Cone and Sphere Class 10 ICSE — A. Das Gupta Chapter 24

Chapter 24, Cone and Sphere, in A. Das Gupta’s ICSE Class 10 Mathematics book teaches mensuration of curved solids. The cone and sphere class 10 ICSE chapter defines each solid, gives its volume and surface area formulas, and applies them to shells and combined shapes. This guide walks through every section in the book’s order, with worked patterns and the chapter PDF for practice.

ICSE Class 10 Mathematics A. Das Gupta Chapter 24 Cone and Sphere PDF

What is inside Chapter 24

24 pages · 7 sections · 55 exercise questions

• EXERCISE 24 (A): 23 questions
• EXERCISE 24 (B): 32 questions

1 Right Circular Cone

A right circular cone is the solid generated when a right-angled triangle is revolved about one of the sides containing the right angle. When a right-angled triangular lamina AOB rotates about OA, it sweeps out the cone. The point A becomes the vertex of the cone.

The book names every part. The length OA = h is the height. The base is a circle with centre O and radius OB = r. The length AB is the slant height, written l. These three lengths form a right-angled triangle AOB, with the slant height as the hypotenuse.

That is why the slant height relation is exactly Pythagoras’ theorem:

l^2 = r^2 + h^2 \quad \text{or} \quad l = \sqrt{r^2 + h^2}

Worked pattern: take a cone with base radius 9 cm and height 12 cm. Then l = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 cm. The slant height is 15 cm. This value feeds into the surface area formulas in the next section.

2 Formulae

For a right circular cone of radius r, height h and slant height l, the book gives three formulas.

• Volume = \frac{1}{3}\pi r^2 h cubic units.
• Curved (lateral) surface area = \pi r l sq units, where l = \sqrt{r^2 + h^2}.
• Total surface area = curved surface + base area = \pi r l + \pi r^2 = \pi r (l + r) sq units.

The hollow right circular cone comes from folding a sector of a circle so the two radii coincide. The centre of the circle becomes the vertex. The radius becomes the slant height. The arc length becomes the circumference of the base. The sector area becomes the curved surface area.

Worked pattern: continue with radius 9 cm and height 12 cm, so slant height is 15 cm. Volume = \frac{1}{3}\pi \times 9^2 \times 12 = 324\pi cm³. Total surface area = \pi \times 9 (15 + 9) = 216\pi cm².

Common mistake: total surface area includes the circular base, but curved surface area does not. If a question asks for the curved surface only, use \pi r l and do not add \pi r^2.

4 Sphere

Objects like a football, volleyball and hockey ball have the shape of a sphere. When a circular lamina is revolved about any of its diameters, the solid generated is called a sphere.

The centre and radius of the revolving circle become the centre and radius of the sphere. Every point of the lamina stays the same distance from the centre as it turns. That is why the surface is round at every point.

A sphere has no flat face, no edge and no vertex. This matters later: its whole surface is described by one formula, unlike a cone or hemisphere which need separate curved and flat terms.

5 Formulae

For a solid sphere of radius r, the book gives two formulas.

• Volume = \frac{4}{3}\pi r^3 cubic units.
• Surface area = 4\pi r^2 sq units.

Why these forms: volume scales with the cube of the radius because a sphere fills three dimensions. Surface area scales with the square of the radius because it is a two-dimensional surface. A sphere has a single surface area value because there is no flat face to add.

Worked pattern: take a sphere of radius 6 cm. Volume = \frac{4}{3}\pi \times 6^3 = 288\pi cm³. Surface area = 4\pi \times 6^2 = 144\pi cm².

Contrast this with the hemisphere in section 7. The sphere uses 4\pi r^2 for its whole surface. A hemisphere uses 3\pi r^2, because it carries only half the curved surface plus one flat circle.

6 Spherical Shell

A spherical shell is the solid enclosed between two concentric spheres. The outer sphere has external radius R and the inner hollow has internal radius r.

The book gives two results.

• Thickness of shell = R - r units.
• Volume of material = \frac{4}{3}\pi (R^3 - r^3) cubic units.

Why: the material is the outer sphere minus the hollow inner sphere. You subtract the two full volumes, which is why each radius is cubed before the subtraction.

Common mistake: (R - r)^3 is not equal to R^3 - r^3. Worked pattern: external radius 5 cm and internal radius 3 cm give material volume \frac{4}{3}\pi (125 - 27) = \frac{4}{3}\pi (98) cubic units. The thickness is 2 cm.

7 Hemisphere

When a plane through the centre of a sphere cuts it into two equal parts, each part is called a hemisphere. For a hemisphere of radius r, the book gives three results.

• Volume = \frac{2}{3}\pi r^3 cubic units.
• Curved surface area = 2\pi r^2 sq units.
• Total surface area = 2\pi r^2 + \pi r^2 = 3\pi r^2 sq units.

Why: the curved part is exactly half of a sphere’s surface, so \frac{1}{2} \times 4\pi r^2 = 2\pi r^2. The total surface area adds the flat circular face of area \pi r^2. The flat face is the part students forget.

Worked pattern: a hemisphere of radius 7 cm with \pi = 22/7 has curved surface 2 \times \frac{22}{7} \times 49 = 308 cm² and total surface 3 \times \frac{22}{7} \times 49 = 462 cm².

Two frequent mistakes: using the diameter instead of the radius, and forgetting the flat circular face. Halve the diameter first. Then add \pi r^2 to the curved surface for the total.

8 Hemispherical Shell

A hemispherical shell is the solid enclosed between two concentric hemispheres. It has external radius R and internal radius r. The book lists five results.

• Thickness = R - r units.
• External curved surface area = 2\pi R^2 sq units.
• Internal curved surface area = 2\pi r^2 sq units.
• Total surface area = 2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) = \pi(3R^2 + r^2) sq units.
• Volume of material = \frac{2}{3}\pi (R^3 - r^3) cubic units.

Why the total surface area has three parts: the outer curved surface 2\pi R^2, the inner curved surface 2\pi r^2, and the flat annular ring between the two circular faces. That ring is the difference of two circle areas, \pi R^2 - \pi r^2 = \pi(R^2 - r^2).

This is the least intuitive formula in the chapter. The flat ring is not a full circle. It is what remains when the inner circle is removed from the outer circle.

Worked pattern: external radius 5 cm and internal radius 3 cm give material volume \frac{2}{3}\pi (125 - 27) = \frac{2}{3}\pi(98) cubic units. Total surface area = \pi(3 \times 25 + 9) = 84\pi sq units.

Here is a summary table of every formula in the chapter, so you can compare the solids side by side.

What to practise from this chapter

The chapter has two exercise blocks, and both carry the standing instruction to take \pi = 22/7 unless a question says otherwise.

• EXERCISE 24 (A): 23 questions on the cone — slant height, curved and total surface area, volume, and combined solids such as tents and pillars.
• EXERCISE 24 (B): 32 questions on the sphere, hemisphere, spherical shell, hemispherical shell and combined solids.

Several questions inside these exercises carry board year marks such as 1993, 1997, 1999, 2003, 2006 and 2009. These problem types have appeared in CISCE papers, so they are worth practising in full. You can verify the current Class 10 syllabus on the official CISCE board website.

For the chapter that introduces the circle formulas these solids build on, see the circumference and area of a circle chapter. The full set of ICSE Class 10 Mathematics chapters is listed on the Class 10 Maths books page.

Frequently asked questions

What is the difference between the curved surface area and the total surface area of a cone?

Curved surface area is just the slanted part, \pi r l. Total surface area adds the circular base, giving \pi r (l + r). Use curved surface for the lateral surface or the canvas of a tent; use total surface when the base is included.

Why is the total surface area of a hemisphere 3πr² and not 2πr²?

The curved part is half a sphere, 2\pi r^2. A hemisphere also has a flat circular face of area \pi r^2. Adding them gives 3\pi r^2. The flat face is the part students forget.

How do you find the volume of material in a spherical shell?

Subtract the inner volume from the outer volume: \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3). Cube each radius first. Never cube the difference (R - r)^3.

When the total surface area of a cone is given in terms of π, how do you find its radius?

Set \pi r (l + r) equal to the given value, divide both sides by \pi, and solve the resulting quadratic for r. Reject the negative root since a radius cannot be negative. Then use l^2 = r^2 + h^2 to find the height.

When should I use π = 22/7 and when should I use π = 3.14?

Use 22/7 unless the question states otherwise. That is the book’s standing instruction for both exercises. If a question explicitly says take \pi = 3.14 or 3.142, use that value instead.

Reference: ICSE Class 10 Mathematics textbook material, chapter 24: Cone and Sphere — CISCE-aligned.

More from this book

• Chapter 1: Compound Interest
• Chapter 8: Ratio
• Chapter 13: Distance and Section Formulae
• Chapter 14: Equation of a Straight Line
• Chapter 16: Similarity As a Size Transformation
• Chapter 17: Similarity of Triangles
• Chapter 20: Angle Properties of a Circle
• Chapter 21: Tangent Properties of Circles
• Chapter 23: Circumference and Area of a Circle
• Chapter 27: Graphical Representation of Statistical Data