Areas Related to Circles | Exercise 11.1

Question 4

A chord of a circle of radius 10 cm10\text{ cm} subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use π=3.14\pi = 3.14)

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Solution

We will find the area of the minor segment and major sector using given radius and angle.

Step 1 — Calculate minor segment area

Let's find the area of the minor sector first. The angle at the centre is 90\mathbf{90^\circ}. The radius of the circle is 10 cm\mathbf{10 \text{ cm}}.

Area of minor sector=θ360×πr2\text{Area of minor sector} = \frac{\theta}{360^\circ} \times \pi r^2

=90360×3.14×(10 cm)2= \frac{90^\circ}{360^\circ} \times 3.14 \times (10 \text{ cm})^2

=14×3.14×100 cm2= \frac{1}{4} \times 3.14 \times 100 \text{ cm}^2

=0.25×314 cm2= 0.25 \times 314 \text{ cm}^2

=78.5 cm2= 78.5 \text{ cm}^2

Next, we find the area of the triangle formed by the chord. The triangle OAB is a right-angled triangle. Its base and height are both equal to the radius.

Area of OAB=12×base×height\text{Area of } \triangle \text{OAB} = \frac{1}{2} \times \text{base} \times \text{height}

=12×10 cm×10 cm= \frac{1}{2} \times 10 \text{ cm} \times 10 \text{ cm}

=12×100 cm2= \frac{1}{2} \times 100 \text{ cm}^2

=50 cm2= 50 \text{ cm}^2

Now, we can find the area of the minor segment. It is the minor sector area minus the triangle area.

Area of minor segment=Area of minor sectorArea of OAB\text{Area of minor segment} = \text{Area of minor sector} - \text{Area of } \triangle \text{OAB}

=78.5 cm250 cm2= 78.5 \text{ cm}^2 - 50 \text{ cm}^2

28.5 cm2\boxed{28.5 \text{ cm}^2}

Diagram 1

Step 2 — Calculate major sector area

Let's find the area of the major sector. We can subtract the minor sector area from the total circle area. The area of the circle is πr2\pi r^2.

Area of circle=πr2\text{Area of circle} = \pi r^2

=3.14×(10 cm)2= 3.14 \times (10 \text{ cm})^2

=3.14×100 cm2= 3.14 \times 100 \text{ cm}^2

=314 cm2= 314 \text{ cm}^2

Now, we subtract the minor sector area.

Area of major sector=Area of circleArea of minor sector\text{Area of major sector} = \text{Area of circle} - \text{Area of minor sector}

=314 cm278.5 cm2= 314 \text{ cm}^2 - 78.5 \text{ cm}^2

235.5 cm2\boxed{235.5 \text{ cm}^2}

Answer

(i) The area of the minor segment is 28.5 cm2\mathbf{28.5 \text{ cm}^2}. (ii) The area of the major sector is 235.5 cm2\mathbf{235.5 \text{ cm}^2}.

More questions in Exercise 11.1

Q1

Unless stated otherwise, use π=227\pi = \frac{22}{7}.

Find the area of a sector of a circle with radius 6 cm6\text{ cm} if angle of the sector is 6060^\circ.

Q2

Find the area of a quadrant of a circle whose circumference is 22 cm22\text{ cm}.

Q3

The length of the minute hand of a clock is 14 cm14\text{ cm}. Find the area swept by the minute hand in 55 minutes.

Q4

A chord of a circle of radius 10 cm10\text{ cm} subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use π=3.14\pi = 3.14)

Q5

In a circle of radius 21 cm21\text{ cm}, an arc subtends an angle of 6060^\circ at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Q6

A chord of a circle of radius 15 cm15\text{ cm} subtends an angle of 6060^\circ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73)

Q7

A chord of a circle of radius 12 cm12\text{ cm} subtends an angle of 120120^\circ at the centre. Find the area of the corresponding segment of the circle.

(Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73)

Q8

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m15\text{ m} by means of a 5 m5\text{ m} long rope (see Fig. 11.8). Find

(i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m10\text{ m} long instead of 5 m5\text{ m}. (Use π=3.14\pi = 3.14)

Q9
  1. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Q10
  1. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Q11
  1. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115115^\circ. Find the total area cleaned at each sweep of the blades.
Q12
  1. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 8080^\circ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π=3.14\pi = 3.14)
Q13
  1. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2\text{cm}^2. (Use 3=1.7\sqrt{3} = 1.7)
Q14
  1. Tick the correct answer in the following :

Area of a sector of angle pp (in degrees) of a circle with radius R is

(A) p180×2πR\frac{p}{180} \times 2\pi\text{R} (B) p180×πR2\frac{p}{180} \times \pi\text{R}^2 (C) p360×2πR\frac{p}{360} \times 2\pi\text{R} (D) p720×2πR2\frac{p}{720} \times 2\pi\text{R}^2

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