ICSE Class 10 Maths Chapter 24 Cone and Sphere

Chapter 24

Cone and Sphere

Points To Remember

1. Right Circular Cone

If a right-angled triangle is revolved about one of the sides containing the right angle, then the solid thus generated, is called a right circular cone.

Thus, when a right-angled triangular lamina AOB is revolved about OA, it generates a cone.

The point A is called the vertex of the cone.

The length OA = h, is called the height of the cone.

The base of a cone is a circle with O as centre and OB as radius.

The length OB is called the radius of its base.

The length AB is called the slant-height of the cone.

In right-angled \(\triangle AOB\), we have \(l^2 = r^2 + h^2\) or \(l = \sqrt{r^2 + h^2}\)

2. Formulae

For a right circular cone of Radius = r, Height = h & Slant Height = l, we have:

(i) Volume of the cone = \(\left(\frac{1}{3}\pi r^2 h\right)\) cubic units

(ii) Area of the curved surface or Lateral surface of the cone = \((\pi rl)\) sq. units, where \(l = \sqrt{r^2 + h^2}\)

(iii) Total surface area of the cone = (Curved Surface Area + Area of Base) = \((\pi rl + \pi r^2)\) sq. units = \(\pi r(l + r)\) sq. units.

3. Hollow Right Circular Cone

Suppose a sector of a circle is folded to make the radii coincide, then we get a hollow right circular cone. In such a cone:

(i) Centre of the circle is vertex of the cone.

(ii) Radius of the circle is slant height of the cone.

(iii) Length of arc AB is the circumference of the base of the cone.

(iv) Area of the sector is the curved surface area of the cone.

Teacher's Note

A cone shape appears in everyday life - from ice cream cones to traffic cones. Understanding its volume and surface area helps us calculate how much material or content it can hold.

4. Sphere

Objects like football, volleyball, hockey ball etc. are said to have the shape of a sphere.

When a circular lamina is revolved about any of its diameters, then the solid generated is called a sphere.

The centre and radius of this circle are respectively the centre and radius of the sphere.

5. Formulae

(a) For a solid sphere of radius = r:

(i) Volume of the sphere = \(\left(\frac{4}{3}\pi r^3\right)\) cubic units.

(ii) Surface area of the sphere = \((4\pi r^2)\) sq. units.

6. Spherical Shell

The solid enclosed between two concentric spheres is called a spherical shell.

(b) For a spherical shell with External Radius = R & Internal Radius = r:

(i) Thickness of shell = (R - r) units

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

7. Hemisphere

When a plane through the centre of a sphere cuts it into two equal parts, then each part is called a hemisphere.

(c) For a Hemisphere of Radius r:

(i) Volume = \(\frac{2}{3}\pi r^3\) cubic units

(ii) Curved surface area = \(2\pi r^2\) sq. units

(iii) Total surface area = \((2\pi r^2 + \pi r^2) = 3\pi r^2\) sq. units.

8. Hemispherical Shell

The solid enclosed between two concentric hemispheres is called a Hemispherical Shell.

(d) For a Hemispherical Shell of External Radius = R and Internal Radius = r:

(i) Thickness of the shell = (R - r) units

(ii) External curved surface area = \((2\pi R^2)\) sq. units

(iii) Internal curved surface area = \((2\pi r^2)\) sq. units

(iv) Total surface area = \(2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) = \pi(3R^2 + r^2)\) sq. units

(v) Volume of the material = \(\frac{2}{3}\pi(R^3 - r^3)\) cubic units.

Teacher's Note

Spheres are found everywhere in nature - from planets to water droplets. Learning about spherical shells helps us understand hollow objects like rubber balls and metal spheres used in engineering.

Exercise 24 (A)

Note: Until and unless mentioned, take \(\pi = \frac{22}{7}\).

Q.1. The height of a right-circular cone is 24 cm and the radius of its base is 7 cm. Calculate:

(i) the slant height of the cone

(ii) the lateral surface area of the cone

(iii) the total surface area of the cone

(iv) the volume of the cone.

Sol. Radius of the base of the cone (r) = 7cm and height (h) = 24 cm.

(i) Slant height (l) = \(\sqrt{r^2 + h^2}\) = \(\sqrt{(7)^2 + (24)^2}\) = \(\sqrt{49 + 576}\) = \(\sqrt{625}\) = 25 cm

(ii) Lateral surface area = \(\pi rl\) = \(\frac{22}{7} \times 7 \times 25\) = 550 cm²

(iii) Total surface area = \(\pi rl + \pi r^2\) = \(\pi r(l + r)\) = \(\frac{22}{7} \times 7(25 + 7)\) cm² = \(22 \times 32\) = 704 cm²

(iv) Volume = \(\frac{1}{3}\pi^2h\) = \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24\) = 1232 cm³ Ans.

Q.2. The height of a right circular cone is 8cm and the diameter of its base is 12 cm. Calculate:

(i) the slant height of the cone

(ii) the total surface area of the cone

(iii) the volume of the cone. (Take π = 3·14)

Sol. Diameter of the base of the cone = 12 cm

Radius (r) = \(\frac{12}{2}\) = 6 cm

Height (h) = 8cm

(i) Slant height (l) = \(\sqrt{r^2 + h^2}\) = \(\sqrt{(6)^2 + (8)^2}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10 cm

(ii) Total surface area = \(\pi rl + \pi r^2\) = \(\pi r(l + r)\) = \(3.14 \times 6(10 + 6)\) cm² = \(3.14 \times 6 \times 16\) = 301.44 cm²

(iii) Volume = \(\frac{1}{3}\pi r^2 h\) = \(\frac{1}{3}(3.14) \times 6 \times 6 \times 8\) cm³ = \(3.14 \times 96\) = 301.44 cm³ Ans.

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