Mensuration is measurement: how much paint covers a dome, how much water a tank holds, how much canvas a tent needs. For the Punjab Police Sub Inspector paper it is one of the most reliable scoring chapters — 83 questions across every SI paper from 2016 to 2023, and almost all of them are solved by putting numbers into a formula you already know. What makes this page different from a school chapter is where it spends its time. The SI does not ask about every shape equally. Nearly two out of every three mensuration questions are about the cylinder, the cone and the sphere. So that is where most of this page lives. You will learn (1) the handful of formulas that are actually tested, (2) the eight question shapes the Commission keeps reusing, and (3) the traps that turn a correct formula into a wrong answer. Start from zero if you need to — nothing here assumes more than multiplication and square roots.
First, the three things that decide your mark
Before any formula — π, units and the one question to ask yourself.
Most mensuration marks are lost after the formula, not before it. Three habits fix that.
Look at the radius. If it is 7, 14, 21, 28, 3.5, 4.2 or any other multiple of 7, the paper wants π = 22/7 — the 7s cancel and the answer comes out clean. If the radius is 5, 10 or the options carry decimals, use π = 3.14. If every option has a π in it (288π cm³), do not substitute at all — leave π standing and save thirty seconds. The paper usually tells you, but when it does not, the radius tells you.
A surface area is in square units, a volume in cube units. Mixing them is the single most common wrong answer. Learn these three conversions cold: 1 m = 100 cm, 1 m² = 10,000 cm², 1 m³ = 1,000,000 cm³. And for liquids, 1 m³ = 1,000 litres, 1,000 cm³ = 1 litre.
| Asked for | Unit looks like | Example from the SI papers |
|---|---|---|
| Length, perimeter, slant height, depth | cm, m | “Find the slant height” → 13 cm |
| Area, surface area, canvas, levelling | cm², m² | “Total surface area” → 1760 cm² |
| Volume, capacity, earth dug out | cm³, m³, ਲਿਟਰ | “Volume of the hemisphere” → 19404 cm³ |
Curved surface area (CSA) is the side only — the label on a tin, the canvas of a tent, the part of a roller that touches the ground. Total surface area (TSA) adds the flat circles: two for a closed cylinder, one for a cone or a hemisphere. Read the stem for the words open, closed, hollow, without lid — they are there to tell you which one.
Flat shapes: 22 questions
Triangle, quadrilateral, polygon and circle.
| Shape | Area | Perimeter |
|---|---|---|
| Square (side a) | a² — or d²/2 from the diagonal | 4a |
| Rectangle (l × b) | l × b | 2(l + b) |
| Triangle (base b, height h) | ½ × b × h | sum of the three sides |
| Right triangle (legs a, b) | ½ × a × b | a + b + √(a²+b²) |
| Equilateral triangle (side a) | (√3/4) × a² | 3a |
| Rhombus (diagonals d₁, d₂) | ½ × d₁ × d₂ | 4 × side |
| Parallelogram | base × height | 2(sum of adjacent sides) |
| Regular hexagon (side a) | (3√3/2) × a² | 6a |
| Circle (radius r) | πr² | 2πr (this is the circumference) |
When you are given three sides and no height, use Heron. First find the semi-perimeter s = (a+b+c)/2, then Area = √[s(s−a)(s−b)(s−c)]. The SI uses it once or twice a year, always with an isosceles triangle so the arithmetic stays kind.
What is the area of the triangle whose sides are 12 cm, 9 cm and 9 cm? (SI 29 Jul 2023)
Show answer
Sides 12, 9, 9. Semi-perimeter s = (12+9+9)/2 = 15. Area = √[15 × (15−12) × (15−9) × (15−9)] = √[15 × 3 × 6 × 6] = √(45 × 36) = 6√45 = 6 × 3√5 = 18√5 cm². Shortcut for an isosceles triangle: the height onto the unequal side is √(9² − 6²) = √45, so area = ½ × 12 × √45 = 6√45 — the same thing, faster.
Area = (√3/4)a² and height = (√3/2)a. They differ only in the 4 and the 2, and the SI has asked the area with side 4 cm and side 12 cm in different years. With √3 ≈ 1.732, a side-12 triangle is (√3/4) × 144 = 36√3 ≈ 62.35 cm².
Rectangles and squares look trivial, and that is exactly why the SI dresses them up. The stem gives you a diagonal, or a difference, or a perimeter shared between two shapes — never the two numbers you want. The move is always the same: name the unknown, write one equation from the sentence, solve, and only then put it into the area formula.
The difference between the length and breadth of a rectangle is 41m. If its perimeter is 270 m, then its area is : (SI 2 Jul 2023)
Show answer
Let the length be l and the breadth b. The sentence gives you two equations: l − b = 41 and 2(l + b) = 270, so l + b = 135. Add them: 2l = 176, so l = 88 and b = 135 − 88 = 47. Area = 88 × 47 = 4136 m².
You are almost never given both diagonals. You are given the perimeter and one diagonal. The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle whose legs are the two half-diagonals. Find the missing half by Pythagoras, double it, then use ½ d₁d₂.
Diya has a rhombus-shaped garden. Its perimeter is 68 m and one of its diagonals is 16 m. Find the cost (in rupees) of levelling the garden at the rate of ₹45.75 per m. (SI 19 Aug 2021)
Show answer
Perimeter 68 m ⇒ side = 68/4 = 17 m. One diagonal is 16 m, so half of it is 8 m. The other half-diagonal = √(17² − 8²) = √(289 − 64) = √225 = 15 m, so the second diagonal is 30 m. Area = ½ × 16 × 30 = 240 m². Cost = 240 × 45.75 = ₹ 10,980.
Everything about a circle comes from r. Circumference = 2πr, Area = πr², Diameter = 2r. A semicircle's perimeter is the curved half πr plus the straight diameter 2r — forgetting the diameter is the classic slip.
- Levelling, fencing, painting, carpeting → area or perimeter? Fencing follows the boundary (perimeter); levelling, turfing and carpeting cover the inside (area).
- A semicircle's perimeter is πr + 2r, not πr.
- “Three sides fenced, the fourth is the house wall” — the fenced length is l + 2b, not the full perimeter.
- An isosceles triangle has two equal sides; an equilateral has three. The formula with √3 belongs only to the second.
Also asked once — open if you have timeTap to open
What will be the area of a rectangular field whose diagonal is 60 m more than its shorter side, and whose longer side is 30 m more than the shorter side? (SI 2 Jul 2023)
Show answer
Let the shorter side be x. Then the longer side is x + 30 and the diagonal is x + 60. Pythagoras: x² + (x+30)² = (x+60)² ⇒ x² − 60x − 2700 = 0 ⇒ (x − 90)(x + 30) = 0 ⇒ x = 90. Sides are 90 and 120, so area = 10800 m².
A rectangular parking area of 64 m² is adjacent to a house. The three sides of this area are fenced and the fourth side is common with the house. If the length of the fenced sides is 24 m in total, then the dimensions of the parking area are : (SI 16 Oct 2022)
Show answer
Three sides fenced: l + 2b = 24 and l × b = 64. Substituting, 2b² − 24b + 64 = 0 ⇒ b² − 12b + 32 = 0 ⇒ (b − 4)(b − 8) = 0. With b = 4, l = 16 — the option offered is 4 m × 16 m.
What is the area of a regular hexagon with each side of it being 2√3 cm long? (SI 24 Aug 2021)
Show answer
A regular hexagon is six equilateral triangles of the same side. Area = 6 × (√3/4)a² = (3√3/2)a². With a = 2√3, a² = 12, so area = (3√3/2) × 12 = 18√3 cm².
The solids: 53 questions
Cylinder, cone, sphere and hemisphere — the heart of the paper.
This is the part worth real study time. Three solids, four formulas each, and the SI has built fifty-three questions out of them. Start by fixing the picture and the letters in your head: r is always the radius of the circular face, h the height measured straight up, and l — only for the cone — the slanting edge from the rim to the tip.
| Solid | Volume | Curved surface (CSA) | Total surface (TSA) |
|---|---|---|---|
| Cylinder | πr²h | 2πrh | 2πr(r + h) |
| Cone | ⅓ πr²h | πrl | πr(r + l) |
| Sphere | ⁴⁄₃ πr³ | 4πr² | 4πr² (same — it has no flat face) |
| Hemisphere | ⅔ πr³ | 2πr² | 3πr² (curved 2πr² + flat πr²) |
Cone is a third of its cylinder; hemisphere is two-thirds of its cylinder. A cone and a cylinder with the same r and the same h — the cone holds exactly one third. That one sentence replaces two formulas and catches the most common SI distractor, which is the answer multiplied or divided by 3.
Look at that triangle once and the cone stops being hard. l² = r² + h². Given any two of r, h, l, you can get the third. The SI asks it in both directions — sometimes you are given r and h and asked for the surface area (which needs l), sometimes you are given the volume and asked for the slant height. Watch for the friendly triples: 3–4–5, 5–12–13, 8–15–17, 7–24–25 and their multiples. If you see a 5 and a 12, the answer is 13 and you have not touched a calculator.
The volume of a right circular cone is 240 π cm³ and its height is 5 cm. What will be its slant height? (SI 2 Jul 2023)
Show answer
Volume = ⅓πr²h = 240π and h = 5, so ⅓ × r² × 5 = 240 ⇒ r² = 144 ⇒ r = 12 cm. Then l = √(r² + h²) = √(144 + 25) = √169 = 13 cm. The 5–12–13 triple again.
- read1. Name the solidCylinder, cone, sphere or hemisphere — and is it solid, hollow or open?
- convert2. Write down r and hHalve any diameter the moment you see it. This is where most marks are lost.
- choose3. Pick the formulaVolume, CSA or TSA? The unit in the options tells you if the stem does not.
- compute4. Choose π lastIf every option carries π, never substitute. Otherwise 22/7 for multiples of 7.
Pattern 1 — Straight substitution
13 of the 53. Free marks.
You are handed r and h and asked for a volume or a surface area. Nothing is hidden. The only thing that can go wrong is a diameter read as a radius, or 22/7 used where the options wanted π left alone.
Find the volume of the cone whose radius is 7 cm and height is 9 cm. (SI 20 Jun 2023)
Show answer
V = ⅓πr²h = ⅓ × (22/7) × 49 × 9. The 7s cancel: (22/7) × 49 = 154. So V = ⅓ × 154 × 9 = 154 × 3 = 462 cm³.
The total surface area of a cylinder whose radius is 14 cm and height is 6 cm is : (Take π=22/7) (SI 9 Jul 2023)
Show answer
TSA = 2πr(r + h) = 2 × (22/7) × 14 × (14 + 6) = 2 × 44 × 20 = 1760 cm². Note (22/7) × 14 = 44 straight away — that is why they chose 14.
- Diameter → radius. Halve it before anything else. Roughly a third of the 53 solid questions begin with a diameter.
- A sphere's TSA equals its CSA (4πr²). A hemisphere's does not — TSA = 3πr², CSA = 2πr².
- If every option ends in π, do not substitute 22/7.
- A “solid hemisphere” painted all over needs 3πr²; a bowl open at the top needs 2πr².
Pattern 2 — Work backwards
11 of the 53. The volume is given; find a dimension.
Here the question hands you the answer to the usual formula and asks for one of its inputs. The method never changes: write the formula, put in what you know, and solve the one-line equation. Keep π as a symbol as long as you can — it usually cancels.
The volume of a cone with radius 7 cm is 735π cm³. Find its height. (SI 20 Jun 2023)
Show answer
⅓πr²h = 735π. The π cancels on both sides at once: ⅓ × 49 × h = 735 ⇒ 49h = 2205 ⇒ h = 45 cm.
If the volume of a 21 cm high right circular cone is 2800π cm³, find its total surface area. (SI 9 Jul 2023)
Show answer
First the radius: ⅓πr² × 21 = 2800π ⇒ 7r² = 2800 ⇒ r² = 400 ⇒ r = 20. Then the slant height: l = √(400 + 441) = √841 = 29. TSA = πr(r + l) = π × 20 × 49 = 980π cm². Two steps, both of them Pattern 2.
Pattern 3 — Melting and recasting
8 of the 53. One idea solves all of them.
When a solid is melted, recast, drawn into a wire, or cut into smaller pieces, the total volume before equals the total volume after. Surface area does not behave this way — it changes, often dramatically. Write the two volume expressions, set them equal, solve.
Metallic spheres of radii 6cm, 8cm and 10cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere. (SI 20 Jun 2023)
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⁴⁄₃πR³ = ⁴⁄₃π(6³ + 8³ + 10³). The ⁴⁄₃π cancels, so R³ = 216 + 512 + 1000 = 1728 and R = ∛1728 = 12 cm. When spheres melt into one sphere the radii simply add in cubes — no π, no fractions.
A solid metallic cylinder has radius 15.4 cm and height 45 cm. It is heated and drawn into a wire of diameter 4 mm. What is the length (to the nearest m) of the wire? (Take π = 22/7) (SI 29 Jul 2023)
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A “wire” is a very long, very thin cylinder. Volume of the block = π × 15.4² × 45. The wire's diameter is 4 mm = 0.4 cm, so its radius is 0.2 cm. π × 15.4² × 45 = π × 0.04 × L ⇒ L = (237.16 × 45)/0.04 = 266,805 cm ≈ 2668 m. The trap is leaving the wire's diameter in millimetres.
Pattern 4 — Ratio and scaling
7 of the 53. No numbers needed at all.
If every length of a shape is multiplied by k, then every area is multiplied by k² and every volume by k³. This holds for any shape whatsoever — you never need the formula, only the ratio.
| If radii are in the ratio | Surface areas are in | Volumes are in |
|---|---|---|
| 2 : 1 | 4 : 1 | 8 : 1 |
| 3 : 1 | 9 : 1 | 27 : 1 |
| 5 : 4 | 25 : 16 | 125 : 64 |
| k : 1 | k² : 1 | k³ : 1 |
If the radius of a sphere is increased by 12%, then the volume will be increased by (correct to two decimal places) : (SI 29 Jun 2023)
Show answer
The radius becomes 1.12 times its old value, so the volume becomes 1.12³ = 1.404928 times. That is an increase of 0.404928, or 40.49% to two decimal places. Percentage questions in disguise: raise (1 + rate) to the power 3 for a volume, 2 for an area.
- Cube root before you square. Given a volume ratio and asked for areas, take ∛ first.
- Check which shape is named first in the answer — the SI reverses the order to catch you.
- A percentage increase in radius compounds: +12% in r is +40.49% in volume, not +36%.
- The ratio method needs the shapes to be the same kind. Two cones with different proportions scale separately in r and h.
The heights of two right circular cones are in the ratio 2 : 3 and the perimeters of their bases are in the ratio 5 : 7. Find the ratio of their volumes. (SI 14 Jul 2023)
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Perimeters of the bases are in 5 : 7, and perimeter = 2πr, so the radii are also in 5 : 7. Volume = ⅓πr²h, so the ratio is (5² × 2) : (7² × 3) = 50 : 147. Here r and h scale by different factors, so you must use the formula, not the k³ shortcut.
Pattern 5 — Tents, canvas and rollers
5 of the 53. Curved surface area, in disguise.
A tent's cloth, a roller's painted stripe, the label on a tin — all of them are curved surface area. These questions are only hard because the useful sentence is buried in a story.
A roller is a cylinder lying on its side. One complete revolution covers a rectangle as long as the circumference and as wide as the roller, which is exactly its CSA = 2πrh. So area levelled in n revolutions = n × 2πrh — and the number of revolutions needed is the playground's area divided by the CSA.
The diameter of a roller is 98 cm and its length is 90 cm. To level a playground it makes 100 complete revolutions. Find the cost of levelling at the rate of ₹10 per m². (SI 15 Jul 2023)
Show answer
Diameter 98 cm ⇒ r = 49 cm. CSA = 2 × (22/7) × 49 × 90 = 27,720 cm². In 100 revolutions: 2,772,000 cm² = 277.2 m² (divide by 10,000). Cost at ₹10 per m² = ₹ 2,772. The whole question turns on that cm² → m² step.
A conical tent's cloth is its CSA = πrl, so you almost always need l from Pythagoras first. Then, because canvas comes off a roll of fixed width, length of canvas = CSA ÷ width. Two steps, in that order.
The radius of the base of a conical tent is 16 m and its height is 12 m. Find the length of a 10 m wide canvas required for this tent. (Use π = 22/7) (SI 2 Jul 2023)
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r = 16, h = 12, so l = √(256 + 144) = √400 = 20 m. CSA = πrl = (22/7) × 16 × 20 = 7040/7 m². The canvas is 10 m wide, so its length = (7040/7) ÷ 10 = 704/7 m.
A conical tent is required for 8 persons to sit in, and each person needs 10 m² of space on the ground and 200 cubic metres of air to breathe. Find the vertical height of the tent. (SI 20 Jun 2023)
Show answer
Ground space: 8 persons × 10 m² = 80 m² = πr². Air: 8 × 200 = 1600 m³ = ⅓πr²h. Substitute the first into the second: ⅓ × 80 × h = 1600 ⇒ h = 4800/80 = 60 m. You never had to find r — πr² was all you needed, and it was handed to you.
Pattern 6 — Capacity, displacement and digging
Litres, water levels and earthwork.
1 m³ = 1,000 litres. 1,000 cm³ = 1 litre. 1 hectare = 10,000 m². A question that asks for litres but gives you metres is testing the first one and nothing else.
Rohan wants to store 120000 litres of oil. So he bought a conical oil tank of a diameter of 7 m and a height of 9 m to store oil. Find how many litres of oil will be left after filling the whole tank? [Use π = 22/7] (SI 16 Oct 2022)
Show answer
Tank volume = ⅓ × (22/7) × 3.5² × 9 = 115.5 m³ = 115,500 litres. Rohan has 120,000 litres, so 4,500 litres are left over. Diameter 7 ⇒ radius 3.5 — the one step the question is built on.
A spherical piece of ice of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere dissolves completely in the water, by how many centimetres will the water level in the vessel rise? (SI 14 Jul 2023)
Show answer
The ice sphere has diameter 6, so r = 3 and its volume is ⁴⁄₃π × 27 = 36π. That water spreads over the vessel's base, radius 6: π × 36 × rise = 36π ⇒ rise = 1 cm. Displacement questions are always “volume added = base area × rise”.
A well of radius 8 m and depth 21 m is dug, and the excavated earth is spread evenly to form a platform measuring 35 m × 15 m. Find the height of the platform. (SI 2 Jul 2023)
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Earth dug out of the well = π × 8² × 21 = (22/7) × 64 × 21 = 4224 m³. Spread over a 35 × 15 = 525 m² platform: height = 4224 ÷ 525 ≈ 8.05 m. Cylinder in, cuboid out — the volume is the bridge.
Roy makes a 14m deep hole with a circumference of 22m. He then digs the soil out from the hole and makes a 1m thick wall of length 14m. Find the width of the wall. [Use π = 22/7] (SI 18 Aug 2021)
Show answer
Circumference 22 m ⇒ 2 × (22/7) × r = 22 ⇒ r = 3.5 m. Soil dug out = πr²h = (22/7) × 12.25 × 14 = 539 m³. That becomes a wall 14 m long and 1 m thick: 14 × 1 × width = 539 ⇒ width = 38.5 m.
Cube and cuboid: 8 questions
Short, and almost always about the volume.
| Cube (side a) | Cuboid (l × b × h) | |
|---|---|---|
| Volume | a³ | l × b × h |
| Total surface area | 6a² | 2(lb + bh + hl) |
| Lateral surface area | 4a² (the four walls) | 2h(l + b) |
| Longest diagonal | a√3 | √(l² + b² + h²) |
A cuboid of size 44 cm × 32 cm × 36 cm contains some curable chemicals and salts. These are to be melted and reshaped in the form of spherical tablets of radius 12 cm, to be given to patients. How many such tablets can be formed? (SI 2 Jul 2023)
Show answer
Cuboid volume = 44 × 32 × 36 = 50,688 cm³. One tablet = ⁴⁄₃ × (22/7) × 12³ = 7241.14 cm³. 50,688 ÷ 7241.14 = exactly 7 tablets — the paper chose its numbers so the count comes out whole. When it does not, always round down: you cannot cast most of a tablet.
Exam craft
The numbers, habits and traps that turn knowledge into marks.
- π = 22/7 when r is a multiple of 7; 3.14 otherwise; leave it as π if the options do
- Cubes: 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 12³=1728
- Triples: 3–4–5, 5–12–13, 8–15–17, 7–24–25
- √2 ≈ 1.414, √3 ≈ 1.732, √5 ≈ 2.236
- 1 m² = 10,000 cm² · 1 m³ = 1,000,000 cm³ · 1 m³ = 1,000 litres
- (22/7) × 7 = 22, × 14 = 44, × 21 = 66, × 49 = 154, × 98 = 308
- Volume of a cone
- ⅓ πr²h — one third of its cylinder
- CSA of a cone
- πrl (needs the slant height, not the height)
- TSA of a cone
- πr(r + l)
- Volume of a sphere
- ⁴⁄₃ πr³
- Surface area of a sphere
- 4πr² — curved and total are the same
- Volume of a hemisphere
- ⅔ πr³
- TSA of a hemisphere
- 3πr² (curved 2πr² + flat circle πr²)
- TSA of a cylinder
- 2πr(r + h)
- Slant height of a cone
- l = √(r² + h²)
- Longest diagonal of a cube
- a√3
- Area of an equilateral triangle
- (√3/4) a²
- Area of a rhombus
- ½ × d₁ × d₂
- Diameter given, radius needed. The most expensive habit in the chapter.
- Height versus slant height. πrl uses l; ⅓πr²h uses h. They are never the same number.
- cm² answers compared with m² options. Divide by 10,000, not 100.
- Rounding a count upward. Seven tablets, not eight — you cannot form a partial solid.
- Reading the ratio backwards. “Second to the first” means you flip your answer.
- Substituting 22/7 when every option has π in it. Pure lost time, and it invites arithmetic slips.
The rare ones — asked once eachTap to open
These have appeared a single time in eight years of SI papers. Read them once so the shape is familiar; do not spend a study session on them.
The radii of the two ends of the frustum of a right circular cone are 7 m and 3 m respectively. If the height of the frustum is 6 m, what will its volume be? (SI 29 Jun 2023)
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Frustum — a cone with its top sliced off. V = ⅓πh(R² + r² + Rr). With R = 7, r = 3, h = 6: ⅓ × π × 6 × (49 + 9 + 21) = 2π × 79 = 158π m³.
A right angled sector of radius 4 cm is rolled up into a cone in such a way that the two binding radii are joined together. The curved surface area of the cone is : (SI 29 Jul 2023)
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A quarter-circle rolled into a cone: the sector's radius becomes the cone's slant height (l = 4) and the sector's arc becomes the base circumference. A right-angled sector is a quarter of the circle, so the arc = ¼ × 2π × 4 = 2π = 2πr ⇒ r = 1. CSA = πrl = π × 1 × 4 = 4π cm².
A plane is drawn parallel to the base of a right circular cone dividing the altitude in the ratio of 2 : 1 to chop off the conical part. The chopped off conical part is taller than the remaining portion. The ratio of the volume of the chopped off part and the remaining part is : (SI 29 Jul 2023)
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A plane parallel to the base makes a similar small cone. The altitude splits 2 : 1 with the chopped-off top taller, so the small cone has ⅔ of the height and therefore ⅔ of every length. Its volume is (⅔)³ = 8/27 of the whole. The remaining frustum is 19/27. Ratio = 8 : 19.
A solid cylinder having a radius of its base as 14 cm and length as 26 cm is bisected from its height to get two identical cylinders. What is the percentage increase in the total surface area? (SI 31 Jul 2023)
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Cutting a cylinder across its height into two makes two new circular faces. Original TSA = 2πr(r + h) = 2π × 14 × 40 = 1120π. The two halves add 2πr² = 392π. Increase = 392/1120 = 35%. Cutting never changes volume, but it always adds surface.
The volume of a sphere is to be reduced to 92.6859% of the initial volume. How much should the radius be reduced to do so? (SI 14 Jul 2023)
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Volume falls to 92.6859% ⇒ the radius falls to ∛0.926859 = 0.975, a drop of 2.5%.
If the numeric value of the volume of a sphere in cm³ is the same as the numeric value of the surface area of another sphere in cm², and the spheres have the same radii, find the common radius. (SI 29 Jun 2023)
Show answer
⁴⁄₃πr³ = 4πr² ⇒ ⅓r = 1 ⇒ r = 3 cm. The numbers match only at r = 3 — and the same trick for a cylinder gives r = 2 when V and TSA agree.
One-page revision
The night before the paper, read only this.
| Shape | Volume | Surface |
|---|---|---|
| Cube a | a³ | TSA 6a² · LSA 4a² · diagonal a√3 |
| Cuboid l,b,h | lbh | 2(lb+bh+hl) · diagonal √(l²+b²+h²) |
| Cylinder | πr²h | CSA 2πrh · TSA 2πr(r+h) |
| Cone | ⅓πr²h | CSA πrl · TSA πr(r+l) · l=√(r²+h²) |
| Sphere | ⁴⁄₃πr³ | 4πr² |
| Hemisphere | ⅔πr³ | CSA 2πr² · TSA 3πr² |
| Frustum | ⅓πh(R²+r²+Rr) | CSA πl(R+r) |
| Circle | — | Area πr² · Circumference 2πr |
| Equilateral △ a | — | (√3/4)a² · height (√3/2)a |
| Rhombus | — | ½ d₁d₂ |
| Regular hexagon a | — | (3√3/2)a² |
Scaling, in one line: length × k → area × k² → volume × k³. Melting, in one line: volume before = volume after. Those two sentences cover 15 of the 53 solid questions on their own.
Every remaining mensuration question
Thirty-five more, grouped by shape.
Plane figures: triangles, rectangles and circlesTap to open
The area of a triangle is 42 cm². Its base is 14 cm. Find its altitude. (SI 2 Jul 2023)
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Area = ½ × base × height, so 42 = ½ × 14 × h and h = 6 cm.
The base of a right-angled triangle is 8 cm and its perpendicular height is 10 cm. Find its area. (SI 29 Jun 2023)
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For a right-angled triangle the two perpendicular sides are the base and the height, so the area is ½ × 8 × 10 = 40 cm². Option A (80) is the product without the half.
The area of a right-angled triangle is 150 m². If its base is 25 m, then find its perpendicular. (SI 29 Jul 2023)
Show answer
150 = ½ × 25 × p, so p = 300/25 = 12 m.
The base and the corresponding height of a triangle are 27 cm and 18 cm respectively. What will be the base of another triangle whose area is twice that of this triangle and whose height is equal? (SI 20 Jun 2023)
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With the same height, the area is proportional to the base. Doubling the area therefore doubles the base: 27 × 2 = 54 cm. The 18 cm height never enters the arithmetic.
Find the area of an equilateral triangle whose side is 12 cm. (SI 20 Jun 2023)
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The area of an equilateral triangle of side a is (√3/4)a². With a = 12 that is (√3/4) × 144 = 36√3 cm².
A triangle whose two sides and two angles are equal, and altitude drawn on non-equal side bisect, is called a/an : (SI 9 Jul 2023)
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Two equal sides and two equal angles, with the altitude to the unequal side bisecting it — that is the definition of an isosceles triangle. An equilateral triangle has three equal sides, so it is a special case rather than the answer the question wants.
The area of an isosceles right triangle whose perimeter is p, is : (SI 2 Jul 2023)
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Let each equal leg be a. The hypotenuse is a√2, so the perimeter is p = a(2 + √2) and a = p/(2 + √2). The area is ½a² = p²/(2(2 + √2)²). Expanding, (2 + √2)² = 6 + 4√2, so the area is p²/(12 + 8√2).
The area of a square field is 625 m². What is the cost of fencing it at ₹ 25 per meter ? (SI 2016)
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Area 625 m² means the side is √625 = 25 m, so the perimeter is 4 × 25 = 100 m. Fencing costs 100 × 25 = ₹2,500.
What will be the area of a rectangle (in m²) if the sum of its length and breadth is 24 m and the ratio of its length and breadth is 5 : 7? (SI 20 Jun 2023)
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The ratio 5 : 7 has twelve parts and the sum is 24 m, so one part is 2 m. The sides are 10 m and 14 m. Area = 10 × 14 = 140 m².
The length of a rectangle is 50cm and its width is three-fifths of its length. What is the area of the square having a perimeter that is equal to the perimeter of the rectangle described here? (SI 29 Jun 2023)
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The width is 3/5 of 50 = 30 cm, so the rectangle's perimeter is 2(50 + 30) = 160 cm. A square with that perimeter has side 40 cm, so its area is 40² = 1600 cm². Option A (2500) is the square of the rectangle's length — the perimeter step skipped.
Alice draws a square. Its perimeter is double the perimeter of a rectangle of length 10 cm and breadth 4 cm. If the diameter of a semicircle is equal to the length of the side of the square, what will its perimeter be (in cm)? [Use π = 22/7] (SI 24 Aug 2021)
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The rectangle's perimeter is 2(10 + 4) = 28 cm, so the square's is 56 cm and its side is 14 cm. The semicircle has diameter 14, so radius 7. Its perimeter is the arc plus the diameter: πr + 2r = (22/7)(7) + 14 = 22 + 14 = 36 cm. Forgetting the diameter is the standard semicircle mistake — the flat edge is part of the boundary.
Cubes and cuboidsTap to open
Find the volume of a cuboid 16 m long, 14 m broad and 7 m high. (SI 2016)
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Volume = l × b × h = 16 × 14 × 7 = 1568 m³.
The length of the sides of a cube is 1.4 cm. What will be its volume? (SI 15 Jul 2023)
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Volume of a cube = a³ = 1.4³ = 2.744 cm³. Option C (2.8) is 2 × 1.4 and option B (1.6) is a rounding of 1.4² — both are there for a rushed cube.
The total of the lengths of the edges of a cube equals 1.5 times the perimeter of a rectangle whose length and breadth are in the ratio 7 : 5. If the numerical value of the volume of the cube is twice the numerical value of the area of the rectangle, then find the length of the rectangle. (SI 24 Aug 2021)
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Let the cube's edge be a and the rectangle's sides be 7k and 5k. A cube has 12 edges, so 12a = 1.5 × 2(7k + 5k) = 36k, giving a = 3k. The volume condition says a³ = 2 × (7k)(5k), so 27k³ = 70k² and k = 70/27. The length is 7k = 490/27 units.
CylindersTap to open
Find the sum of the curved surface area and the total surface area of a cylinder whose diameter of the base is 14 cm and height is 20 cm. (SI 14 Jul 2023)
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r = 7 and h = 20. CSA = 2πrh = 2 × (22/7) × 7 × 20 = 880 cm². TSA = CSA + 2πr² = 880 + 2 × (22/7) × 49 = 880 + 308 = 1,188 cm². Their sum is 880 + 1,188 = 2068 cm². Options C and D are the two parts offered separately.
A solid cylinder has total surface area of 1848 cm². The curved surface area is one-third of its total surface area. What is the radius of the base of this cylinder? [Use π = 22/7] (SI 18 Aug 2021)
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CSA is one third of TSA, so CSA = 616 and the two circular ends account for 1,848 − 616 = 1,232 cm². Those ends are 2πr², so 2 × (22/7) × r² = 1,232, giving r² = 196 and r = 14 cm. Splitting the total surface into curved and flat parts is what makes this a one-line question.
The sum of the radius of the base and the height of a cylinder is 42 m. If the total surface area of that cylinder is 2838 m², find the curved surface area of the cylinder. (SI 20 Jun 2023)
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TSA = 2πr(r + h) = 2,838, and r + h = 42 is given, so 2 × (22/7) × r × 42 = 2,838. That gives 264r = 2,838 and r = 10.75 m, so h = 42 − 10.75 = 31.25 m. CSA = 2πrh = 2 × (22/7) × 10.75 × 31.25 = 2111.61 m². Notice that TSA factorises as 2πr(r + h) — that is what lets the given sum do all the work.
The curved surface area and the volume of a cylindrical pillar are 184 m² and 488 m³ respectively. Find the height of the pillar. (SI 20 Jun 2023)
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Divide the volume by the curved surface area: V/CSA = (πr²h)/(2πrh) = r/2. So r = 2 × 488/184 = 5.304 m. Then h = CSA/(2πr) = 184 ÷ (2 × (22/7) × 5.304) = 5.52 m. Dividing the two given quantities removes h in one step — far quicker than solving simultaneously.
A cylindrical can is designed to have volume 2750/7 cubic units. The bottom costs double the top which in turn costs double the sides. If the height of the can is 4 times the diameter and the cost of the side is ₹14 square units, what is the cost of the can? (SI 29 Jul 2023)
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Let the radius be r. The height is 4 diameters = 8r, so the volume is πr²(8r) = (22/7)(8r³) = 2750/7, giving r³ = 15.625 and r = 2.5. Then h = 20. The areas are: side 2πrh = 2 × (22/7) × 2.5 × 20 = 2200/7, and each circular face πr² = 275/14. Costs: side ₹14 per unit, top ₹28, bottom ₹56. Total = (2200/7)(14) + (275/14)(28) + (275/14)(56) = 4,400 + 550 + 1,100 = ₹6,050.
ConesTap to open
The height of a right circular cone is 63 cm and its base radius is 4 cm. Find its volume. (SI 2 Jul 2023)
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V = (1/3)πr²h = (1/3) × (22/7) × 16 × 63 = 1056 cm³. 63 is a multiple of 7, so the π cancels cleanly — the examiner chose it for that reason.
The volume of a cone having radius 4 cm and slant height 5 cm is : (SI 14 Jul 2023)
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With r = 4 and slant l = 5, the height comes from the right triangle: h = √(25 − 16) = 3. So V = (1/3)π(16)(3) = 16π cm³. The 3-4-5 triangle is the examiner's favourite cone.
The volume of a cone whose base diameter is 12 cm is 84π cm³. Find its height. (SI 29 Jun 2023)
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The base diameter is 12, so r = 6 and 84π = (1/3)π(36)h = 12h, giving h = 7 cm. Halving the diameter before squaring is the step most often skipped.
Find the radius of the largest right circular cone that can be cut out of a cube of side 7.2 cm. (SI 2 Jul 2023)
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The largest cone that fits inside a cube has its base inscribed in one face, so its diameter equals the cube's edge. The radius is therefore 7.2 ÷ 2 = 3.6 cm.
The volume of a conical vessel is 12936 cm³ and its height is 28 cm. What is its curved surface area (in cm²)? (Take π = 22/7) (SI 29 Jun 2023)
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From the volume, 12,936 = (1/3)(22/7)r²(28), so r² = 441 and r = 21 cm. The slant height is √(21² + 28²) = √1225 = 35 — another 3-4-5 triangle scaled by 7. CSA = πrl = (22/7)(21)(35) = 2310 cm².
The volume of a right-circular cone is 314 2/7 cm³. If the ratio of the base radius to the height is 5 : 12, then what is the slant height? (SI 29 Jul 2023)
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Take r = 5k and h = 12k. Then V = (1/3)(22/7)(25k²)(12k) = (2200/7)k³. The given volume is 314 2/7 = 2200/7, so k³ = 1 and k = 1. That makes r = 5 and h = 12, and the slant height √(25 + 144) = 13 cm. 5-12-13 is the other Pythagorean triple the SI paper leans on.
Find what length of canvas 5/6 m wide is required to make a conical tent 48 m in diameter and 7 m high. [Use π = 3.14] (SI 29 Jul 2023)
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The tent has r = 24 and h = 7, so the slant height is √(576 + 49) = √625 = 25 m. The canvas needed is the curved surface area: πrl = 3.14 × 24 × 25 = 1,884 m². At 5/6 m wide, the length required is 1,884 ÷ (5/6) = 2260.8 m.
A solid right circular conical empty tank has radius r feet and height h feet. The tank is filled with water at the rate of W cubic feet per second. In terms of r, h and w, which of the following expressions represents the number of minutes taken for the tank to be completely filled? (SI 14 Jul 2023)
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The cone holds πr²h/3 cubic feet, and water arrives at w cubic feet per second, so filling takes πr²h/(3w) seconds. Dividing by 60 converts to minutes: πr²h/(180w). The whole question is the unit change; option C is the answer in seconds.
If the slant length of a cone is 3.5 times the diameter, then the ratio of the value of the volume of the cone and slant height is ________ if the diameter is 2 (SI 31 Jul 2023)
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Diameter 2 gives r = 1, and the slant length is 3.5 × 2 = 7. The height is √(49 − 1) = √48 = 4√3, so the volume is (1/3)π(1)(4√3) = 4√3π/3. The ratio volume : slant height is (4√3π/3) : 7 = 4√3π : 21. Dividing both sides by √3 gives 4π : 7√3.
Spheres and hemispheresTap to open
If the diameter of a football is 12 cm, then its volume is (in cu. cm) : (SI 29 Jun 2023)
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Diameter 12 gives r = 6, so V = (4/3)π(216) = 288π cu. cm. Option C (1152π) is what you get by using the diameter instead of the radius.
Find the volume of a sphere of radius 7 cm. (SI 29 Jun 2023)
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V = (4/3)πr³ = (4/3)(22/7)(343) = 1437.3 cm³. r = 7 cancels the 7 in 22/7, which is why the paper keeps choosing it.
Find the volume of a sphere whose diameter is 84cm. (SI 31 Jul 2023)
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Diameter 84 gives r = 42, so V = (4/3)(22/7)(42³) = (4/3)(22/7)(74,088) = 310464 cm³.
Find the volume of a hemisphere of radius 2 cm. (SI 14 Jul 2023)
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A hemisphere is half a sphere, so V = (2/3)πr³ = (2/3)π(8) = 16π/3 cm³.
A solid sphere of diameter 42 cm is cut into two equal parts. Find the surface area of the hemisphere so obtained. (Use π = 22/7) (SI 19 Aug 2021)
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Diameter 42 gives r = 21. A hemisphere's total surface area is the curved half plus the flat circle: 2πr² + πr² = 3πr². So the area is 3 × (22/7) × 441 = 4158 cm². Cutting a sphere in two creates a new flat face — leaving it out gives 2772, which is the classic error here.
Three spheres of radii 15 m, 9 m and 12 m are melted to form a new sphere. Find the radius of the new sphere. (SI 31 Jul 2023)
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Melting conserves volume, and volume goes with r³, so the new radius satisfies R³ = 15³ + 9³ + 12³ = 3,375 + 729 + 1,728 = 5,832. Since 18³ = 5,832, the new radius is 18 m. The (4/3)π cancels from both sides — melting questions are always a sum of cubes.
A hollow sphere whose outer and inner diameters are 12 cm and 6 cm respectively is melted down to make a cone with a base diameter of 12 cm. Find the height of the cone. (SI 29 Jun 2023)
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The hollow sphere's outer radius is 6 and inner radius 3, so its volume of metal is (4/3)π(6³ − 3³) = (4/3)π(189) = 252π. The cone has base diameter 12, so r = 6 and its volume is (1/3)π(36)h = 12πh. Setting 12πh = 252π gives h = 21 cm.
Before you write anything down: halve the diameter, note the unit, decide π. Three seconds, and most of the traps on this page stop working.