), front-back ( $l h$ twice), left-right ( $bh$ twice). Add them all: total surface $= 2 (l b + bh + h l)$ . Take only the four walls (leaving top and bottom) and open them out like a strip — length $2 (l + b)$ (the perimeter), height $h$ , so lateral surface $= 2 (l + b) h$ .

The diagonal runs from one corner to the exactly opposite corner. First the floor diagonal: $l^{2} + b^{2}$ (Pythagoras). This floor diagonal and the upright height $h$ make another right angle, so $d = l^{2} + b^{2} + h^{2}$ . In a cube all edges are $a$ , so the formulas become $a^{3}$ , $4 a^{2}$ , $6 a^{2}$ , $a 3$ .

◭ Diagram

Cuboid: length

l

, breadth

b

, height

h

, diagonal

d

◭ Diagram

Cube: every edge

a

, diagonal

a 3

Measure	Cuboid	Cube
Volume	$l bh$	$a^{3}$
Lateral surface (4 walls)	$2 (l + b) h$	$4 a^{2}$
Total surface	$2 (l b + bh + h l)$	$6 a^{2}$
Diagonal	$l^{2} + b^{2} + h^{2}$	$a 3$
Largest sphere inside / sphere around	—	radius $\frac{a}{2}$ / $\frac{3}{2} a$

∑ Formula

V = x yz (l + b + h)^{2} = d^{2} + TSA

Two shortcuts: volume from the areas

x, y, z

of three adjacent faces; and total surface from the edge sum and the diagonal

Spot itRecognise: “longest rod/bamboo/pencil” → diagonal. “areas of three faces” →

x yz

. “

l = 3 b = 4 h

” or “ratio” → edges

p x, q x, r x

(for the ratio take the LCM:

l = 3 b = 4 h

⟹

l : b : h = 12 : 4 : 3

). “sum of 12 edges” →

4 (l + b + h)

.

Write every measure in one unit.
Decide what is asked: space (volume), covering (lateral or total surface), or the longest object (diagonal).
If an edge is unknown, call it $x$ (with a ratio, $p x, q x, r x$ ) and write the equation for the given quantity.
Find $x$ , apply the formula for the asked quantity and check the unit (cubic/square).

Example 1 · Basic

A pencil box is 24 cm × 8 cm × 6 cm. Find its volume, lateral surface, total surface and the longest pencil that fits in it.

Solution, step by step

1Volume tells how much space the box has: $V = l bh = 24 \times 8 \times 6 = 1152$ cu cm.
2Open the four walls into one strip: length = perimeter $2 (24 + 8) = 64$ , height 6, so lateral surface $= 64 \times 6 = 384$ sq cm.
3Add the top and bottom faces ( $24 \times 8 = 192$ each): total surface $= 384 + 2 \times 192 = 768$ sq cm. Check: $2 (192 + 48 + 144) = 768$ ✓.
4The longest pencil runs corner to opposite corner: $d = 576 + 64 + 36 = 676 = 26$ cm.

Answer 1152 cu cm; 384 sq cm; 768 sq cm; 26 cm

Notice a 26 cm pencil fits in a 24 cm long box — because it lies slanting.

Example 2 · Standard

Three adjacent faces of a cuboid cake box have areas 48, 60 and 80 sq cm. Find the volume and the three edges.

Solution, step by step

1Three adjacent faces are $l b$ , $bh$ , $h l$ . Multiplying all three uses every edge twice: $(l b) (bh) (h l) = (l bh)^{2}$ .
2So $V^{2} = 48 \times 60 \times 80 = 230400$ , and $V = 230400 = 480$ cu cm.
3Each edge = volume ÷ the face made by the other two edges: $480 \div 48 = 10$ , $480 \div 60 = 8$ , $480 \div 80 = 6$ .

Answer 480 cu cm; edges 10, 8, 6 cm

The long way (solving three equations) gives the same, but

x yz

gets it in one line.

Example 3 · Standard

A cubical block has a total surface of 726 sq cm. Find its volume and diagonal.

Solution, step by step

1A cube has 6 equal faces: $6 a^{2} = 726$ , so one face $a^{2} = 121$ and $a = 11$ cm.
2Volume $= a^{3} = 1 1^{3} = 1331$ cu cm.
3Diagonal $= a 3 = 113$ cm.

Answer 1331 cu cm; $113$ cm

Example 4 · Advanced

In a cuboid

l = 2 b = 3 h

and the volume is 2304 cu cm. Find its total surface.

Solution, step by step

1All three equal one number: $l = 2 b = 3 h = k$ . Then $l = k$ , $b = \frac{k}{2}$ , $h = \frac{k}{3}$ ; to clear fractions take the LCM of 2 and 3, i.e. 6: $l : b : h = 6 : 3 : 2$ .
2Edges $6 x, 3 x, 2 x$ : volume $= 36 x^{3} = 2304$ ⟹ $x^{3} = 64$ ⟹ $x = 4$ . Edges 24, 12, 8 cm.
3Total surface $= 2 (24 \cdot 12 + 12 \cdot 8 + 8 \cdot 24) = 2 (288 + 96 + 192) = 1152$ sq cm.

Answer 1152 sq cm

Trap:

2 (l + b) h = 576

is only the four walls. “Total” adds the top and bottom too.

BasicTry it yourself first

Paper is to be pasted on all the faces of a cubical box of edge 9 cm. How much paper (sq cm) is needed?

BasicTry it yourself first

A godown is 16 m long, 12 m wide and 15 m high. What is the longest straight bamboo that can be kept in it?

BasicTry it yourself first

A cubical water tank has an inner edge of 13 dm. How many litres does it hold? (1 cu dm = 1 litre)

StandardTry it yourself first

The areas of three adjacent faces of a brick are 60, 35 and 84 sq cm. Find the volume of the brick (cu cm).

StandardTry it yourself first

The diagonal of a cube is

83

cm. Find its total surface (sq cm).

StandardTry it yourself first

The length, breadth and height of a cuboid are in the ratio 4 : 3 : 2 and its total surface is 1,300 sq cm. Find the volume (cu cm).

AdvancedTry it yourself first

The length, breadth and height of a cuboid are in the ratio 12 : 3 : 2 and its volume is 4,608 cu cm. Find the total surface (sq cm).

AdvancedTry it yourself first

The 12 edges of a cuboid add up to 76 cm and its diagonal is

13

cm. Find its total surface (sq cm).

Rooms, boxes and cost

Walls · paper · thickness

Cost of painting or papering walls, carpets, a box from a sheet with corners cut, and a chest with thick walls.

A room is a cuboid seen from inside. The four walls are its lateral surface $2 (l + b) h$ ; the floor and the ceiling are each $l b$ . Doors and windows get no paint or paper, so their area is subtracted. Cost is always area × rate (per sq m).

A carpet or wallpaper is one long strip of fixed width. Strip area = length × width, so length = area ÷ width; if the price is “per metre”, multiply by the length. Box from a sheet: cut a square of side $x$ at every corner and each side loses $2 x$ , while the folded flaps become the height $x$ . Chest with thick walls: inner measure = outer − thickness on both sides; wood = outer volume − inner volume.

∑ Formula

4 walls = 2 (l + b) h V_{box} = (L - 2 x) (B - 2 x) x

Four walls, and the box made by cutting corners

Chest	Inner length, breadth	Inner height
closed (with lid)	$l - 2 t, b - 2 t$	$h - 2 t$
open (no lid)	$l - 2 t, b - 2 t$	$h - t$

Spot itRecognise: “paint/plaster the walls” →

2 (l + b) h

; “ceiling too” →

+ l b

; “width of carpet/paper” → length = area ÷ width; “squares cut from the corners” →

(L - 2 x) (B - 2 x) x

; “how much wood” → outer − inner.

Draw a sketch and decide which surfaces are covered (walls? ceiling? floor?).
Add their areas and subtract the openings.
If the rate is per sq m, multiply directly; if a strip is priced per metre, first length = area ÷ width.
In thickness questions write the inner measures first, then subtract the volumes.

Example 5 · Basic

A hall is 12 m × 9 m × 5 m. It has one door of 2 m × 1.5 m. Find the cost of painting the four walls and the ceiling at ₹25 per sq m.

Solution, step by step

1Four walls $= 2 (l + b) h = 2 (12 + 9) \times 5 = 210$ sq m.
2The door is not painted: subtract $2 \times 1.5 = 3$ sq m, leaving 207 sq m.
3The ceiling is the same as the floor: add $12 \times 9 = 108$ sq m; total $207 + 108 = 315$ sq m.
4Cost $= 315 \times 25$ = ₹7875.

Answer ₹7,875

Example 6 · Standard

A room is 7 m × 5 m × 3.5 m with two windows of 1.5 m × 1.2 m and one door of 2 m × 1 m. Paper 70 cm wide costs ₹12 per metre. Find the cost of papering the walls.

Solution, step by step

1Walls $= 2 (7 + 5) \times 3.5 = 84$ sq m.
2Windows $2 \times 1.8 = 3.6$ and the door 2, total 5.6 sq m to subtract: $84 - 5.6 = 78.4$ sq m.
3The paper strip is 0.7 m wide, so its length $= 78.4 \div 0.7 = 112$ m.
4Cost $= 112 \times 12$ = ₹1344.

Answer ₹1,344

Do not forget to change the width to metres — dividing by 70 gives an answer 100 times too small.

Example 7 · Standard

An open box is made from a 40 cm × 30 cm sheet by cutting 5 cm squares at the corners. How many litres of water will it hold?

Solution, step by step

1Each side loses 5 cm at both ends: base $(40 - 10) \times (30 - 10) = 30 \times 20$ .
2The folded flaps are 5 cm high, so volume $= 30 \times 20 \times 5 = 3000$ cu cm.
31000 cu cm = 1 litre, so 3 litres.

Answer 3 litres

Example 8 · Advanced

A closed wooden box measures 60 cm × 40 cm × 30 cm outside and the wood is 2.5 cm thick. Find the volume of the wood.

Solution, step by step

1A closed box has a wall on both sides in every direction, so each measure loses $2 \times 2.5 = 5$ : inside $55 \times 35 \times 25$ .
2Outer volume $= 60 \times 40 \times 30 = 72000$ ; inner $= 55 \times 35 \times 25 = 48125$ .
3Wood $= 72000 - 48125 = 23875$ cu cm.

Answer 23,875 cu cm

For an open box the height would lose only 2.5 (no lid).

BasicTry it yourself first

A room is 10 m × 8 m × 4 m. Find the cost of painting the four walls at ₹30 per sq m.

BasicTry it yourself first

The floor of a 9 m × 6 m hall is to be covered with carpet 75 cm wide, sold at ₹40 per metre (of length). Find the total cost.

StandardTry it yourself first

A room is 6 m × 5 m × 4 m. It has 1 opening of 2 m × 1.5 m and 2 openings of 1.5 m × 1 m. Find the cost of painting the four walls and the ceiling at ₹45 per sq m.

StandardTry it yourself first

Plastering the four walls of an auditorium 15 m long and 10 m wide cost ₹5,400 at ₹18 per sq m. Find its height.

StandardTry it yourself first

Squares of side 4 cm are cut from the four corners of a 36 cm × 24 cm tin sheet and the sides are folded up to make an open tray. Find the capacity of the tray (cu cm).

AdvancedTry it yourself first

A open (lidless) wooden chest has outer measures 50 cm × 40 cm × 32 cm and the wood is 2 cm thick. Find the volume of wood used (cu cm).

AdvancedTry it yourself first

Wallpaper 50 cm wide, costing ₹15 per metre, is to be put on the walls of a 8 m × 6 m × 4 m room. The doors and windows together have an area of 6 sq m. Find the cost.

Right prism (any base)

Base × height

A solid whose cross-section is the same from bottom to top: volume, lateral and total surface, and the base-area formulas.

A prism is a solid that shows the same shape wherever you cut it (parallel to the base) — a loaf of bread, a pillar, a canal, a triangular glass prism. A cuboid (rectangle base) and a cylinder (circle base) are prisms too. So one rule: volume = base area × height — stack thin copies of the base up to the height.

The side faces are rectangles; open them out in a row and they form one big rectangle whose width = base perimeter and height = prism height. So lateral surface = perimeter × height, and total = lateral + 2 × base. The real work is the base area — remember the table below.

◭ Diagram

Triangular prism: the same base at both ends, height (length)

H

∑ Formula

V = base \times H LSA = perimeter \times H TSA = LSA + 2 \times base

For every right prism

Base	Area
right triangle (legs $a, b$ )	$\frac{1}{2} ab$
any triangle (Heron)	$s (s - a) (s - b) (s - c)$
triangle, inradius $r$	$r \times s$ ( $s$ = semi-perimeter)
equilateral triangle	$\frac{3}{4} a^{2}$
trapezium	$\frac{1}{2} (a + b) h$
regular hexagon	$\frac{3 3}{2} a^{2}$
regular octagon	$2 (1 + 2) a^{2}$

Spot itRecognise: “base is a triangle/trapezium/hexagon” or a canal, pillar, drain → prism. Careful: the height of the triangle and the height of the prism are different things. For an equation like $a (a + k) = N$ guess

a

from the unit digit: in

a (a + 30) = 304

the product ends in 4,

a = 8

(

8 \times 38 = 304

) ✓.

Identify the base shape and find its area (table).
For the volume multiply by the height; for the height divide the volume by the base.
Lateral surface = base perimeter × height; for the total add two bases.

Example 9 · Basic

A triangular prism has a base triangle with sides 9 cm, 12 cm, 15 cm and a length of 20 cm. Find the volume, lateral and total surface.

Solution, step by step

1 $9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$ , so the base is a right triangle (the right angle between 9 and 12). Area $= \frac{1}{2} \times 9 \times 12 = 54$ .
2Volume = base × length $= 54 \times 20 = 1080$ cu cm.
3Perimeter $= 9 + 12 + 15 = 36$ ; lateral $= 36 \times 20 = 720$ sq cm.
4Total = lateral + both ends $= 720 + 2 \times 54 = 828$ sq cm.

Answer 1080 cu cm; 720 sq cm; 828 sq cm

Example 10 · Standard

A prism stands on a triangle with sides 13, 14, 15 cm and has a volume of 1260 cu cm. Find its height and total surface.

Solution, step by step

1The triangle is not right-angled ( $1 3^{2} + 1 4^{2} \neq = 1 5^{2}$ ), so use Heron: $s = \frac{13 + 14 + 15}{2} = 21$ .
2Area $= 21 \times 8 \times 7 \times 6 = 7056 = 84$ sq cm.
3Height $= 1260 \div 84 = 15$ cm.
4Total $= 42 \times 15 + 2 \times 84 = 630 + 168 = 798$ sq cm.

Answer 15 cm; 798 sq cm

Example 11 · Standard

A prism on a square base is 12 cm high with a total surface of 512 sq cm. Find its volume.

Solution, step by step

1Base side $a$ : the four side faces give $4 \times a \times 12 = 48 a$ and the two squares $2 a^{2}$ . So $2 a^{2} + 48 a = 512$ ⟹ $a^{2} + 24 a = 256$ ⟹ $a (a + 24) = 256$ .
2The product ends in 6; try $a = 8$ : $8 \times 32 = 256$ ✓.
3Volume $= a^{2} \times H = 64 \times 12 = 768$ cu cm.

Answer 768 cu cm

The quadratic formula gives the same, but in the exam “guess from factors” is faster.

Example 12 · Advanced

A regular hexagonal prism has a base side of 4 cm and a height of

53

cm. Find its volume and lateral surface.

Solution, step by step

1A regular hexagon splits from its centre into 6 equilateral triangles: area $= 6 \times \frac{3}{4} \times 16 = 243$ .
2Volume $= 243 \times 53 = 24 \times 5 \times 3 = 360$ cu cm.
3Lateral = perimeter × height $= 24 \times 53 = 1203$ sq cm.

Answer 360 cu cm; $1203$ sq cm

BasicTry it yourself first

A canal (water channel) has a trapezium cross-section: 1.2 m wide at the top, 0.8 m at the bottom and 0.5 m deep. How much water (cu m) does 6 m of it hold?

BasicTry it yourself first

A glass prism stands on a right triangle whose legs are 8 cm and 15 cm; the prism is 20 cm long. Find its lateral surface (sq cm).

StandardTry it yourself first

A concrete pillar (prism) has a triangular base with sides 10 m, 17 m, 21 m and a height of 9 m. How much concrete (cu m) is used?

StandardTry it yourself first

A tent-shaped box (triangular prism) has a base triangle of sides 7, 24, 25 cm and a length of 10 cm. Find its total surface (sq cm).

StandardTry it yourself first

A prism stands on a triangle with perimeter 36 cm and inradius 3 cm. The prism's volume is 540 cu cm. Find its total surface (sq cm).

AdvancedTry it yourself first

A prism (pillar) on a square base is 10 cm high and has a total surface of 312 sq cm. Find its volume (cu cm).

AdvancedTry it yourself first

A hexagonal pencil (regular hexagonal prism) has a base side of 4 mm and a lateral surface of 144 sq mm. Find its volume (cu mm).

2

Round solids

Cylinder · pipe · sphere

Cylinder, hollow cylinders and pipes, winding wire, sphere and hemisphere.

Cylinder

π r^{2} h

·

2 π r h

Volume, curved and total surface of a cylinder; a cylinder from a rolled sheet; rollers; CSA–TSA relations.

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Hollow cylinders, pipes and winding

π (R^{2} - r^{2}) h

Metal in a pipe, the whole surface of a hollow cylinder, radii from differences, winding wire and a slanting string.

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Sphere and hemisphere

\frac{4}{3} π r^{3}

·

4 π r^{2}

Sphere and hemisphere formulas, hollow ball, bowl, cutting a sphere into wedges and ratios of radii.

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3

Pointed solids

Cone · frustum · pyramid

Cone, cutting a cone and the frustum, pyramid and tetrahedron.

Cone

\frac{1}{3} π r^{2} h

·

π r l

Volume and surface of a cone, slant height, a cone from a sector, tents and the semi-vertical angle.

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Cutting a cone; the frustum

Length

k

→ volume

k^{3}

A cone cut parallel to the base: the similar small cone, equal-height pieces (1:7:19), frustum formulas and buckets.

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Pyramid and tetrahedron

\frac{1}{3} \times

base

\times H

The three heights of a pyramid (vertical, slant height, slant edge), square/triangle/hexagon/rectangle bases and the regular tetrahedron.

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4

Solids together

Inside · joined · hollowed

One solid inside another, joined solids, hollowed cavities and the solid made by turning a triangle.

Solids inside solids

Find the shared length

Sphere/cylinder/cone in a cube, cube in a sphere, cube or sphere in a hemisphere, sphere in a cone, percentage of wood wasted.

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Composite solids and cavities

Add volumes · think surfaces

Tents, toys, capsules, hollowed cones/hemispheres, and solids made by turning a right triangle.

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5

Recasting, water and flow

The volume stays the same

Melting into a new solid, water rising when something is dropped in, flow through pipes and canals, rain and dug-out earth.

Melting and recasting

Old volume = new volume

Melting into another shape, how many pieces, wire, many spheres into one, loss, cutting cubes and painted cubes.

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Rise of water level

Rise = volume ÷ base

How far water rises when a solid is dropped in, counting marbles, people diving, just covering a ball and a tilted aquarium.

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Flow through pipes, rain and digging

Per hour = section × speed

Filling and emptying through pipes/canals, the speed needed, rainwater, spreading dug earth, a platform round a well and a sloping pool.

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6

Compare and change

Ratios · % change

Comparing two solids by ratios, and the % change in volume/surface when measures change.

Ratio method

Cancel constants, plug ratios

Comparing volumes/surfaces of two solids: $r^{2} h$ , $r^{3}$ , $r h$ , equal volumes, and $r : h$ from a relation between formulas.

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% change in dimensions

Multiply the factors

The % change in volume and surface when radius/height/edge change, the successive % formula, and the “add $x$ to the radius or the height” shortcut.

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7

Summary, mixed practice, revision

Every method in one place

A one-page table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

When · what · how

Every method: how to recognise the question and which formula or shortcut to use.

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Mixed practice (exam style)

All methods mixed

Mixed questions like SSC CGL/CHSL/CPO — first spot the method, then solve.

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Revision

Cards · common traps

Cards to memorise formulas and shortcuts, and the most common exam mistakes.

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This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

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Also in these passes: RAS Pre 2026 — 40 Day Study Plan + Complete Notes (₹300.00) · Competitive Maths (English Medium) (₹90.00) · RRB NTPC Graduation Level 2026 (₹300.00) · RRB NTPC Under Graduate Level 2026 (₹300.00)