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Q13:
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 cm$^2$. (Use $\sqrt{3} = 1.7$)
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 cm$^2$. (Use $\sqrt{3} = 1.7$)
Solution :
Given:
1. A circular table cover with radius $r = 28$ cm.
2. There are six equal designs on the table cover, which are segments of the circle.
3. The rate of making the designs is ₹ $0.35$ per cm$^2$.
4. The value of $\sqrt{3} = 1.7$.
To Find:
The total cost of making the six designs.
Step 1: Determine the central angle of each sector.
Since there are 6 equal designs, the circle is divided into 6 equal sectors. The central angle $\theta$ for each sector is:
$\theta = \frac{360^\circ}{6} = 60^\circ$
Step 2: Calculate the area of one design (segment).
The area of a segment is given by the formula: Area of sector - Area of the triangle formed by the two radii and the chord.
Area of one sector = $\frac{\theta}{360^\circ} \times \pi r^2$
Using $\pi = \frac{22}{7}$ and $r = 28$:
Area of one sector = $\frac{60}{360} \times \frac{22}{7} \times 28 \times 28$
Area of one sector = $\frac{1}{6} \times 22 \times 4 \times 28 = \frac{2464}{6} = \frac{1232}{3} \text{ cm}^2$
Since the central angle is $60^\circ$ and the two sides are equal (radii), the triangle is equilateral. The area of an equilateral triangle is $\frac{\sqrt{3}}{4} \times \text{side}^2$.
Area of triangle = $\frac{\sqrt{3}}{4} \times 28 \times 28 = \sqrt{3} \times 7 \times 28 = 196\sqrt{3}$
Given $\sqrt{3} = 1.7$, Area of triangle = $196 \times 1.7 = 333.2 \text{ cm}^2$
Area of one design = $\frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47 \text{ cm}^2$
Step 3: Calculate the total area of the six designs.
Total Area = $6 \times (\text{Area of one design})$
Total Area = $6 \times (\frac{1232}{3} - 333.2) = 2 \times 1232 - 6 \times 333.2$
Total Area = $2464 - 1999.2 = 464.8 \text{ cm}^2$
Step 4: Calculate the total cost.
Cost = Total Area $\times$ Rate
Cost = $464.8 \times 0.35$
Cost = $162.68$
Final Answer: The total cost of making the designs is ₹ 162.68.
More Questions from Class 10 Mathematics Areas Related to Circles EXERCISE 11.1
- Q1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q2: Find the area of a quadrant of a circle whose circumference is 22 cm. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q4(i): A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (Use $\pi = 3.14$)
- Q4(ii): A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector. (Use $\pi = 3.14$)
- Q5(i): In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q5(ii): In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (ii) area of the sector formed by the arc (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q5(iii): In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (iii) area of the segment formed by the corresponding chord (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q6: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
- Q7: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
- Q8(i): A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find (i) the area of that part of the field in which the horse can graze. (Use $\pi = 3.14$)
- Q8(ii): A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\pi = 3.14$)
- Q9(i): 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. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q9(ii): 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 : (ii) the area of each sector of the brooch. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q10: 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. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q11: A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (Unless stated otherwise, use $\pi = \frac{22}{7}$)
- Q12: To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)
- Q14: Tick the correct answer in the following : Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is
CBSE Solutions for Class 10 Mathematics Areas Related to Circles
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