ICSE Class 10 Maths Chapter 23 Circumference and Area of a Circle

Unit 5 Mensuration

Chapter 23 Circumference and Area of a Circle

Points To Remember

Some Important Formulae

(i) For a circle of radius = r units, we have

(i) Circumference of the circle = \(2\pi r\) units = \(\pi d\) units, where d is the diameter.

(ii) Area of the circle = \(\pi r^2\) sq. units.

(ii) For a semi-circle of radius = r units, we have

(i) Area of the semi-circle = \(\left(\frac{1}{2}\pi r^2\right)\) sq. units

(ii) Perimeter of the semi-circle = \(\pi r + 2r\) units.

(iii) Area of a Circular Ring:

If R and r be the outer and inner radii of a ring, then Area of the ring = \(\pi(R^2 - r^2)\) sq. units.

Results On Sectors and Segments

Suppose an arc ACB makes an angle \(\theta^{\circ}\) at the centre O of a circle of radius = r units. Then:

(i) Length of arc ACB = \(\left(\frac{2\pi r\theta}{360}\right)\) units

(ii) Area of sector OACBO = \(\left(\frac{\pi r^2\theta}{360}\right)\) sq. units

= \(\frac{1}{2} \times r \times \left(\frac{2\pi r\theta}{360}\right)\) sq. units

= \(\left(\frac{1}{2} \times \text{radius} \times \text{arc length}\right)\) sq. units

(iii) Perimeter of sector OACBO = length of arc ACB + OA + OB

= \(\left(\frac{2\pi r\theta}{360} + 2r\right)\) units

(iv) Area of segment ACBA = (Area of sector OACBO) - (Area of \(\triangle\)OAB)

= \(\left(\frac{\pi r^2\theta}{360} - \frac{1}{2}r^2 \sin\theta\right)\) sq. units.

(v) Perimeter of segment ACBA = (arc ACB + chord AB) units.

(vi) Area of Major segment BDAB = (Area of circle) - (Area of minor segment ACBA).

Rotations Made By a Wheel

(i) Distance moved by a wheel in 1 revolution = Circumference of the wheel.

(ii) Number of rotations made by a wheel in unit time = \(\frac{\text{Distance moved by it in unit time}}{\text{Circumference of the wheel}}\)

Angles Described By Clock Hands

(i) Angle described by minute hand in 60 minutes = \(360^{\circ}\).

(ii) Angle described by minute hand in 5 minutes = \(\left(\frac{360}{60} \times 5\right)^{\circ} = 30^{\circ}\).

(iii) Angle described by hour hand in 12 hours = \(360^{\circ}\).

(iv) Angle described by hour hand in 1 hour = \(30^{\circ}\).

In An Equilateral Triangle Of Side a Units

(i) Height of the triangle, \(h = \frac{\sqrt{3}}{2}a\) units.

(ii) Area of the triangle = \(\left(\frac{\sqrt{3}}{4}a^2\right)\) sq. units.

(iii) Radius of incircle, \(r = \frac{1}{3}h = \left(\frac{1}{3} \cdot \frac{\sqrt{3}}{2}a\right) = \left(\frac{a}{2\sqrt{3}}\right)\) units.

(iv) Radius of circumcircle, \(R = \frac{2}{3}h = \left(\frac{2}{3} \cdot \frac{\sqrt{3}}{2}a\right) = \left(\frac{a}{\sqrt{3}}\right)\) units.

Thus, \(r = \frac{a}{2\sqrt{3}}\) and \(R = \frac{a}{\sqrt{3}}\).

Note: Until and unless stated otherwise take \(\pi = \frac{22}{7}\)

Teacher's Note

Understanding circles and their measurements is fundamental to real-world applications like designing wheels, constructing circular pools, or calculating the area of round tables in a restaurant.

Exercise 23

Note: Take \(\pi = \frac{22}{7}\), unless mentioned otherwise.

Q. 1. A sheet is 11 cm long and 2 cm wide. Circular pieces 0-5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared. (2004)

Sol. Length of sheet = 11 cm

Width of sheet = 2 cm

First of all, we have to cut the sheet in squares of side 0-5 cm.

No. of squares = \(\frac{11}{0.5} \times \frac{2}{0.5}\)

= \(\frac{11 \times 10}{5} \times \frac{2 \times 10}{5}\)

= \(22 \times 4 = 88\)

No. of discs will be equal to number of squares cut out = 88 Ans.

Q. 2. Find the circumference and area of a circle of radius 17-5 cm.

Sol. Radius (r) of the circle = 17.5 cm

Circumference (C) = \(2\pi r\)

= \(2 \times \frac{22}{7} \times 17.5 = 110\) cm

And area (A) = \(\pi r^2 = \frac{22}{7} \times (17.5)^2\) cm\(^2\)

= \(\frac{22}{7} \times \frac{175}{10} \times \frac{175}{10} = 962.5\) cm\(^2\) Ans.

Q. 3. Find the circumference and area of a circle of diameter 91 cm.

Sol. Diameter of a circle = 91 cm.

Radius (r) = \(\frac{91}{2}\) cm.

Circumference (C) = \(2\pi r\)

= \(2 \times \frac{22}{7} \times \frac{91}{2}\) cm \(= 286\) cm

And Area (A) = \(\pi r^2 = \frac{22}{7} \times \frac{91}{2} \times \frac{91}{2}\) cm\(^2\)

= \(\frac{26026}{4}\) cm\(^2\) = \(6506.5\) cm\(^2\) Ans.

Q. 4. Find the circumference and area of a circle of radius 15 cm. (Take \(\pi = 3.14\))

Sol. Radius of a circle (r) = 15 cm

Circumference (C) = \(2\pi r\)

= \(2 \times 3.14 \times 15 = 94.2\) cm

And Area (A) = \(\pi r^2 = 3.14 \times (15)^2\) cm\(^2\)

= \(3.14 \times 15 \times 15 = 706.5\) cm\(^2\) Ans.

Q. 5. The circumference of a circle is 123-2 cm. Taking \(\pi = \frac{22}{7}\), calculate:

(i) The radius of the circle in cm;

(ii) The area of the circle in cm\(^2\), correct to the nearest cm\(^2\);

(iii) The effect on the area of the circle if the radius is doubled.

Sol. The circumference of a circle = 123.2 cm

(i) Let, radius of the circle be r, then

\(2\pi r = 123.2\)

\(\Rightarrow 2 \times \frac{22}{7} r = 123.2\)

\(r = \frac{123.2 \times 7}{2 \times 22} = 19.6\) cm

(ii) Circumference = \(2\pi r\)

= \(2 \times \frac{22}{7} \times 11.2\) cm

= \(70.4\) cm Ans.

(iii) If the radius is doubled,

Then effect of the area of the circle

= \(\frac{\pi r^2}{\pi(2r)^2} = \frac{\pi r^2}{4\pi r^2} = \frac{1}{4}\)

Area of the resulting circle is four times the area of the original circle.

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