Class 10 Mathematics Chapter 11 Areas Related To Circles

Key Concept: Application Problems

d) 439.6 m
[Solution Description] To find the circumference, which represents the length of one lap, we use the formula $C = \pi$, d where d is the diameter. Given d = 140 meters, we have:

$$C = \pi \times 140$$

Approximating $\pi$ as 3.14, we calculate:

$$C = 3.14 \times 140 = 439.6 \text{ meters}$$

Hence, the length of one lap around the track is 439.6 meters.

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Key Concept: Direct Formula Application

c) $6.75\pi cm²$
[Solution Description]
Apply the direct formula to calculate the area of a sector:

$$
\text{Area} = \frac{\theta}{360} \times \pi r^2
$$

Substitute $ \theta = 30^\circ $ and  r = 9 cm:

$$
\text{Area} = \frac{30}{360} \times \pi \times 9^2 = \frac{1}{12} \times 81\pi = 6.75\pi
$$

The area of the sector is $ 6.75\pi  cm²$.

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Key Concept: Arc Length Formula

c) 10.47 cm
[Solution Description]
Using the formula for the arc length:
$$
\text{Arc Length} = \frac{\theta}{360} \times 2\pi r
$$
Substituting values:
$$
\text{Arc Length} = \frac{60}{360} \times 2 \times 3.14 \times 10 = 10.47 \, \text{cm}
$$

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Key Concept: Sector Angle Calculation

d) 180°
[Solution Description]
$$
\text{Area of the sector} = \frac{\theta}{360} \times \pi r^2 = 77
$$

Rearranging gives:
$$
\theta = \frac{77 \times 360}{\pi \times 7^2} = \frac{27720}{49\pi}
$$

Using $\pi \approx 3.14$, we find:
$$
\theta \approx \frac{27720}{153.86} \approx 180°
$$

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Key Concept: Segment Area

b) 28.5 square cm
[Solution Description]
To find the area of the minor segment, we need to calculate both the area of the sector and the area of the triangle formed by the chord.

The area of the sector is given by:
$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \pi \times 10^2 = \frac{1}{4} \times \pi \times 100 = 25\pi \text{ square cm}.$$

The area of the triangle can be calculated using the formula for the area of an isosceles right triangle, which is half the product of its legs:
$$\text{Area of triangle} = \frac{1}{2} \times 10 \times 10 = 50 \text{ square cm}.$$

Therefore, the area of the minor segment is:
$$\text{Area of minor segment} = 25\pi – 50 \text{ square cm}.$$

Approximating $\pi$ as 3.14, we have:
$$25 \times 3.14 – 50 = 78.5 – 50 = 28.5 \text{ square cm}.$$

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Key Concept: Chord Length

b) 14 cm
[Solution Description]
Using the formula for chord length:
$$
\text{Chord length} = 2 \times r \times \sin\left(\frac{\theta}{2}\right)
$$
Here, r = 14 and $\theta = 60^\circ$. Therefore:
$$
\text{Chord length} = 2 \times 14 \times \sin(30^\circ) = 28 \times 0.5 = 14 \, \text{cm}
$$

Key Concept: Unit Conversion

a) $78.5 cm^2$
[Solution Description] The radius is half of the diameter. Thus, if the diameter is 10 cm, the radius $r = \frac{10}{2} = 5$ cm. Using the formula $A = \pi r^2$:

$$
A = 3.14 \times (5)^2 = 3.14 \times 25 = 78.5 \text{ cm}^2
$$

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Key Concept: Sector and Segment Identification

d) Sector
[Solution Description] A sector of a circle is the region enclosed by two radii and an arc.

Key Concept: Sector and Segment Differentiation

a) 5.80 cm²
[Solution Description]
First, calculate the area of the sector:
$$
\text{Area of sector} = \frac{60}{360} \times \pi \times (8)^2
$$

Simplify:
$$
= \frac{1}{6} \times \pi \times 64
$$

$$
= \frac{64\pi}{6} \approx 33.51\, \text{cm}^2
$$

Now calculate the area of the equilateral triangle (side length = 8 cm):
$$
\text{Area of triangle} = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64
$$

$$
= 16\sqrt{3} \approx 27.71 \, \text{cm}^2
$$

Finally, the area of the segment is:
$$
\text{Area of segment} = 33.51 – 27.71 = 5.80\, \text{cm}^2
$$

Thus, the area of the segment is approximately 5.80 cm².

Key Concept: Application in Design

a) $28\pi mm^2$
[Solution Description] The radius r of the brooch is half of the diameter:
$$
r = \frac{28}{2} = 14 \text{ mm}
$$

Since the brooch is divided into 7 equal sectors, the central angle $\theta$ of each sector is:
$$
\theta = \frac{360}{7} \text{ degrees}
$$

Using the formula for the area of a sector:
$$
\text{Area of each sector} = \frac{\theta}{360} \times \pi r^2
$$

Calculate the specific area:
$$
\text{Area of each sector} = \frac{360/7}{360} \times \pi \times 14^2 = \frac{1}{7} \times \pi \times 196 = \frac{196}{7}\pi = 28\pi \text{ mm}^2
$$

Therefore, the area of each sector is $28\pi mm^2$

Key Concept: Units Conversion

b) 2 m
[Solution Description]
To convert cm to meters, divide by 100. Thus, 200 cm is:

$$
\frac{200}{100} = 2 \, \text{meters}.
$$

Hence, the circumference is 2 meters.

Key Concept: Minor vs Major Segment

b) Minor segment
[Solution Description] The smaller region formed by a chord and its arc is known as the minor segment.

Key Concept: Comparative Area

a) The area of the major sector is 75π greater than the minor sector.
[Solution Description]
The area of the minor sector is calculated using the formula:
$$\text{Area of minor sector} = \frac{45^\circ}{360^\circ} \times \pi \times (10)^2 = \frac{1}{8} \times \pi \times 100 = 12.5\pi\ \text{cm}^2$$

The major sector has an angle of $360^\circ – 45^\circ = 315^\circ$. The area is:
$$\text{Area of major sector} = \frac{315^\circ}{360^\circ} \times \pi \times (10)^2 = \frac{7}{8} \times \pi \times 100 = 87.5\pi\ \text{cm}^2$$

Therefore, the area of the major sector is greater than the minor sector by:
$$87.5\pi – 12.5\pi = 75\pi\ \text{cm}^2$$

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Key Concept: Major Sector

c) $75.36 \, \text{cm}^2$
[Solution Description]
First, calculate the area of the entire circle:
$$
\text{Area of Circle} = \pi \times (6)^2 = 36\pi \, \text{cm}^2
$$
Next, determine the area of the minor sector where $\theta = 120^\circ$:
$$
\text{Area of Minor Sector} = \frac{120}{360} \times \pi \times (6)^2 = \frac{1}{3} \times 36\pi = 12\pi \, \text{cm}^2
$$
The area of the major sector is:
$$
\text{Area of Major Sector} = 36\pi – 12\pi = 24\pi \, \text{cm}^2
$$
Approximating using $\pi \approx 3.14$,
$$
\text{Area of Major Sector} \approx 75.36 \, \text{cm}^2
$$

Key Concept: Circumference Comparison

b) Circle Y
[Solution Description]
Calculate the circumference of Circle X using its diameter d = 10 cm:

$$C_X = \pi \times 10 = 10\pi \, \text{cm}$$

Calculate the circumference of Circle Y using its radius $r = 6$ cm:

$$C_Y = 2\pi \times 6 = 12\pi \, \text{cm}$$

Compare 10πcm with 12π cm. Since $12\pi > 10\pi$, Circle Y has a larger circumference.

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Key Concept: Real-World Application

d) $24.5\pi$ sq. m

[Solution Description]

We calculate the area of the sector using the formula:

$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$

Given:

$\theta = 45^\circ$, $r = 14$ m.

Substitute these values:

$$\text{Area of sector} = \frac{45}{360} \times \pi \times 14^2$$

Simplify:

$$\frac{45}{360} = \frac{1}{8}, \text{so,}$$

$$\text{Area of sector} = \frac{1}{8} \times \pi \times 196 = 24.5\pi$$

Therefore, the area of the sector is $24.5\pi$ square meters.

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Key Concept: Unit Conversion

b) 314 cm
[Solution Description] First, convert the diameter from meters to centimeters:
$$
1 \text{ m} = 100 \text{ cm}
$$
Now calculate the circumference using
$$
C = \pi d = 3.14 \times 100 = 314 \text{ cm}
$$
Therefore, the circumference is $ 314 \text{ cm} $.

Key Concept: Real-world Application

d) 10.5 m²
[Solution Description]
First, calculate the area of the sector:
$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$

Substituting the values:
$$\text{Area of sector} = \frac{90}{360} \times 3.14 \times 30^2 = 235.5 \text{ m}^2$$

Now, calculate the area of the triangle:
$$\text{Area of } \Delta OAB = \frac{1}{2} r^2 \sin(\theta)$$

With $\theta = 90^\circ$, $\sin(90^\circ) = 1$, so:
$$\text{Area of } \Delta OAB = \frac{1}{2} \times 30^2 = 450 \div 2 = 225 \text{ m}^2$$

Finally, compute the area of the segment:
$$\text{Area of segment} = 235.5 – 225 = 10.5 \text{ m}^2$$

Thus, the area of the garden segment is 10.5 m².

Key Concept: Conceptual Understanding

d) It quadruples
[Solution Description] The area of a sector is proportional to the square of its radius $(r^2)$. If the radius is doubled, the area becomes four times larger because $(2r)^2 = 4r^2$.

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Key Concept: Umbrella Rib Area, Real-World Application

c) 1131 square centimeters
[Solution Description]
The angle subtended by each rib at the center is:
$$
\theta = \frac{360^\circ}{10} = 36^\circ
$$

The area of the sector formed by two ribs is:
$$
\text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{36}{360} \times \pi \times (60)^2
$$

Calculating further:
$$
\text{Area of sector} = \frac{1}{10} \times \pi \times 3600 = 360\pi \, \text{square centimeters}
$$

So, the area between two consecutive ribs is approximately $1131 \, \text{square centimeters}$.

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Key Concept: Comparative Analysis

a) First sector
[Solution Description]
To find the area of each sector, use the formula for the area of a sector:
$$\text{Area} = \frac{\theta}{360} \times \pi \times r^2$$

For the first sector:
$$
\text{Area}_1 = \frac{60}{360} \times 3.14 \times 5^2 = 13.0833\, \text{cm}^2
$$

For the second sector:
$$
\text{Area}_2 = \frac{30}{360} \times 3.14 \times 7^2 = 12.84\, \text{cm}^2
$$

The first sector has a larger area.

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Key Concept: Minor Segment Area

b) $10.27 cm^2$
[Solution Description] Calculate the area of the sector first:
$$
\text{Area of the sector} = \frac{90}{360} \times \pi \times 6^2 = \frac{1}{4} \times \pi \times 36 = 9\pi
$$
Then calculate the area of triangle formed by two radii and the chord using:
$$
\text{Area of the triangle} = \frac{1}{2} \times 6 \times 6 \times \sin(90^\circ) = 18
$$
Now, the area of the minor segment is:
$$
\text{Area of the segment} = 9\pi – 18 \approx 10.27 \text{ cm}^2
$$

Key Concept: Application of π

c) 20.93 m
[Solution Description]
The formula for the length of an arc is given by:
$$
\text{Length of arc} = \frac{\theta}{360} \times 2\pi r
$$
where $ r = 20 \, \text{m} $ and $ \theta = 60^\circ $.

Substituting the values:
$$
\text{Length of arc} = \frac{60}{360} \times 2 \times 3.14 \times 20
$$

Calculating step by step:
$$
\frac{60}{360} = \frac{1}{6}
$$
$$
2 \times 3.14 \times 20 = 125.6
$$
$$
\text{Length of arc} = \frac{1}{6} \times 125.6 = 20.93 \, \text{m}
$$
Hence, the length of the arc is approximately 20.93 m.

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Key Concept: Basic Formula

a) Area of sector – Area of triangle
[Solution Description] The area of a segment of a circle can be calculated by finding the area of the sector and subtracting the area of the triangle formed by the chord and the radii. Hence, the formula is:
$$
\text{Area of Segment} = \frac{\theta}{360} \times \pi r^2 – \frac{1}{2} \times r^2 \times \sin(\theta)
$$

Key Concept: Arc Length and Sector Area

a) $Arc length: 78.54 m; Area: 3926.99 m^2$
[Solution Description] To find the length of the arc, use the formula $\frac{\theta}{360} \times 2\pi r$. Here, $\theta = 45^\circ$ and $r = 100$ meters.

$$
\text{Length of arc} = \frac{45}{360} \times 2\pi \times 100 = \frac{1}{8} \times 200\pi = 25\pi \approx 78.54 \text{ meters}
$$

Next, to find the area of the sector, use the formula $\frac{\theta}{360} \times \pi r^2$.

$$
\text{Area of sector} = \frac{45}{360} \times \pi \times 100^2 = \frac{1}{8} \times 10000\pi = 1250\pi \approx 3926.99 \text{ square meters}
$$

Hence, the length of the arc is approximately 78.54 meters, and the area of the sector is approximately 3926.99 square meters.

Key Concept: Segment Area Calculation

a) $66.54 \, \text{cm}^2$
[Solution Description]
First, calculate the area of the sector:

$$
\text{Area of sector} = \frac{135}{360} \times 3.14 \times 9^2 = \frac{3}{8} \times 3.14 \times 81 = 95.175 \, \text{cm}^2
$$

The area of the triangle can be calculated as follows:

$$
\text{Area of triangle} = \frac{1}{2} \times 9^2 \times \sin(135^\circ) = \frac{1}{2} \times 81 \times \frac{\sqrt{2}}{2} = 28.64 \, \text{cm}^2
$$

Finally, the area of the segment:
$$
\text{Area of segment} = 95.175 – 28.64 = 66.535 \, \text{cm}^2
$$