RRB NTPC (Railways) · Quantitative Aptitude

Mensuration (2D & 3D)

Area and perimeter of 2D figures and volume and surface area of 3D solids like prism, cone, cylinder and sphere.

Seven concepts covering the 2D areas and the 3D solids that SSC and banking papers actually ask — rectangles and triangles, cuboids and cubes, the cylinder's CSA/TSA trap, the cone's slant height, sphere and hemisphere, the k / k² / k³ scaling rule, and melting–recasting as volume conservation. The arithmetic is the same at every exam that clones this topic; what changes is how deep the blueprint cuts.

RRB NTPC (Railways)
Medium level
7 concepts
44 practice questions

1Area of a rectangle, a triangle and a circle

Area of a rectangle is l\times b, of a triangle \frac12\times\text{base}\times\text{height}, and of a circle \pi r^2 with circumference 2\pi r. Those four formulas are the whole of plane mensuration that competitive papers lean on; every composite figure is one of them glued to another.

The trap is not the formula but the quantity. A fencing question wants the perimeter; a carpeting question wants the area; a circular track question wants the circumference. Read the unit and the verb before reaching for \pi.

Figure. The rectangle carries length versus area. Circle area \pi r^2 and the triangle \tfrac12 bh sit in the topic formula sheet — a circle has no arc primitive here.

How it works

Name the shapeSplit a composite figure into rectangles, triangles and circles before writing a single product.
Pick area or lengthFencing, edging and a lap of a track are lengths; flooring, painting a face and a shaded region are areas.
Choose \piUse \pi=\frac{22}{7} when r is a multiple of 7; otherwise \pi=3.14 keeps the decimals honest.

Park area and fencing cost

A rectangular park is 40\,m long and 25\,m wide. Find its area, and the cost of fencing it at ₹12 per metre.

Area = l\times b = 40\times 251000\,\mathrm{m}^2
Perimeter = 2(l+b) = 2(40+25)130\,\mathrm{m}
Fencing cost = 130\times 12₹1560
Check: cost uses perimeter, not areanot 1000\times 12

Pro tip. The distractor on every fencing item is the area times the rate. Row four is the whole point of the example: the number you just computed for area is sitting there waiting to be misused.

A circle has radius 7\,cm. Using \pi=\frac{22}{7}, its area is

154\,\mathrm{cm}^2
44\,\mathrm{cm}^2
154\pi\,\mathrm{cm}^2

\pi r^2 = \frac{22}{7}\times 49 = 154. 44 is the circumference 2\pi r, the length answer to an area question. Leaving the answer as 154\pi means \pi was never substituted — the question that handed you \frac{22}{7} expected you to use it.

2Volume and surface area of a cuboid and a cube

A cuboid of edges l, b, h has volume lbh and total surface area 2(lb+bh+hl). Its space diagonal is \sqrt{l^2+b^2+h^2}. A cube is the special case l=b=h=a: volume a^3, TSA 6a^2, diagonal a\sqrt3.

Painting and wrapping questions want TSA; packing and capacity want volume; a fly walking on the surface still wants a face diagonal or a net, not the space diagonal. Match the formula to the path.

Figure. Unfold the cuboid into its net: three face-pairs lb, bh and lh, each appearing twice. Total surface area is 2(lb+bh+hl); volume lbh is packing, not wrapping. Faces are schematic, not to scale.

How it works

List the three edgesWrite l, b, h (or a single a for a cube) before multiplying anything.
Choose volume or surfaceCapacity, melting and metal used for a solid need volume; paint and gift-wrap need TSA.
Expand TSA as three face-pairs2(lb+bh+hl) is three rectangular faces, each counted twice — one for each opposite pair.

Cuboid versus cube
Quantity	Cuboid	Cube (a)
Volume	lbh	a^3
TSA	2(lb+bh+hl)	6a^2
Space diagonal	\sqrt{l^2+b^2+h^2}	a\sqrt3

Cuboid volume and TSA

Find the volume and total surface area of a cuboid 10\,cm by 8\,cm by 6\,cm.

Volume = lbh = 10\times 8\times 6480\,\mathrm{cm}^3
Face-pair sum = lb+bh+hl = 80+48+60188
TSA = 2\times 188376\,\mathrm{cm}^2
Units checkcm³ for volume, cm² for TSA

Pro tip. Adding lb+bh+hl once and doubling is faster and less error-prone than computing six faces. The common slip is to forget the leading 2 and report 188 cm² as the TSA.

A cube has total surface area 150\,\mathrm{cm}^2. Its volume is

125\,\mathrm{cm}^3
150\,\mathrm{cm}^3
25\,\mathrm{cm}^3

6a^2=150\Rightarrow a^2=25\Rightarrow a=5, so volume a^3=125. Reporting 150 reuses the surface-area number as a volume. 25 is a^2, one step short.

3Cylinder: curved surface versus total surface

For a solid cylinder, the curved surface area is \mathrm{CSA}=2\pi rh but the total surface area is \mathrm{TSA}=2\pi r(h+r), which adds the two circular ends. Mixing these up is a classic error — and open cylinders, pipes and tanks without lids drop one or both end-circles on purpose.

Volume is the easy sibling: V=\pi r^2 h. Recasting and capacity questions use volume; paint on the curved wall alone uses CSA; a closed tin uses TSA.

Figure. Unrolled curved wall only. The two end-circles that turn CSA into TSA cannot be drawn without an arc primitive.

How it works

Read open or closedClosed tin → TSA; label on the curved wall → CSA; open-top tank → CSA + one base =\pi r(2h+r).
Write the factor 2\pi rBoth CSA and TSA start from the circumference 2\pi r; CSA multiplies by h, TSA by (h+r).
Cancel when r is a multiple of 7With \pi=\frac{22}{7} and r=7k, the 7 cancels before any other arithmetic.

Total surface area of a closed cylinder

Find the total surface area of a closed cylinder with radius 7\,cm and height 10\,cm (use \pi=\frac{22}{7}).

TSA = 2\pi r(h+r)2\pi\cdot 7\cdot(10+7)
= 2\times\frac{22}{7}\times 7\times 172\times 22\times 17
2\times 22\times 17748\,\mathrm{cm}^2
CSA alone would be 2\pi rh440\,\mathrm{cm}^2 (not asked)

Pro tip. Choosing \pi=\frac{22}{7} when r=7 makes the 7 cancel, avoiding decimals entirely. The 440 in the last row is the CSA distractor — same cylinder, wrong surface.

An open-top cylinder has r=7\,cm and h=10\,cm. Using \pi=\frac{22}{7}, the surface to be painted (outside curved wall and base) is

594\,\mathrm{cm}^2
748\,\mathrm{cm}^2
440\,\mathrm{cm}^2

Open top means CSA + one base = 2\pi rh+\pi r^2=\pi r(2h+r)=\frac{22}{7}\times 7\times 27=594. 748 is the closed TSA; 440 is CSA alone and drops the base the question still wants.

4Cone: slant height links radius and height

The slant height of a cone is l=\sqrt{r^2+h^2} — the hypotenuse of the right triangle formed by the radius and the vertical height. Curved surface area is \pi r l and volume is \frac13\pi r^2 h. The standard slip is to put vertical height into the curved-surface formula in place of slant height.

A 3-4-5 or 5-12-13 triple keeps the slant an integer and the arithmetic clean. When the question hands you the slant and the radius, recover the vertical height by Pythagoras before touching the volume.

Figure. Axial cross-section only: the right triangle whose hypotenuse is the slant height. The circular base needs an arc primitive and is not drawn.

How it works

Draw the right triangleRadius on the base, height upright, slant as hypotenuse — l is never an independent third length.
Find the missing edgeFrom any two of r, h, l, Pythagoras gives the third.
Match formula to surfaceCSA uses l; volume uses h. A closed cone's TSA is \pi r(l+r).

Cone from a 3-4-5 triple

A cone has radius 3\,cm and vertical height 4\,cm. Find its slant height, curved surface area and volume (leave answers in terms of \pi where they involve \pi).

l=\sqrt{r^2+h^2}=\sqrt{3^2+4^2}5\,\mathrm{cm}
CSA =\pi r l=\pi\cdot 3\cdot 515\pi\,\mathrm{cm}^2
Volume =\frac13\pi r^2 h=\frac13\pi\cdot 9\cdot 412\pi\,\mathrm{cm}^3
If CSA wrongly used h: \pi\cdot 3\cdot 412\pi\,\mathrm{cm}^2 (trap)

Pro tip. Row four is the mark-losing answer: swapping h for l in CSA produces a number that equals the volume's coefficient, which is why it looks plausible in an options list.

A cone has r=5\,cm and h=12\,cm. Its slant height is

13\,cm
17\,cm
12\,cm

l=\sqrt{25+144}=\sqrt{169}=13. Adding instead of Pythagoras gives 17; taking h itself as the slant is the CSA trap waiting for the next line.

5Sphere and hemisphere surface and volume

A sphere has volume \frac43\pi r^3 and surface area 4\pi r^2. A hemisphere's curved surface is 2\pi r^2, and its total surface area is 3\pi r^2 — curved part plus the flat circular face \pi r^2. Dropping the flat face when the question says "total" is the hemisphere trap.

A thin hemispherical bowl painted outside only wants the curved 2\pi r^2; the same bowl with a lid wants 3\pi r^2. Read whether the flat face is exposed.

Figure. Heights are coefficients of \pi r^2: a full sphere is 4, a hemisphere's curved surface is 2, and its total surface is 3 once the flat face is included. Dropping that face when the question says total is the hemisphere trap — the figure cannot draw the sphere itself.

How it works

Sphere or hemisphereFull sphere → 4\pi r^2 and \frac43\pi r^3; cut in half and decide if the flat face counts.
Curved versus totalHemisphere CSA =2\pi r^2; hemisphere TSA =3\pi r^2.
Cancel \pi or substituteLeave answers in terms of \pi unless a value is given; with r=7 and \pi=\frac{22}{7} the 7 cancels.

Sphere versus hemisphere
Solid	Volume	Curved SA	Total SA
Sphere	\frac43\pi r^3	4\pi r^2	4\pi r^2
Hemisphere	\frac23\pi r^3	2\pi r^2	3\pi r^2

Hemisphere total surface area

Find the total surface area of a hemisphere of radius 7\,cm (use \pi=\frac{22}{7}).

TSA = 3\pi r^23\pi\cdot 49
= 3\times\frac{22}{7}\times 493\times 22\times 7
3\times 22\times 7462\,\mathrm{cm}^2
CSA alone = 2\pi r^2308\,\mathrm{cm}^2 (flat face dropped)

Pro tip. 308 is the answer when the question wants only the curved surface. TSA is one flat face more: add \pi r^2=154 to 308 and you recover 462.

A hemispherical bowl of radius 7\,cm is to be painted on the curved outside only. Using \pi=\frac{22}{7}, the area is

308\,\mathrm{cm}^2
462\,\mathrm{cm}^2
154\,\mathrm{cm}^2

Curved only → 2\pi r^2=\frac{22}{7}\times 2\times 49=308. 462 is TSA and paints the flat rim-face as well. 154 is a single flat circle \pi r^2.

6Scale lengths by k, areas by k², volumes by k³

If every linear dimension of a solid scales by k, its surface area scales by k^2 and its volume by k^3. Double every edge and the paint needed becomes four times, while the material needed becomes eight times — used constantly in melting, model-to-statue and "similar solid" problems.

The converse is how you recover k: if volumes are in the ratio 27:8, then k=3/2, not 27/8. Always take the root that matches the dimension you were given.

Figure. For k=2, the three bars are 2, 4 and 8 — length, area and volume scale factors drawn to one common scale.

How it works

Find k from the given ratioLength ratio is k; area ratio is k^2 so k=\sqrt{\text{area ratio}}; volume ratio is k^3 so k=\sqrt[3]{\text{volume ratio}}.
Raise k to the power you needAsked for a surface → k^2; asked for a volume or a weight of similar material → k^3.
Refuse to mix powersAn area ratio applied to a volume, or a volume ratio applied raw to a length, is the mark-losing move.

Cube edges doubled

A cube has edge 4\,cm. A second cube has edge 8\,cm. By what factor do the total surface area and the volume increase?

Linear scale k = 8/42
Surface-area factor = k^2 = 2^24
Volume factor = k^3 = 2^38
Check: TSA 6\times 16\to 6\times 64; V\, 64\to 512factors 4 and 8

Pro tip. The check in row four uses the absolute formulas only to confirm the powers. Once you trust k, you never need the old surface or volume again — multiply the factor and move on.

Two similar solids have volumes in the ratio 27:1. Their surface areas are in the ratio

9:1
27:1
3:1

k^3=27\Rightarrow k=3, so the area ratio is k^2=9. Copying 27:1 across to surface area skips the root; 3:1 stops at the linear ratio.

7Melting and recasting conserve volume

In melting and recasting problems, equate volumes only — mass is conserved for the same metal at the same density, which for a full solid means volume is conserved. Cancel the common \pi before solving; it is never the unknown.

A sphere melted into a cylinder, a cube into a set of smaller cubes, or n cones into one sphere are all the same equation: V_{\mathrm{old}}=V_{\mathrm{new}}. Surface areas are not conserved — the metal is reshaped, so TSA usually changes.

Figure. Equal bars for the cancelled-\pi volumes in the worked example (288\pi each). The sphere and cylinder shapes need arc/3D primitives and are not drawn — only the equality is.

How it works

Write both volume formulasSphere \frac43\pi R^3, cylinder \pi r^2 h, cube a^3 — whichever pair the question names.
Cancel \pi and common factorsDo it before expanding; recasting arithmetic is almost always an integer after the cancel.
Solve for the missing lengthUsually a height or a new radius; check the unit matches the inputs.

Recasting a sphere into a cylinder

A metal sphere of radius 6\,cm is melted and recast into a solid cylinder of radius 4\,cm. Find the height of the cylinder.

Volume conserved: \frac43\pi R^3 = \pi r^2 hcancel \pi
\frac43(6)^3 = (4)^2 h\frac43\times 216 = 16h
288 = 16hh=18
Height of the cylinder18\,\mathrm{cm}

Pro tip. Cancel \pi on both sides immediately — recasting problems are pure volume equations. Equating surface areas instead is a different (wrong) problem.

A metal cube of edge 6\,cm is melted into a cuboid of base 9\,cm by 6\,cm. The height of the cuboid is

4\,cm
6\,cm
9\,cm

Volume 6^3=216 equals 9\times 6\times h, so h=216/54=4. Keeping height 6 assumes the metal was not reshaped; 9 reuses a base edge as the height.

Notes

2D Area Basics: Area of a rectangle is l\times b, of a triangle \frac12\times\text{base}\times\text{height}, and of a circle \pi r^2 with circumference 2\pi r.
Curved vs Total Surface Area: For a solid cylinder, CSA =2\pi rh but TSA =2\pi r(h+r) adds the two circular ends — mixing these up is a classic error.
Cone Slant Height: The slant height l=\sqrt{r^2+h^2} links radius and vertical height; CSA of a cone is \pi r l and volume is \frac13\pi r^2 h.
Sphere & Hemisphere: A sphere has volume \frac43\pi r^3 and surface area 4\pi r^2; a hemisphere's TSA is 3\pi r^2 (curved 2\pi r^2 plus flat circle \pi r^2).
Scaling Effect: If every linear dimension of a solid scales by k, its surface area scales by k^2 and volume by k^3 — used in melting/recasting problems.

Formulas

Circle: Area =\pi r^2, Circumference =2\pi r
Cylinder: Volume =\pi r^2 h, CSA =2\pi rh, TSA =2\pi r(h+r)
Cone: Volume =\frac13\pi r^2 h, CSA =\pi r l, l=\sqrt{r^2+h^2}
Sphere: Volume =\frac43\pi r^3, Surface area =4\pi r^2
Cuboid: Volume =lbh, TSA =2(lb+bh+hl), diagonal =\sqrt{l^2+b^2+h^2}
Cube: Volume =a^3, TSA =6a^2, diagonal =a\sqrt3

Exam traps & shortcuts

In melting/recasting problems, equate volumes only (mass is conserved) and cancel the common \pi before solving.
Read carefully whether a question wants CSA/LSA or TSA — for open tanks and pipes drop the end-circle terms.
Use \pi=\frac{22}{7} when the radius is a multiple of 7 to keep arithmetic clean; otherwise \pi=3.14.

Reference tables

Night-before reference. CSA means curved (lateral) surface; TSA means every exposed face.

Mensuration formula sheet
Figure	Volume / area	Surface / perimeter
Rectangle	Area =lb	Perimeter =2(l+b)
Triangle	Area =\frac12 bh	—
Circle	Area =\pi r^2	Circumference =2\pi r
Cuboid	lbh	TSA =2(lb+bh+hl)
Cube	a^3	TSA =6a^2
Cylinder	\pi r^2 h	CSA =2\pi rh; TSA =2\pi r(h+r)
Cone	\frac13\pi r^2 h	CSA =\pi rl with l=\sqrt{r^2+h^2}
Sphere	\frac43\pi r^3	Surface =4\pi r^2
Hemisphere	\frac23\pi r^3	CSA =2\pi r^2; TSA =3\pi r^2

Recap

Read only this the night before.

2D: Rectangle lb, triangle \frac12 bh, circle \pi r^2 / 2\pi r. Fencing is perimeter; carpeting is area.
Cuboid / cube: Volume lbh or a^3; TSA 2(lb+bh+hl) or 6a^2. Diagonal needs Pythagoras in 3D.
CSA ≠ TSA: Cylinder CSA =2\pi rh, TSA =2\pi r(h+r). Open top drops one end: \pi r(2h+r).
Cone slant: l=\sqrt{r^2+h^2}. CSA uses l; volume uses h. Never swap them.
Hemisphere: Curved 2\pi r^2, total 3\pi r^2. The flat face is the difference.
Scaling: Lengths \times k, areas \times k^2, volumes \times k^3. Recover k with the matching root.
Recast: Equate volumes, cancel \pi, solve. Surface area is not conserved.

Practise Mensuration (2D & 3D)

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