CBSE Class 10 Maths — Surface Areas and Volumes

Chapter Overview & Weightage

Surface Areas and Volumes is a high-scoring chapter in CBSE Class 10 Maths. It’s almost entirely numerical — if you know the formulas and practice enough varieties, you can score full marks.

Weightage: This chapter typically carries 6-8 marks in the CBSE board exam. You can expect 1 long-answer question (4 marks) involving a combination of solids, and possibly 1 short question. The frustum of a cone is the most frequently tested concept in the 4-mark section.

Year	Marks	Topics
2024	7	Frustum, Combination of solids
2023	6	Sphere + cone combination, Surface area
2022	7	Frustum, Cylinder + cone
2019	8	Combination, Conversion

Key Concepts You Must Know

1. Surface area and volume of basic solids — cube, cuboid, cylinder, cone, sphere, hemisphere. These are prerequisites from Class 9.

2. Surface area and volume of combined solids — when one shape is attached to or cut from another. Real-world objects are rarely single shapes.

3. Frustum of a cone — the piece left when a smaller cone is cut from the top of a larger cone. New in Class 10, high weightage.

4. Conversion of solids — melting a solid and recasting it in another shape (volume stays constant, surface area changes).

5. Right circular cone problems — calculating slant height $l = \sqrt{r^2 + h^2}$ is often a required intermediate step.

Important Formulas

Curved Surface Area (CSA) $= 2\pi rh$

Total Surface Area (TSA) $= 2\pi r(r + h)$

Volume $= \pi r^2 h$

Slant height: $l = \sqrt{r^2 + h^2}$

CSA $= \pi r l$

TSA $= \pi r(r + l)$

Volume $= \frac{1}{3}\pi r^2 h$

Sphere: Surface area $= 4\pi r^2$ , Volume $= \frac{4}{3}\pi r^3$

Hemisphere: CSA $= 2\pi r^2$ , TSA $= 3\pi r^2$ , Volume $= \frac{2}{3}\pi r^3$

$R$ = radius of larger base, $r$ = radius of smaller base, $h$ = height

Slant height: $l = \sqrt{h^2 + (R-r)^2}$

CSA $= \pi(R + r)l$

TSA $= \pi(R + r)l + \pi R^2 + \pi r^2$

Volume $= \frac{\pi h}{3}(R^2 + Rr + r^2)$

The TSA of a frustum includes two circular ends (top and bottom). If the frustum is open at one end (like a bucket), use only one circular end. Read the question carefully.

Solved Previous Year Questions

PYQ 1 — 2024 Style (4 marks)

A bucket is in the shape of a frustum with top radius 30 cm, bottom radius 20 cm, and height 24 cm. Find the volume and cost of painting the outer surface at ₹5 per 100 cm². (Use $\pi = 3.14$ )

Slant height: $l = \sqrt{24^2 + (30-20)^2} = \sqrt{576 + 100} = \sqrt{676} = 26$ cm

Volume $= \frac{\pi h}{3}(R^2 + Rr + r^2) = \frac{3.14 \times 24}{3}(900 + 600 + 400)$

$= 3.14 \times 8 \times 1900 = 3.14 \times 15200 = 47728$ cm³

CSA (outer surface, no top) $= \pi(R + r)l + \pi r^2$ (only bottom)

Wait — for a bucket, the outer surface is the lateral surface plus the bottom:

$= \pi(R+r)l + \pi r^2 = 3.14 \times 50 \times 26 + 3.14 \times 400$

$= 3.14(1300 + 400) = 3.14 \times 1700 = 5338$ cm²

Cost $= \frac{5338}{100} \times 5 = \text{₹}266.90$

PYQ 2 — Conversion Problem (3 marks)

A cone of radius 6 cm and height 8 cm is melted and recast into a sphere. Find the radius of the sphere.

Volume of cone $= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 36 \times 8 = 96\pi$ cm³

Volume of sphere $= \frac{4}{3}\pi R^3 = 96\pi$

$R^3 = 96 \times \frac{3}{4} = 72$

$R = \sqrt[3]{72} = \sqrt[3]{8 \times 9} = 2\sqrt[3]{9} \approx 4.16$ cm

PYQ 3 — Combination (2+2 marks)

A toy is in the form of a cone mounted on a hemisphere of the same radius 3.5 cm. Total height of toy = 15.5 cm. Find the total surface area.

Height of cone $= 15.5 - 3.5 = 12$ cm

Slant height $= \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm

TSA of toy = CSA of cone + CSA of hemisphere (they share a circular base — it’s internal)

$= \pi r l + 2\pi r^2 = \pi \times 3.5 \times 12.5 + 2\pi \times 3.5^2$

$= \pi(43.75 + 24.5) = 68.25\pi = 68.25 \times \frac{22}{7} = 214.5$ cm²

Difficulty Distribution

Level	% of Questions	Topics
Easy	30%	Direct formula application, cylinder, cone
Medium	40%	Combinations of solids, conversion
Hard	30%	Frustum problems, multi-step combinations

Expert Strategy

Step 1: Memorize all formulas perfectly — there’s no shortcut here. Write them out from memory each morning during revision week.

Step 2: For combination problems, always draw a diagram first. Label all dimensions. Identify which surfaces are internal (not to be counted in surface area).

Step 3: Use $\pi = \frac{22}{7}$ when the radius is a multiple of 7. Use $\pi = 3.14$ otherwise. The question usually specifies.

Step 4: For conversion problems, set volumes equal. For surface area problems, identify which faces are exposed.

Step 5: Cross-check your answer by estimating. If a bucket holds about 48 litres, the volume should be roughly 48,000 cm³. If your answer is wildly off, recheck.

The frustum formula for volume is the hardest to remember. Recall it as: “one-third pi times h times (three terms in the bracket).” The bracket has $R^2$ , $Rr$ , and $r^2$ — think of it as a “product expansion pattern.”

Common Traps

Trap 1: When a sphere is placed in a cylinder and the question asks for the CSA of the cylinder not covered by the sphere — this requires finding the surface area carefully. The sphere touches the cylinder in a circle, not over an area.

Trap 2: In combination problems, students add TSA of both shapes. Wrong — the joined face is internal and should not be counted. Subtract both circular bases at the join.

Trap 3: Forgetting to find slant height when asked for CSA of cone or frustum. CSA uses $l$ , not $h$ . Always compute $l = \sqrt{h^2 + r^2}$ or $l = \sqrt{h^2 + (R-r)^2}$ first.

Trap 4: In “how many small balls can be made from a large ball” type questions, the answer is: Volume(large) ÷ Volume(small) — assuming no wastage. If wastage percentage is given, account for it.