Areas Related to Circles | Exercise 11.1

Question 9

  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.

Question diagram 1
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Solution

We need to find the total wire length and the area of one sector.

Step 1 — Find the total wire length

Let's first note the given diameter. The diameter of the circle is 35 mm. So, the radius is half of the diameter. Radius r=35/2r = 35/2 mm.

The total wire includes the circle's circumference. It also includes the length of the 5 diameters.

Let's calculate the circumference. Circumference C=π×diameterC = \pi \times \text{diameter}

C=227×35C = \frac{22}{7} \times 35

C=22×5C = 22 \times 5

C=110 mmC = 110 \text{ mm}

Next, we find the length of the 5 diameters. Length of 5 diameters =5×diameter= 5 \times \text{diameter}

=5×35= 5 \times 35

=175 mm= 175 \text{ mm}

Now, we add these lengths together. Total length of wire = Circumference + Length of 5 diameters

=110+175= 110 + 175

285 mm\boxed{285 \text{ mm}}

Diagram 1

Step 2 — Find the area of each sector

The circle is divided into 10 equal sectors. We need to find the angle of each sector. Total angle in a circle is 360∘360^\circ.

Angle of each sector θ=360∘10\theta = \frac{360^\circ}{10}

θ=36∘\theta = 36^\circ

Now, we use the formula for the area of a sector. Area of a sector =θ360∘×πr2= \frac{\theta}{360^\circ} \times \pi r^2

=36360×227×(352)2= \frac{36}{360} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2

=110×227×352×352= \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}

=110×22×52×352= \frac{1}{10} \times 22 \times \frac{5}{2} \times \frac{35}{2}

=110×11×5×352= \frac{1}{10} \times 11 \times 5 \times \frac{35}{2}

=110×55×352= \frac{1}{10} \times 55 \times \frac{35}{2}

=12×111×352= \frac{1}{2} \times \frac{11}{1} \times \frac{35}{2}

=3854 mm2= \frac{385}{4} \text{ mm}^2

96.25 mm2\boxed{96.25 \text{ mm}^2}

Answer

(i) The total length of the silver wire required is 285 mm. (ii) The area of each sector of the brooch is 385/4 mm² (or 96.25 mm²).

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 60∘60^\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 60∘60^\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 60∘60^\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 120∘120^\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 115∘115^\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 80∘80^\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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