Exercise Set 6.3 Maths Class 9 Ganita Manjari Chapter 6

 

Exercise Set 6.3 - Step-by-Step Solutions

Unless stated otherwise, use the approximation \(\pi = \frac{22}{7}\).

Q1. Find the area of a sector of a circle with radius \(7\text{ cm}\) if the angle of the sector is \(60^\circ\).

Solution:

Given: Radius \(r = 7\text{ cm}\), Angle \(\theta = 60^\circ\)

Formula: \(\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2\)

\(\text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2\)

\(\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 49 = \frac{1}{6} \times 22 \times 7 = \frac{154}{6} = \frac{77}{3}\text{ cm}^2\)

Answer: \(\frac{77}{3}\text{ cm}^2\) (or \(25.67\text{ cm}^2\))

Q2. Find the area of a quadrant of a circle whose circumference is \(44\text{ cm}\).

Solution:

Let the radius of the circle be \(r\).

Circumference \(= 2\pi r = 44\text{ cm}\)

\(2 \times \frac{22}{7} \times r = 44 \implies \frac{44}{7} \times r = 44 \implies r = 7\text{ cm}\)

A quadrant is a sector with angle \(\theta = 90^\circ\).

\(\text{Area of quadrant} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 7^2 = \frac{1}{4} \times 154 = \frac{77}{2}\text{ cm}^2\)

Answer: \(\frac{77}{2}\text{ cm}^2\) (or \(38.5\text{ cm}^2\))

Q3. The length of the minute hand of a clock is \(7\text{ cm}\). Find the area swept by the minute hand in 10 minutes.

Solution:

Length of minute hand \(r = 7\text{ cm}\)

Angle swept in 60 minutes \(= 360^\circ\)

Angle swept in 1 minute \(= \frac{360^\circ}{60} = 6^\circ\)

Angle swept in 10 minutes \(\theta = 10 \times 6^\circ = 60^\circ\)

\(\text{Area swept} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2\)

\(\text{Area} = \frac{1}{6} \times 154 = \frac{77}{3}\text{ cm}^2\)

Answer: \(\frac{77}{3}\text{ cm}^2\) (or \(25.67\text{ cm}^2\))

Q4. A chord of a circle of radius \(10\text{ cm}\) subtends \(90^\circ\) at the centre. Find the area of the corresponding:
(i) minor sector (that subtends \(90^\circ\) at the centre), and
(ii) major sector (that subtends \(270^\circ\) at the centre).
(Use \(\pi \approx 3.14\))

Solution:

(i) Minor sector (\(\theta = 90^\circ\)):

\(\text{Area} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 10^2 = \frac{1}{4} \times 314 = 78.5\text{ cm}^2\)

(ii) Major sector (\(\theta = 270^\circ\)):

\(\text{Area} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times 3.14 \times 10^2 = \frac{3}{4} \times 314 = 235.5\text{ cm}^2\)

Answer: (i) \(78.5\text{ cm}^2\), (ii) \(235.5\text{ cm}^2\)

Q5. A chord of a circle of radius \(15\text{ cm}\) subtends an angle of \(60^\circ\) at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle.
(Use \(\pi \approx 3.14\) and \(\sqrt{3} \approx 1.73\))

Solution:

Given: \(r = 15\text{ cm}\), \(\theta = 60^\circ\)

\(\text{Area of sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = 117.75\text{ cm}^2\)

Since angle is \(60^\circ\), the triangle formed by radius and chord is equilateral with side \(r=15\text{ cm}\).

\(\text{Area of triangle} = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125\text{ cm}^2\)

Minor segment area \(= \text{Area of sector} - \text{Area of triangle}\)

\(= 117.75 - 97.3125 = 20.4375\text{ cm}^2\)

Major segment area \(= \text{Area of circle} - \text{Area of minor segment}\)

\(\text{Area of circle} = \pi r^2 = 3.14 \times 225 = 706.5\text{ cm}^2\)

\(\text{Major segment area} = 706.5 - 20.4375 = 686.0625\text{ cm}^2\)

Answer: Minor segment: \(20.4375\text{ cm}^2\), Major segment: \(686.0625\text{ cm}^2\)

Q6. A car has two wipers which do not overlap. Each wiper has a blade of length \(28\text{ cm}\) and sweeps through an angle of \(120^\circ\). Find the total area cleaned at each sweep of the blades.

Solution:

Given: \(r = 28\text{ cm}\), \(\theta = 120^\circ\), Number of wipers \(= 2\)

\(\text{Area cleaned by 1 wiper} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 28^2\)

\(= \frac{1}{3} \times \frac{22}{7} \times 784 = \frac{1}{3} \times 22 \times 112 = \frac{2464}{3}\text{ cm}^2\)

\(\text{Total area cleaned by 2 wipers} = 2 \times \frac{2464}{3} = \frac{4928}{3}\text{ cm}^2\)

Answer: \(\frac{4928}{3}\text{ cm}^2\) (or \(1642.67\text{ cm}^2\))

*7. A chord of a circle of radius \(r\) subtends an angle of \(60^\circ\) at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to \(\pi r^2 \left( \frac{1}{6} - \frac{\sqrt{3}}{4\pi} \right)\) or \(r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Proof:

1. \(\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \pi r^2 = \frac{1}{6} \pi r^2\)

2. The triangle formed at the center has sides \(r, r\) and angle \(60^\circ\), which makes it an equilateral triangle.

\(\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} r^2\)

3. \(\text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle}\)

\(= \frac{1}{6} \pi r^2 - \frac{\sqrt{3}}{4} r^2\)

\(= \pi r^2 \left( \frac{1}{6} - \frac{\sqrt{3}}{4\pi} \right)\)

Hence Proved.

*8. An equilateral triangle is inscribed in a circle of radius \(r\). Show that the ratio of the area of the triangle to the area of the circle is equal to \(\frac{3\sqrt{3}}{4\pi} \approx 0.413\).

Proof:

1. For an equilateral triangle inscribed in a circle of radius \(r\), the circumradius formula is:

\(r = \frac{a}{\sqrt{3}} \implies a = r\sqrt{3}\) (where \(a\) is the side length of the triangle).

2. \(\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (r\sqrt{3})^2 = \frac{\sqrt{3}}{4} (3r^2) = \frac{3\sqrt{3}}{4} r^2\)

3. \(\text{Area of circle} = \pi r^2\)

4. \(\text{Ratio} = \frac{\text{Area of triangle}}{\text{Area of circle}} = \frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi}\)

Using \(\pi \approx 3.1416\) and \(\sqrt{3} \approx 1.732\):

\(\text{Ratio} \approx \frac{3 \times 1.732}{4 \times 3.1416} = \frac{5.196}{12.566} \approx 0.413\)

Hence Proved.

*9. A square is inscribed in a circle of radius \(r\). Show that the ratio of the area of the square to the area of the circle is equal to \(\frac{2}{\pi} \approx 0.637\).

Proof:

1. The diagonal of the inscribed square equals the diameter of the circle \(= 2r\).

2. If \(x\) is the side of the square, then diagonal \(= x\sqrt{2} = 2r \implies x = \frac{2r}{\sqrt{2}} = r\sqrt{2}\).

3. \(\text{Area of square} = x^2 = (r\sqrt{2})^2 = 2r^2\)

4. \(\text{Area of circle} = \pi r^2\)

5. \(\text{Ratio} = \frac{\text{Area of square}}{\text{Area of circle}} = \frac{2r^2}{\pi r^2} = \frac{2}{\pi}\)

Numerical value: \(\frac{2}{3.14159} \approx 0.637\)

Hence Proved.

*10. A hexagon is inscribed in a circle of radius \(r\). Show that the ratio of the area of the hexagon to the area of the circle is equal to \(\frac{3\sqrt{3}}{2\pi} \approx 0.827\). Can you see why the answer is exactly twice the answer to Question 8?

Proof & Explanation:

1. A regular hexagon inscribed in a circle consists of 6 congruent equilateral triangles with side length equal to radius \(r\).

2. \(\text{Area of 1 equilateral triangle} = \frac{\sqrt{3}}{4} r^2\)

3. \(\text{Area of hexagon} = 6 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{2} r^2\)

4. \(\text{Area of circle} = \pi r^2\)

5. \(\text{Ratio} = \frac{\text{Area of hexagon}}{\text{Area of circle}} = \frac{\frac{3\sqrt{3}}{2} r^2}{\pi r^2} = \frac{3\sqrt{3}}{2\pi}\)

Numerical value: \(\frac{3 \times 1.732}{2 \times 3.1416} \approx 0.827\)

Why is it twice the answer to Question 8?

An inscribed equilateral triangle can be split into 3 congruent isosceles triangles from the center, whereas an inscribed regular hexagon is formed by 6 identical equilateral triangles from the center. Its total area is precisely twice the area of the inscribed triangle, making its ratio to the circle's area exactly double:

\(2 \times \left(\frac{3\sqrt{3}}{4\pi}\right) = \frac{3\sqrt{3}}{2\pi}\).

Hence Proved.