Quick Summary: Mensuration and Geometry contribute 1 to 3 marks in GATE General Aptitude 2027 across every paper (DA, RA, CS, ME, EE, CE, EC). GATE asks for areas, perimeters, volumes and surface areas of triangles, quadrilaterals, circles, cylinders, cones and spheres, often in a “cut, fold, revolve or melt” word problem. This guide covers every formula GATE uses, three labelled diagrams, 8 solved previous-year and GATE-style questions, a 5-question practice set, a one-page formula sheet and the traps that cost a mark you had already earned.
📋 Table of Contents
- Why this topic matters in GATE GA
- What the official GA syllabus says
- Triangles: area, Pythagoras, similarity
- Quadrilaterals and polygons
- Circles, sectors and inscribed figures
- Solids: cube, cuboid, cylinder, cone, sphere
- Shortcuts and traps
- Solved PYQs and GATE-style questions
- Practice set
- One-page formula sheet
- Common mistakes
- FAQs
Why this topic matters in GATE GA
Mensuration is the chapter GATE uses to make a 1-mark question feel like a 2-mark question. The formulas are Class 10 level, but the question wraps them in a story: a sheet is rolled into a tube, a triangle is revolved into a cone, a sphere is melted into a wire, a square has a circle cut out. If you can translate the story into the right formula in 20 seconds, the mark is yours. If you cannot, you spend four minutes and still guess.
The GA section is common to DA, RA, CS, ME, EE, CE and EC, so everybody faces the same mensuration question. Engineering students have a hidden advantage here: they already visualise solids. The disadvantage is overconfidence, which is how πr²h becomes 2πrh in the heat of the exam.
Here is roughly how much this chapter has contributed. Candidates targeting 15/15 in GA should be able to finish any mensuration question inside 90 seconds.
| Year | Marks from this topic (approximate, across papers) | Typical question type |
|---|---|---|
| 2019 | 1 to 2 | Area of a shaded region, triangle inequality |
| 2020 | 1 to 2 | Circle inscribed in a square, cylinder volume |
| 2021 | 2 to 3 | Triangle revolved into a cone, folded sheet |
| 2022 | 1 to 3 | Area ratio of inscribed figures, hexagon |
| 2023 | 2 to 3 | Sheet rolled into a tube, cube surface area |
| 2024 | 1 to 2 | Sector area, similar triangles |
| 2025 | 1 to 2 | Sphere melted into cylinders, polygon angles |
| 2026 | 1 to 3 | Path around a rectangular field, cone frustum |
What the official GA syllabus says
“Mensuration and geometry” is the exact phrase, and it is deliberately broad. In practice GATE stays within plane figures (triangle, quadrilateral, polygon, circle) and the five standard solids. The complete GA syllabus and weightage are in the GATE General Aptitude 2027 hub. The previous post covered Permutation, Combination and Probability.
Triangles: area, Pythagoras, similarity
The area of any triangle is half the base times the perpendicular height. The height must be perpendicular to the chosen base, which is the detail GATE hides in a slanted figure.
- Area = (1/2) × base × height = (1/2) ab sin C = √[s(s − a)(s − b)(s − c)] (Heron, with s = (a + b + c)/2).
- Equilateral side a: area = (√3/4) a², height = (√3/2) a. Right isosceles with legs a: hypotenuse a√2, area a²/2.
- Pythagoras: a² + b² = c². Triples to recognise on sight: 3-4-5, 5-12-13, 8-15-17, 7-24-25 and their multiples (6-8-10, 9-12-15).
- Similar triangles: corresponding sides are in the same ratio k; areas are in ratio k². A line parallel to one side cuts the other two sides proportionally.
- Triangle inequality: the sum of any two sides exceeds the third. Angles add to 180°; the exterior angle equals the sum of the two opposite interior angles.
Solution: s = (13 + 14 + 15)/2 = 21. Area = √[21 × 8 × 7 × 6] = √7056 = 84 cm². (Check: the height on the 14 cm side is 12 cm, and (1/2) × 14 × 12 = 84.)
Solution: AD/AB = 2/5, so triangles ADE and ABC are similar with ratio 2/5. Area ratio = (2/5)² = 4/25.
Quadrilaterals and polygons
| Figure | Area | Perimeter / other |
|---|---|---|
| Square, side a | a² = d²/2 | 4a; diagonal d = a√2 |
| Rectangle l × b | lb | 2(l + b); diagonal √(l² + b²) |
| Parallelogram | base × height | opposite sides equal and parallel |
| Rhombus, diagonals d1, d2 | (1/2) d1 d2 | side = (1/2)√(d1² + d2²) |
| Trapezium, parallel sides a, b, height h | (1/2)(a + b) h | median = (a + b)/2 |
| Regular hexagon, side a | (3√3/2) a² | 6a; made of 6 equilateral triangles |
| Regular n-gon | interior angle sum (n − 2) × 180° | each exterior angle 360°/n; diagonals n(n − 3)/2 |
A path of uniform width w around a rectangle l × b has area (l + 2w)(b + 2w) − lb when outside, and lb − (l − 2w)(b − 2w) when inside. Do not try to add four strips; the corners get counted twice.
Solution: Outer rectangle = 64 × 44 = 2816 m². Lawn = 2400 m². Path = 2816 − 2400 = 416 m².
Circles, sectors and inscribed figures
Circumference = 2πr, area = πr². A sector with central angle θ (degrees) is the fraction θ/360 of the whole circle: arc length = (θ/360) × 2πr and sector area = (θ/360) × πr². In radians: arc = rθ, area = (1/2) r²θ.
- Angle in a semicircle is 90°. The angle subtended at the centre is twice the angle at the circumference on the same arc.
- A tangent is perpendicular to the radius at the point of contact. Two tangents from an external point are equal.
- Chord of length c at distance d from the centre: (c/2)² + d² = r².
- Circle inscribed in a square of side a: r = a/2, circle area = πa²/4, so the circle covers π/4 ≈ 78.5% of the square and the four corners together are (1 − π/4) a² ≈ 0.2146 a².
- Square inscribed in a circle of radius r: the diagonal is 2r, so side = r√2 and square area = 2r². The square covers 2/π ≈ 63.7% of the circle.
Solution: Arc = (90/360) × 2 × (22/7) × 14 = (1/4) × 88 = 22 cm. Perimeter = arc + two radii = 22 + 14 + 14 = 50 cm. (The sector area would be (1/4) × 616 = 154 cm².)
Solution: Circle radius = 5 cm, area = 25π ≈ 78.54 cm². Remaining = 100 − 25π ≈ 21.46 cm². In exact form: 100 − 25π = 25(4 − π).
Solids: cube, cuboid, cylinder, cone, sphere
| Solid | Volume | Curved / lateral surface | Total surface |
|---|---|---|---|
| Cube, side a | a³ | 4a² | 6a²; space diagonal a√3 |
| Cuboid l, b, h | lbh | 2h(l + b) | 2(lb + bh + hl); diagonal √(l² + b² + h²) |
| Cylinder r, h | πr²h | 2πrh | 2πr(r + h) |
| Cone r, h, slant l = √(r² + h²) | (1/3)πr²h | πrl | πr(r + l) |
| Sphere r | (4/3)πr³ | 4πr² | 4πr² |
| Hemisphere r | (2/3)πr³ | 2πr² | 3πr² |
Three “story” conversions GATE loves:
- Rolling a sheet into a tube: the edge that is joined becomes the circumference; the other edge becomes the height. A sheet l × b joined along the l edges gives a cylinder with 2πr = b and h = l.
- Revolving a right triangle about a leg: that leg is the height, the other leg is the radius, the hypotenuse is the slant height.
- Melting and recasting: volume is conserved. Surface area is not.
Solution: Cone with h = 8, r = 6, slant l = √(36 + 64) = 10. Volume = (1/3)π × 36 × 8 = 96π cm³ (about 301.6). Curved surface = π × 6 × 10 = 60π cm² (about 188.5).
Each of these conversions appears as a timed question in the GATE Aptitude Test Series topic test on mensuration, with worked video solutions.
Shortcuts and traps
❌ Using the diameter where the formula needs the radius (quadruples the area).
❌ Adding the base area to the curved surface when the question asks for curved surface only, or forgetting it when the question asks for total.
❌ Taking the joined edge of a rolled sheet as the height instead of the circumference.
❌ Conserving surface area in a melting problem. Only volume is conserved.
❌ Forgetting the two radii when asked for the perimeter of a sector.
❌ Writing the regular hexagon area as 6a² instead of (3√3/2) a².
Solved previous-year and GATE-style questions
Solution: (√3/4) a² = √3, so a² = 4 and a = 2. Perimeter = 3 × 2 = 6. Answer: (C) 6.
Solution: The triangle has legs 1 and 1 and hypotenuse √2. Revolving about a leg gives a cone with r = 1 and h = 1. Volume = (1/3)π(1)²(1) = π/3 ≈ 1.047.
Solution: Sheet area = 216 cm². Cube: 6a² = 216, a = 6, volume 216. Tube: joining the 54 cm edges makes the 4 cm edge the circumference, so 2πr = 4, r = 2/π, height 54. Volume = π × (4/π²) × 54 = 216/π. Ratio = (216/π)/216 = 1/π.
Solution: The square’s diagonal is the diameter 2r, so its side is r√2 and its area is 2r². Circle area = πr². Ratio = πr² : 2r² = π : 2.
Solution: V = πr²h. New volume = π(2r)²(h/2) = 2πr²h. The volume doubles.
Solution: Slant height l = √(9 + 16) = 5 cm. Curved surface = πrl = 15π. Base = πr² = 9π. Total = 24π cm² ≈ 75.4 cm².
Solution: Sphere volume = (4/3)π × 216 = 288π cm³. Wire is a cylinder: π × (0.2)² × L = 0.04πL. So L = 288/0.04 = 7200 cm = 72 m.
Solution: A regular hexagon is six equilateral triangles of side 4: 6 × (√3/4) × 16 = 24√3 ≈ 41.57 cm².
★ PiyushAI Test Series
Best GATE Aptitude Test Series 2027
General Aptitude is 15 marks in every GATE paper and the easiest 15 marks to lose to silly mistakes. The PiyushAI Aptitude Test Series by Piyush Wairale (IIT Madras) gives you topic-wise tests on exactly this chapter, full-length GA mocks in the real GATE interface, detailed video and text solutions, and All-India rank analysis.
Practice set
Attempt these in 7 minutes before opening the key.
- The diagonals of a rhombus are 10 cm and 24 cm. Its perimeter is (A) 48 cm (B) 52 cm (C) 60 cm (D) 68 cm
- A trapezium has parallel sides 8 cm and 12 cm and height 5 cm. Its area is (A) 40 cm² (B) 50 cm² (C) 60 cm² (D) 100 cm²
- The circumference of a circle is 44 cm (take π = 22/7). Its area is (A) 154 cm² (B) 144 cm² (C) 616 cm² (D) 77 cm²
- The space diagonal of a cube is 6√3 cm. Its volume is (A) 108 cm³ (B) 216 cm³ (C) 36 cm³ (D) 648 cm³
- Each interior angle of a regular octagon measures (A) 120° (B) 135° (C) 140° (D) 144°
Show answer key
- (B) Side = √(5² + 12²) = 13; perimeter = 52 cm.
- (B) (1/2)(8 + 12) × 5 = 50 cm².
- (A) 2 × (22/7) × r = 44 gives r = 7; area = (22/7) × 49 = 154 cm².
- (B) a√3 = 6√3 gives a = 6; volume = 216 cm³.
- (B) Exterior angle = 360/8 = 45°; interior = 180 − 45 = 135°.
One-page formula sheet
2. Pythagoras a² + b² = c²; triples 3-4-5, 5-12-13, 8-15-17, 7-24-25.
3. Similar figures: sides ratio k, areas k², volumes k³.
4. Square a², diagonal a√2. Rectangle lb, diagonal √(l² + b²). Rhombus (1/2) d1d2. Trapezium (1/2)(a + b)h.
5. Regular hexagon (3√3/2) a². n-gon: interior sum (n − 2)180°, exterior 360°/n, diagonals n(n − 3)/2.
6. Circle: 2πr, πr². Sector: arc (θ/360) 2πr, area (θ/360) πr² = (1/2) r × arc.
7. Circle in square of side a: πa²/4. Square in circle of radius r: 2r².
8. Cube a³, 6a², diagonal a√3. Cuboid lbh, 2(lb + bh + hl), diagonal √(l² + b² + h²).
9. Cylinder πr²h, CSA 2πrh, TSA 2πr(r + h).
10. Cone (1/3)πr²h, CSA πrl, TSA πr(r + l), l = √(r² + h²).
11. Sphere (4/3)πr³, 4πr². Hemisphere (2/3)πr³, CSA 2πr², TSA 3πr².
12. Rolled sheet: joined edge = circumference. Revolved triangle: axis leg = height. Melting: volume conserved.
Common mistakes
❌ Reporting a NAT answer with π left inside when a number is required (write 1.047, not π/3).
❌ Confusing the slant height with the vertical height of a cone.
❌ Using the interior-angle formula for an irregular polygon’s individual angles (only the sum is fixed).
❌ Forgetting that a hemisphere’s total surface includes the flat circular face.
❌ Computing the path area as 2w(l + b), which misses the four corner squares.
FAQs
How many mensuration questions come in GATE GA?
Usually one question worth 1 or 2 marks per paper, sometimes two. It is often a 2-mark question dressed as a story (rolling, revolving, melting, cutting), so it rewards the formula sheet and a quick sketch.
Which value of π should I use in GATE?
If the options have π in them, do not substitute at all. For a NAT, use 3.1416; the accepted range is built around 3.14. Use 22/7 only when the question explicitly says so.
Do I need coordinate geometry or trigonometry for GA?
Only the basics: distance between two points, the 30-60-90 and 45-45-90 triangle ratios, and sin, cos, tan of standard angles. Full coordinate geometry belongs to the core Engineering Mathematics section, not GA.
Is the PiyushAI Aptitude Test Series enough for GA?
Yes. The Best GATE Aptitude Test Series 2027 has a topic test on mensuration and geometry, tests on every other GA chapter, full-length GA mocks in the GATE interface, solved PYQs from 2010 to 2026 and All-India rank analysis. With the free guides on this site, it covers everything the GA section asks.
Should I draw a figure for every mensuration question?
Yes, a ten-second sketch on the rough sheet. Almost every mensuration error in GATE comes from misreading which edge is joined, which leg is the axis, or which region is shaded. The sketch removes all three.
Stop losing easy marks. Practise this topic under exam pressure.
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Related guides: GATE General Aptitude 2027 hub | All GATE General Aptitude articles | GATE DA Syllabus 2027 | GATE RA Syllabus 2027
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