Mensuration: Area, Volume and Surface Area for CTET Paper II (Class 6–8)

intermediate 18 min read

Concept

Mensuration is the branch of mathematics concerned with measuring geometric figures — how much space they cover (area), how much boundary they have (perimeter), how much space they enclose in three dimensions (volume), and how much outer surface they expose (surface area).

Think of it this way. You are tiling a kitchen floor — you need area. You are fencing a garden — you need perimeter. You are filling a water tank — you need volume. You are painting a box — you need surface area. These four ideas cover practically every mensuration question you will encounter in CTET Paper II.

The subject divides cleanly into two worlds:

2D (flat) shapes — triangles, rectangles, squares, parallelograms, trapeziums, circles, and sectors. Here you work with area and perimeter only.

3D (solid) shapes — cubes, cuboids, cylinders, cones, spheres, and hemispheres. Here you work with volume, total surface area (TSA), and curved/lateral surface area (CSA/LSA).

The conceptual leap that confuses most candidates is this: surface area is still a 2D measurement (it counts square units of outer skin), while volume counts cubic units of interior space. When you paint a box, you care about surface area. When you fill it with water, you care about volume. Keep that distinction sharp and half the confusion dissolves.

A useful analogy for the classroom: imagine wrapping a gift box. The wrapping paper you need corresponds to total surface area. The space inside the box that the gift occupies corresponds to volume. CTET questions on teaching methodology sometimes test whether a teacher can explain this distinction — so knowing it conceptually, not just formula-mechanically, matters.


Deep Dive

2D Shapes: Area and Perimeter Formulas

Master these cold. No derivation time in the exam hall.

| Shape | Area | Perimeter | |---|---|---| | Rectangle | l × b | 2(l + b) | | Square | | 4a | | Triangle | (1/2) × b × h | sum of three sides | | Parallelogram | b × h | 2(a + b) | | Trapezium | (1/2)(a + b) × h | sum of all sides | | Circle | πr² | 2πr | | Sector | (θ/360°) × πr² | 2r + (θ/360°) × 2πr |

For CTET, trapezium appears frequently. The formula (1/2)(sum of parallel sides)(height) is the one you must know forwards and backwards — specifically, you must be comfortable rearranging it to find a missing side when area and height are given.

3D Shapes: Surface Area and Volume

Here is the master reference. Every formula below has appeared directly or indirectly in CTET PYQs.

Cube (side a):

Cuboid (length l, breadth b, height h):

Cylinder (radius r, height h):

Hollow Cylinder (outer radius R, inner radius r, height h):

Cone (radius r, height h, slant height l):

Sphere (radius r):

Hemisphere (radius r):

The Scaling Rule — A Concept CTET Loves

When you scale all linear dimensions by a factor k:

Look — if every dimension of a cuboid is doubled (k = 2), volume becomes 2³ = 8 times the original. This is a recurring CTET question type. Don't compute both volumes; just apply the cube of the scale factor.

Working Backwards from Formulas

Many CTET questions give you surface area or volume and ask for a missing dimension. The approach is always the same: write the formula, substitute known values, isolate the unknown, solve. The algebraic manipulation is usually linear (first-degree), not quadratic. If you find yourself solving a quadratic, re-read the question — you have probably misidentified the unknown.

For cuboid TSA problems (a CTET favourite), the key rearrangement is:

2(lb + bh + lh) = TSA

Substitute l and b, combine the h terms, solve. This is a one-step linear equation once you expand. Do not panic — it always simplifies cleanly in CTET problems because the numbers are chosen to give integer answers.

π = 22/7 vs π = 3.14

CTET questions always specify which value of π to use. When radius or diameter is a multiple of 7, use 22/7. When it is not, use 3.14. If the question says nothing, default to 22/7 because the numbers in CTET problems are constructed to cancel cleanly with 22/7. Using 3.14 when 22/7 is intended gives a messy decimal that does not match any option.


Memory Tricks & Shortcuts

patternTSA-to-Volume Cuboid Chain

When a cuboid TSA question gives you two dimensions and asks for volume, use the chain: expand TSA formula → collect h terms → solve for h → multiply lbh.

Example: TSA = 194, l = 8, b = 6. Step 1: 2(48 + 6h + 8h) = 194 → 96 + 28h = 194 → h = 3.5 Step 2: V = 8 × 6 × 3.5 = 168

This two-step chain takes about 25 seconds once practised. The standard approach of trial-and-error with options takes 60–90 seconds. Save 40+ seconds per such question.

substitutionTrapezium Reverse: Assume Variables from the Difference

When a trapezium problem says "one parallel side is longer by k cm," immediately set the shorter side as a and longer as a + k. Plug into (1/2)(2a + k)(h) = Area, solve for a, then report a + k.

Example: Area = 105, h = 7, difference = 6. (1/2)(2a + 6)(7) = 105 → (7/2)(2a + 6) = 105 → 2a + 6 = 30 → a = 12, longer = 18.

Standard guess-and-check from options: 4 substitutions, ~60 seconds. This algebraic path: ~20 seconds.

patternScaling Exponent Rule for Dimension Changes

Dimensions scaled by factor k → Area scales by k², Volume by k³. No calculation needed.

If all sides double: Volume = 2³ = 8× original. If all sides triple: Volume = 3³ = 27× original. If all sides halve: Volume = (1/2)³ = 1/8 original.

Standard method (compute both volumes and divide): ~45 seconds for a cuboid. This rule: 5 seconds — read the scale factor, cube it, done.

eliminationCone CSA Shortcut: Cancel π Early

For CSA of cone = πrl, when π = 22/7, cancel it before multiplying.

Example: CSA = 550, l = 25. Find r. 550 = (22/7) × r × 25 → 550 = (550r)/7 → r = 7.

Notice: 22 × 25 = 550, so (22/7 × 25) = 550/7. The 550 cancels immediately: r = 550 × 7/550 = 7.

Standard method (multiply 22/7 × 25 first, then divide): 3 steps, ~30 seconds. Cancellation method: 1 step, ~8 seconds.

patternHeight-from-Volume: The Area-Divisor

For any prism or cylinder: Volume = Base Area × Height. So Height = Volume ÷ Base Area. This works for rectangular tanks, cylindrical tanks, any uniform cross-section solid.

Example: Volume = 140 m³, Base Area = 700 m². Height = 140/700 = 0.2 m = 20 cm.

Don't reach for the full formula. Just remember: V = A × h → h = V/A. Two-second setup, five-second division.


Fast-Solving Framework

When a mensuration question appears, run this decision tree in your head before writing anything:

Step 1 — Identify the shape. 2D or 3D? Name it exactly (trapezium, cuboid, hollow cylinder, etc.).

Step 2 — Identify what is asked. Area / Perimeter / CSA / TSA / Volume? Pick exactly one formula.

Step 3 — Check what is given vs. what is needed. Are all variables in the formula provided, or do you need to find a missing one first? (E.g., in cuboid TSA → volume, you must find h first.)

Step 4 — Handle unit conversion before substituting. If area is in m² and volume in m³, the height comes out in metres — convert to cm only at the end if the answer choices are in cm.

Step 5 — Cancel before you multiply. With π = 22/7, look for 7s in the radius or area to cancel. With scaling questions, apply the or rule directly — never compute both values.

If two answer options are close in value (say, 168 vs 126), do not guess — these are designed to trap arithmetic errors. Recheck your algebra for sign or coefficient mistakes in the h-term.


Solved PYQs

Why this question: The simplest application of V = A × h in reverse. Every CTET batch has at least one "find the height of water" question.

Previous Year Questionपिछले वर्ष का प्रश्न2018
To fill a rectangular tank of area 700 m², 140 m³ of water is required. What will be the height of the water level in the tank?
  1. 10 cm
  2. 20 cm
  3. 30 cm
  4. 40 cm
Solutionसमाधान
Volume of water = Area of base × Height. So Height = Volume/Area = 140 m³/700 m² = 0.2 m = 20 cm. The height of water level in the tank is 20 cm.

Solving path: Write V = Base Area × h. Rearrange: h = V/A = 140/700 = 0.2 m. Convert: 0.2 m = 20 cm. Match option B. Total time: under 15 seconds.


Why this question: Classic two-step cuboid: TSA → find missing dimension → compute volume. If you do not recognise the chain, this question eats 3 minutes. If you do, it takes 30 seconds.

Previous Year Questionपिछले वर्ष का प्रश्न2019
The total surface area of a cuboid is 194 m². If its length is 8 m and breadth is 6 m, then what is its volume (in m³)?
  1. 112
  2. 126
  3. 168
  4. 224
Solutionसमाधान
TSA of cuboid = 2(lb + bh + lh) = 194. With l=8, b=6: 2(48 + 6h + 8h) = 194 → 96 + 28h = 194 → 28h = 98 → h = 3.5 m. Volume = lbh = 8 × 6 × 3.5 = 168 m³.

Solving path: Expand TSA: 2(8×6 + 6h + 8h) = 194 → 2(48 + 14h) = 194 → 96 + 28h = 194 → 28h = 98 → h = 3.5. Volume = 8 × 6 × 3.5 = 168 m³. Option C.


Why this question: Trapezium reverse — given area and height, find a side. The "one side longer by k" phrasing is a CTET signature.

Previous Year Questionपिछले वर्ष का प्रश्न2019
The area of a trapezium is 105 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 6 cm, then the length of the longer side, in cm, is
  1. 18
  2. 16
  3. 15
  4. 12
Solutionसमाधान
Area of trapezium = (1/2)(sum of parallel sides)(height). Let shorter side = a, longer = a+6. So 105 = (1/2)(a + a+6)(7) → 105 = (7/2)(2a+6) → 30 = 2a+6 → 2a = 24 → a = 12. Longer side = 12+6 = 18 cm.

Solving path: Let shorter side = a, longer = a + 6. Area formula: 105 = (1/2)(a + a + 6)(7) = (7/2)(2a + 6). Multiply both sides by 2/7: 30 = 2a + 6 → a = 12. Longer side = 18 cm. Option A.


Why this question: Hollow cylinder tests whether you know V = π(R² − r²)h. Many candidates incorrectly subtract volumes as π(R−r)²h.

Previous Year Questionपिछले वर्ष का प्रश्न
A hollow cylinder has outer radius 7 cm, inner radius 5 cm, and height 10 cm. What is the volume of the material used to make the cylinder? (Use π = 22/7)
एक खोखले बेलन की बाहरी त्रिज्या 7 सेमी, आंतरिक त्रिज्या 5 सेमी और ऊँचाई 10 सेमी है। बेलन बनाने में प्रयुक्त सामग्री का आयतन क्या है? (π = 22/7 लें)
  1. 754 cm³
  2. 680 cm³
  3. 440 cm³
  4. 1540 cm³
  1. 754 सेमी³
  2. 680 सेमी³
  3. 440 सेमी³
  4. 1540 सेमी³
Solutionसमाधान
Volume of material = π(R² − r²) × h = (22/7) × (49 − 25) × 10 = (22/7) × 24 × 10 = (22 × 240)/7 = 5280/7 ≈ 754.28 ≈ 754 cm³.
सामग्री का आयतन = π(R² − r²) × h = (22/7) × (49 − 25) × 10 = (22/7) × 24 × 10 = 5280/7 ≈ 754 सेमी³।

Solving path: V = (22/7)(7² − 5²)(10) = (22/7)(49 − 25)(10) = (22/7)(24)(10) = (22 × 240)/7 = 5280/7 ≈ 754 cm³. Option A. Note: subtract the squares of the radii, not the square of their difference.


Why this question: Tests scaling intuition. The answer is k³ = 2³ = 8 and requires zero computation if you know the rule.

Previous Year Questionपिछले वर्ष का प्रश्न
A cuboid has dimensions 15 cm × 10 cm × 8 cm. If each dimension is doubled, by what factor does the volume increase?
एक घनाभ की विमाएँ 15 सेमी × 10 सेमी × 8 सेमी हैं। यदि प्रत्येक विमा को दोगुना कर दिया जाए, तो आयतन कितने गुना बढ़ जाएगा?
  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times
  1. 2 गुना
  2. 4 गुना
  3. 6 गुना
  4. 8 गुना
Solutionसमाधान
Original volume = 15 × 10 × 8 = 1200 cm³. New volume = 30 × 20 × 16 = 9600 cm³. Factor = 9600/1200 = 8. When all three dimensions are doubled, volume increases by 2³ = 8 times.
मूल आयतन = 15 × 10 × 8 = 1200 सेमी³। नया आयतन = 30 × 20 × 16 = 9600 सेमी³। अनुपात = 9600/1200 = 8। जब तीनों विमाएँ दोगुनी होती हैं तो आयतन 2³ = 8 गुना बढ़ता है।

Solving path: All three dimensions double → k = 2. Volume scales by k³ = 8. Confirm: original = 1200 cm³, new = 9600 cm³, ratio = 8. Option D. Using the rule takes 5 seconds; computing both volumes takes 30–40 seconds.


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