CUET PG /Quantitative Aptitude

Mensuration

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CUET PG – Quantitative Aptitude

Mensuration – Study Notes

Mensuration is the branch of mathematics that deals with the measurement of geometric figures such as lengths, areas, and volumes.
In CUET‑PG, you’ll encounter a mix of basic formulas and application questions that require quick substitutions and sometimes multiple steps.


1. Simple Explanation

  • Perimeter / Circumference – Total distance around a figure.

    • Perimeter of a polygon = sum of all side lengths.
    • Circumference of a circle = (2\pi r) (or (\pi d)).
  • Area – Measure of the surface covered by a figure.

    • 2‑D shapes: use the “base × height” or “radius × radius” methods.
    • 3‑D shapes: surface area = sum of areas of all faces.
  • Volume – Measure of space occupied by a solid.

    • 3‑D shapes: usually “area of base × height” or “(\frac{1}{3}) × base area × height” for pyramids/cones.

2. Key Points & Formulas

Shape Parameters Area Perimeter / Circumference Volume Surface Area
Square side = (s) (s^2) (4s) – –
Rectangle length (l), breadth (b) (lb) (2(l+b)) – –
Triangle base (b), height (h) (\frac{1}{2}bh) – –
Circle radius (r) (\pi r^2) (2\pi r) – –
Parallelogram base (b), height (h) (bh) – –
Trapezium bases (a, b), height (h) (\frac{1}{2}(a+b)h) – –
Cube side (a) – – (a^3) (6a^2)
Cuboid length (l), breadth (b), height (h) – – (lbh) (2(lb+bh+hl))
Cylinder radius (r), height (h) – – (\pi r^2 h) (2\pi r(r+h))
Cone radius (r), height (h), slant height (l) – – (\frac{1}{3}\pi r^2 h) (\pi r(r+l))
Sphere radius (r) – – (\frac{4}{3}\pi r^3) (4\pi r^2)
Right Pyramid base area (A_b), height (h) – – (\frac{1}{3}A_b h) –
Right Ellipsoid semi‑axes (a,b,c) – – (\frac{4}{3}\pi abc) – (no simple closed form)

Remember:
Use ( \pi = 3.14 ) or ( \pi = \frac{22}{7} ) as required by the question.
Always check the units (cm, m, etc.) – they usually cancel out in ratios.


3. Quick Revision Checklist

Category Formula Key Notes
Perimeter (P = 2(l+b)) (rect.) Add all sides.
Circumference (C = 2\pi r) (r = \frac{d}{2})
Area (2‑D) (A = \frac{1}{2}bh) (tri.) Use base × height.
Area (circle) (A = \pi r^2) (r = \frac{d}{2})
Volume (cylinder) (V = \pi r^2 h) Base area × height.
Volume (cone) (V = \frac{1}{3}\pi r^2 h) 1/3 of cylinder.
Volume (sphere) (V = \frac{4}{3}\pi r^3) 4/3 of cylinder with same radius and height.
Surface Area (cylinder) (S = 2\pi r(r+h)) Lateral + top & bottom.
Surface Area (cone) (S = \pi r(r+l)) Lateral + base.
Surface Area (sphere) (S = 4\pi r^2) No lateral part.

4. Worked Example

Problem (CUET‑PG style):
A right circular cone has a height of 12 cm and a slant height of 15 cm. Find

  1. the radius of the base
  2. the volume of the cone
  3. the total surface area of the cone.

Solution

Step 1 – Find base radius (r)
In a right cone, the slant height (l), radius (r), and height (h) satisfy the Pythagorean relation:

[ l^2 = r

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