Chapter 12 · Class X Mathematics · NCERT Exercises

Surface Areas and Volumes — Exercises

Cube · Cylinder · Cone · Sphere — All 36 SAV Problems Solved

📂 2 Exercises 📝 17 Questions 🎓 Moderate-High

▶ Start Exercises 📖 Revise Notes

Exercise Index

①

Exercise 12.1

Surface area of combined solids: cone+cylinder, hemisphere+cylinder etc.

9 Q Moderate

②

Exercise 12.2

Volume of combined solids; frustum of cone; conversion word problems

8 Q Moderate-Hard

2 exercise files · 17 total questions

Chapter at a Glance

CBSE BoardsNTSE

12 Concepts

18 Formulas

Moderate-High Difficulty

8–10% Weightage

Before You Begin

Prerequisites

✓Surface area and volume of basic solids (Class IX)
✓Pythagoras for slant height
✓Ratio and proportion

Have Ready

🔧Formulae card for all 6 solids
🔧Calculator
🔧π = 22/7 as directed

Exercise Topic Map

Exercise 12.1 Combined SA = sum of visible curved/flat surfaces; subtract hidden junction circles

Exercise 12.2 Combined Volume = simple sum; Frustum: V = πh/3(R²+r²+Rr); CSA = πl(R+r)

Key Formulae — Recall Before Solving

\(\text{Cylinder: } V=\pi r^2 h,\ CSA=2\pi rh,\ TSA=2\pi r(h+r)\)

\(\text{Cone: } V=\tfrac{1}{3}\pi r^2 h,\ l=\sqrt{r^2+h^2},\ CSA=\pi rl\)

\(\text{Sphere: } V=\tfrac{4}{3}\pi r^3,\ SA=4\pi r^2;\quad \text{Hemisphere: } V=\tfrac{2}{3}\pi r^3\)

\(\text{Frustum: } V=\tfrac{\pi h}{3}(R^2+r^2+Rr),\ CSA=\pi(R+r)l\)

NCERT Solving Method

Step 1 — Identify each component solid; list all required dimensions. Step 2 — SA of combined solid: add all EXPOSED surfaces; subtract area of any circular base that is hidden between two joined solids. Step 3 — Volume of combined solid: just add component volumes (no subtraction needed). Step 4 — Conversion problems: Volume₁ = n × Volume₂; solve for n or unknown dimension. Step 5 — Always compute slant height l = √(r²+h²) for cones before applying any formula.