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Mensuration Questions for SBI PO

Free, AI-curated practice for the Mensuration section of SBI PO. We have 20+ verified questions in this bank. Below: 5 sample questions. Sign up free to unlock unlimited practice + AI explanations + per-topic analytics.

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📍 Mensuration is also tested in:
SSC CGL (47)SSC MTS (44)SSC GD (40)RRB NTPC (30)
Why this topic matters · 9 min read
Mensuration appears in almost every SBI PO Prelims and Mains quant section, contributing 2-4 questions directly and supporting DI calculations. SBI PO tests it slightly harder than IBPS — expect combined-shape problems, surface area of hollow cylinders, or questions where you derive one dimension from another. Speed matters here: knowing formulas cold saves 30-40 seconds per question.

Key 2D Shapes: Area and Perimeter

Two-dimensional mensuration covers flat shapes. You need both area (space inside) and perimeter (boundary length). SBI PO frequently combines two shapes — like a rectangle with semicircles on its ends — so treat each sub-shape separately and add up.

  • Rectangle: Area = l x b, Perimeter = 2(l + b)
  • Square: Area = side squared, Perimeter = 4 x side, Diagonal = side x root(2)
  • Triangle: Area = 0.5 x base x height; for equilateral = (root3 / 4) x side squared
  • Circle: Area = pi x r squared, Circumference = 2 x pi x r
  • Trapezium: Area = 0.5 x (sum of parallel sides) x height
  • Rhombus: Area = 0.5 x d1 x d2 (product of diagonals halved)
Key formulas
Heron's Formula (Triangle)
Area = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2
When: Use when all three sides are given but height is not
Sector Area
Area = (theta/360) x pi x r squared
When: For a slice of circle with central angle theta degrees
Arc Length
Arc = (theta/360) x 2 x pi x r
When: Boundary of the curved part of a sector
Worked examples

A park is shaped like a rectangle (80m x 40m) with semicircles on both shorter ends. Find total area. Rectangle area = 80 x 40 = 3200 sq m. Two semicircles with radius 20m = one full circle = pi x 20 x 20 = 1257 sq m (approx). Total = 3200 + 1257 = 4457 sq m.

A rhombus has diagonals 12 cm and 16 cm. Area = 0.5 x 12 x 16 = 96 sq cm. Side = sqrt(6 squared + 8 squared) = sqrt(100) = 10 cm.

Key 3D Shapes: Volume and Surface Area

Three-dimensional mensuration tests volume (space inside) and surface area (total outer skin). SBI PO loves cylinder and cone problems, often asking you to compare volumes when dimensions change. A common trick: if radius doubles, volume becomes 4 times (because r is squared in cylinder formula).

  • Cube: Volume = a cubed, Total Surface Area (TSA) = 6 x a squared, Diagonal = a x root(3)
  • Cuboid: Volume = l x b x h, TSA = 2(lb + bh + lh)
  • Cylinder: Volume = pi x r squared x h, CSA = 2 x pi x r x h, TSA = 2 x pi x r x (r + h)
  • Cone: Volume = (1/3) x pi x r squared x h, CSA = pi x r x l, where slant l = sqrt(r squared + h squared)
  • Sphere: Volume = (4/3) x pi x r cubed, Surface Area = 4 x pi x r squared
  • Hemisphere: Volume = (2/3) x pi x r cubed, CSA = 2 x pi x r squared, TSA = 3 x pi x r squared
Key formulas
Cylinder Volume
V = pi x r squared x h
When: Tanks, pipes, circular wells — most common 3D shape in PO papers
Cone Volume
V = (1/3) x pi x r squared x h
When: Cone = one-third of a cylinder with same base and height — use this ratio to save time
Slant Height of Cone
l = sqrt(r squared + h squared)
When: Always needed before calculating cone CSA or TSA
Worked examples

A cylinder has radius 7 cm and height 10 cm. Volume = (22/7) x 7 x 7 x 10 = 1540 cubic cm. CSA = 2 x (22/7) x 7 x 10 = 440 sq cm.

A cone and cylinder have the same base radius 6 cm and same height 9 cm. Ratio of their volumes = (1/3 x pi x 36 x 9) : (pi x 36 x 9) = 1:3. Cone is always one-third the cylinder.

Combined and Conversion Problems

SBI PO especially favours problems where a shape is melted or converted into another — a sphere melted into small spheres, or water flowing through a pipe filling a tank. The key principle: volume stays constant in melting or filling problems. Set up an equation equating volumes.

  • Melting/recasting: Volume of original shape = total volume of new shapes
  • Water flow: Volume of water = cross-section area of pipe x speed x time
  • Painting/fencing problems: calculate only the relevant surface (TSA vs CSA vs base excluded)
  • Cost problems: multiply area or volume by rate per unit
  • If n small spheres are made from one big sphere: n x (4/3 pi r-small cubed) = (4/3 pi R-big cubed)
Key formulas
Number of small spheres from big sphere
n = (R / r) cubed
When: Big sphere melted into small spheres of radius r
Water flow volume
Volume = pi x r squared x speed x time
When: Pipe filling a tank — treat pipe as cylinder
Worked examples

A big sphere of radius 6 cm is melted into small spheres of radius 2 cm. Number of spheres = (6/2) cubed = 3 cubed = 27 spheres.

A pipe of radius 3.5 cm flows water at 5 m/s for 2 minutes. Volume = pi x 3.5 squared x 500 cm x 120 s... Tip: convert all units to same before calculating.

Ratio and Scaling Shortcuts

When dimensions change by a factor, areas and volumes change by the square or cube of that factor. SBI PO loves these shortcut questions because aspirants who do not know the scaling rules waste time re-calculating from scratch.

  • If side of square doubles: area becomes 4 times (2 squared)
  • If radius of circle doubles: area becomes 4 times
  • If all dimensions of cuboid double: volume becomes 8 times (2 cubed)
  • If radius of sphere triples: volume becomes 27 times (3 cubed), surface area becomes 9 times
  • Percentage increase in area when side increases by x percent: use (100 + x) squared / 100 squared - 1
Key formulas
Area scaling
New Area = Old Area x (scale factor) squared
When: Any 2D shape when linear dimensions are scaled
Volume scaling
New Volume = Old Volume x (scale factor) cubed
When: Any 3D shape when all linear dimensions are scaled equally
⚠ Common mistakes to avoid
  • Using diameter instead of radius in circle formulas — always halve the diameter before plugging into pi x r squared
  • Forgetting to use slant height (not vertical height) when calculating cone CSA — slant l = sqrt(r squared + h squared) must be computed first
  • Confusing CSA and TSA for cylinders and hemispheres — in open tank problems use only CSA plus base; read the question carefully about what surface is being painted or covered
  • In melting problems, equating surface areas instead of volumes — volume is conserved in melting, not surface area
  • Not converting units before calculating — mixing metres and centimetres leads to wrong answers in pipe and tank problems
🧠 Memory aids
  • Cone = Cylinder divided by 3 — a cone always holds exactly one-third of a cylinder with the same base and height. Visualise pouring a cone three times to fill the cylinder.
  • Acronym SCCT for sphere surface areas: S = 4 pi r squared (sphere), C = 3 pi r squared (closed hemisphere TSA = 3), C = 2 pi r squared (curved hemisphere only = 2). Numbers go 4, 3, 2.
  • For scaling: AREA squares, VOLUME cubes — if something is scaled by k, area goes k-squared, volume goes k-cubed. Think: 2D needs 2 multiplications, 3D needs 3.
  • Diagonal memory: Square diagonal = side x root(2), Cuboid diagonal = root(l squared + b squared + h squared), Cube diagonal = side x root(3). Dimensions keep adding under the root.
🎯 SBI PO exam tips
  • SBI PO Prelims typically has 2-3 direct mensuration questions; Mains DI sets often embed volume or area calculations inside data tables, so formula recall must be instant.
  • Expect at least one combined-shape or melting-and-recasting problem in Mains — these take 2-3 minutes if you set up the equation correctly from the start, but 5+ minutes if you go by trial.
  • SBI PO sets questions where a cylinder is open at the top (exclude one base from TSA) or a hollow pipe (subtract inner cylinder volume from outer) — read every word of the problem.
  • Use pi = 22/7 when radius is a multiple of 7 (7, 14, 21); use pi = 3.14 otherwise. Mixing these is a time-waster.
  • If a question gives cost of painting per sq metre, immediately calculate only the surface relevant (walls of room = CSA of four walls, floor excluded unless stated) — do not auto-compute full TSA.

Sample questions

Q1 · medium · AI-verified
A cone has a base radius of 14 cm and slant height of 25 cm. What is the curved surface area of the cone? (Use π = 22/7)
  1. 1100 sq cm
  2. 1000 sq cm
  3. 1200 sq cm
  4. 900 sq cm
Q2 · medium · PYQ 2024
The area of rectangle is 920 cm² more than the area of a circle. The ratio of length and breadth of the rectangle is 3 : 2. The ratio of breadth of rectangle and the radius of circle is 16 : 7, then find the radius of circle.
  1. 14 cm
  2. 12 cm
  3. 36 cm
  4. 28 cm
Q3 · medium · PYQ 2019
If perimeter of the base of a cylinder is 88 cm. Then find volume of cylinder if height of cylinder is 0.08 m
  1. 4186 cm³
  2. 4928 cm³
  3. 1111 cm³
  4. 2046 cm³
Q4 · medium · AI-verified
A cuboid has length 15 cm, breadth 12 cm, and height 8 cm. What is the length of its diagonal?
  1. √457 cm
  2. √389 cm
  3. √433 cm
  4. √401 cm
Q5 · medium · AI-verified
A trapezium has parallel sides of lengths 24 cm and 16 cm, and the distance between them is 15 cm. Find the area of the trapezium.
  1. 300 sq cm
  2. 240 sq cm
  3. 360 sq cm
  4. 280 sq cm
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