Karnataka TET – Quantitative Aptitude
Topic: Mensuration
Mensuration deals with finding the area, perimeter, volume, and surface area of various 2‑ and 3‑dimensional shapes.
1. Simple Explanation
| 2‑D Figures | 3‑D Figures |
|---|---|
| Area – the amount of space inside a shape. | Volume – the amount of space a solid occupies. |
| Perimeter (or Circumference) – the total length of the boundary. | Surface Area – total area of all the external faces. |
Why it matters in TET:
Questions often mix different shape properties (e.g., find the perimeter of a rectangle if its area is known). Knowing formulas and being able to rearrange them is key.
2. Key Points & Formulas
| Figure | Symbol | Area | Perimeter | Volume | Surface Area |
|---|---|---|---|---|---|
| Rectangle | – | (A = l \times w) | (P = 2(l + w)) | – | – |
| Square | – | (A = s^2) | (P = 4s) | – | – |
| Triangle | – | (\displaystyle A = \frac{1}{2}\times \text{base}\times\text{height}) | (P = a+b+c) | – | – |
| Circle | – | (A = \pi r^2) | (C = 2\pi r) | – | – |
| Parallelogram | – | (A = \text{base}\times\text{height}) | (P = 2(\text{base}+\text{side})) | – | – |
| Trapezium | – | (\displaystyle A = \frac{1}{2}(a+b)\times h) | (P = a+b+c+d) | – | – |
| Cylinder | – | – | – | (V = \pi r^2 h) | (S = 2\pi r(h+r)) |
| Sphere | – | – | – | (V = \frac{4}{3}\pi r^3) | (S = 4\pi r^2) |
| Cube | – | – | – | (V = s^3) | (S = 6s^2) |
| Cuboid | – | – | – | (V = lwh) | (S = 2(lw+wh+lh)) |
| Cone | – | – | – | (V = \frac{1}{3}\pi r^2 h) | (S = \pi r(r+s)) where (s=\sqrt{r^2+h^2}) |
Commonly used constants
- (\pi \approx 3.14) or (22/7) (use the one given, if any).
- 1 square unit = 1 unit², 1 cubic unit = 1 unit³.
Quick Tips
- Perimeter of rectangle = 2 × sum of length & breadth.
- Area of trapezium = average of parallel sides × height.
- Volume of cone = 1/3 × base area × height.
- Remember the factor 1/3 for cones and pyramids, 4/3 for spheres.
3. Worked Example (Typical TET Question)
Question
A right circular cylinder has a height of 12 cm and a volume of 452 π cm³.
Find the radius of the cylinder.
Solution
-
Write the volume formula for a cylinder
[ V = \pi r^2 h ] -
Substitute known values
[ 452\pi = \pi r^2 \times 12 ] -
Cancel (\pi) from both sides
[ 452 = 12,r^2 ] -
Solve for (r^2)
[ r^2 = \frac{452}{12} = \frac{113}{3} ] -
Take square root
[ r = \sqrt{\frac{113}{3}} \approx \sqrt{37.67} \approx 6.14 \text{ cm} ]
Answer: ( r \approx 6.14 \text{ cm} )
4. Quick Revision Summary
| Concept | Key Formula | Quick Check |
|---|---|---|
| Rectangle Area | (A = l \times w) | Remember “length × breadth”. |
| Rectangle Perimeter | (P = 2(l + w)) | “Twice the sum of sides”. |
| Triangle Area | (A = \frac{1}{2} \times \text{base} \times \text{height}) | “Half the base × height”. |
| Circle Area | (A = \pi r^2) | “Pi times radius squared”. |
| Circle Circumference | (C = 2\pi r) | “Twice pi × radius”. |
| Cylinder Volume | (V = \pi r^2 h) | “Base area × height”. |
| Sphere Volume | (V = \frac{4}{3}\pi r^3) | “Four‑thirds pi × radius cubed”. |
| Cone Volume | (V = \frac{1}{3}\pi r^2 h) | “One‑third base area × height”. |
| Cube Surface Area | (S = 6s^2) | “Six faces, each side²”. |
| Cuboid Surface Area | (S = 2(lw+wh+lh)) | “Twice sum of three face areas”. |
5. Quick Practice Problem
A right circular cone has a slant height of 13 cm and a base radius of 5 cm.
Find the total surface area of the cone.
Try solving before looking at the solution.
Solution Outline
- Find the height using Pythagoras: (h = \sqrt{s^2 - r^2}).
- Compute lateral area (= \pi r s).
- Compute base area (= \pi r^2).
- Add them: (S