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AREAS RELATED TO CIRCLES - Q&A

EXERCISE 11.1

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Solution:
Given:
Radius of the circle (r) = 6 cm
Angle of the sector (θ) = 60°

Formula for Area of a sector = (θ/360°) × πr²
Substituting the values:
Area = (60/360) × (22/7) × 6²
= (1/6) × (22/7) × 36
= (1 × 22 × 6) / 7
= 132/7 cm²

Answer: The area of the sector is 132/7 cm² (or approx 18.86 cm²).

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:
Let the radius of the circle be 'r'.
Given Circumference = 22 cm
2πr = 22
2 × (22/7) × r = 22
r = (22 × 7) / (2 × 22)
r = 7/2 = 3.5 cm

A quadrant of a circle subtends an angle of 90° at the centre.
Area of quadrant = (90/360) × πr² = (1/4)πr²
Area = (1/4) × (22/7) × (7/2) × (7/2)
= (1/4) × 22 × (1/2) × (7/2)
= (22 × 7) / (4 × 2 × 2)
= 154 / 16
= 77/8 cm²

Answer: The area of the quadrant is 77/8 cm² (or 9.625 cm²).

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:
Length of minute hand (radius, r) = 14 cm
The minute hand completes a full circle (360°) in 60 minutes.
Angle swept in 1 minute = 360° / 60 = 6°
Angle swept in 5 minutes (θ) = 6° × 5 = 30°

Area swept = Area of sector with angle 30°
Area = (θ/360) × πr²
= (30/360) × (22/7) × 14 × 14
= (1/12) × 22 × 2 × 14
= (1/12) × 616
= 154/3 cm²

Answer: The area swept by the minute hand is 154/3 cm² (or approx 51.33 cm²).

4. 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 (ii) major sector. (Use π = 3.14)

Solution:
Given: Radius (r) = 10 cm, Angle (θ) = 90°.

(i) Area of minor segment
Area of minor sector = (90/360) × πr²
= (1/4) × 3.14 × 10 × 10
= (1/4) × 314 = 78.5 cm²

Area of triangle formed by radii and chord (Right angled triangle):
Area of Δ = (1/2) × base × height = (1/2) × 10 × 10 = 50 cm²

Area of minor segment = Area of minor sector - Area of triangle
= 78.5 - 50 = 28.5 cm²

(ii) Area of major sector
Area of circle = πr² = 3.14 × 100 = 314 cm²
Area of major sector = Area of circle - Area of minor sector
= 314 - 78.5 = 235.5 cm²
(Alternatively, Angle of major sector = 360° - 90° = 270°. Area = (270/360) × 314 = 235.5 cm²)

Answer: (i) 28.5 cm² (ii) 235.5 cm²

5. In a circle of radius 21 cm, an arc subtends an angle of 60° 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

Solution:
Given: Radius (r) = 21 cm, Angle (θ) = 60°.

(i) Length of the arc
Length = (θ/360) × 2πr
= (60/360) × 2 × (22/7) × 21
= (1/6) × 2 × 22 × 3
= (1/6) × 132 = 22 cm

(ii) Area of the sector
Area = (θ/360) × πr²
= (60/360) × (22/7) × 21 × 21
= (1/6) × 22 × 3 × 21
= 231 cm²

(iii) Area of the segment
Area of segment = Area of sector - Area of corresponding triangle.
Since the angle is 60° and two sides are radii (equal), the triangle is equilateral.
Area of equilateral triangle = (√3 / 4) × side²
= (√3 / 4) × 21 × 21
= (441√3) / 4 cm²

Area of segment = (231 - 441√3/4) cm²

Answer: (i) 22 cm, (ii) 231 cm², (iii) (231 - 441√3/4) cm²

6. 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 π = 3.14 and √3 = 1.73)

Solution:
Given: Radius (r) = 15 cm, Angle (θ) = 60°.

Minor Segment:
Area of minor sector = (60/360) × 3.14 × 15 × 15
= (1/6) × 3.14 × 225
= 117.75 cm²

Since the angle is 60°, the triangle formed is equilateral.
Area of triangle = (√3 / 4) × r²
= (1.73 / 4) × 225
= 97.3125 cm²

Area of minor segment = Area of sector - Area of triangle
= 117.75 - 97.3125 = 20.4375 cm²

Major Segment:
Area of circle = πr² = 3.14 × 225 = 706.5 cm²
Area of major segment = Area of circle - Area of minor segment
= 706.5 - 20.4375 = 686.0625 cm²

Answer: Minor segment area = 20.4375 cm², Major segment area = 686.0625 cm²

7. 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 π = 3.14 and √3 = 1.73)

Solution:
Given: Radius (r) = 12 cm, Angle (θ) = 120°.
Area of sector = (120/360) × 3.14 × 12 × 12
= (1/3) × 3.14 × 144
= 3.14 × 48 = 150.72 cm²

To find the area of the triangle OAB (where O is centre, AB is chord):
Draw OM perpendicular to AB. This bisects angle θ (making it 60°) and chord AB.
In ΔOMA, OM = r cos 60° = 12 × 0.5 = 6 cm.
AM = r sin 60° = 12 × (√3/2) = 6√3 cm.
Base AB = 2 × AM = 12√3 cm.
Area of ΔOAB = (1/2) × Base × Height
= (1/2) × 12√3 × 6
= 36√3
= 36 × 1.73 = 62.28 cm²

Area of segment = Area of sector - Area of triangle
= 150.72 - 62.28 = 88.44 cm²

Answer: The area of the segment is 88.44 cm².

8. 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.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Solution:
The horse can graze in the shape of a quadrant of a circle (since the corner of a square is 90°).
(i) Area with 5 m rope:
Radius (r) = 5 m, θ = 90°.
Area = (90/360) × πr²
= (1/4) × 3.14 × 5 × 5
= (1/4) × 78.5
= 19.625 m²

(ii) Increase in area with 10 m rope:
New Radius (R) = 10 m.
New Area = (1/4) × 3.14 × 10 × 10
= (1/4) × 314 = 78.5 m²
Increase in area = New Area - Old Area
= 78.5 - 19.625
= 58.875 m²

Answer: (i) 19.625 m² (ii) 58.875 m²

9. 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.

Solution:
Diameter (d) = 35 mm, Radius (r) = 35/2 mm.
(i) Total length of silver wire:
Wire is used for the circumference and 5 diameters.
Length = Circumference + (5 × diameter)
= πd + 5d
= (22/7) × 35 + 5 × 35
= 110 + 175
= 285 mm

(ii) Area of each sector:
The circle is divided into 10 equal sectors.
Area of one sector = (1/10) × Area of circle
= (1/10) × πr²
= (1/10) × (22/7) × (35/2) × (35/2)
= (1/10) × 22 × (5/2) × (35/2)
= (11 × 5 × 35) / 20
= 385 / 4 mm²

Answer: (i) 285 mm (ii) 385/4 mm² (or 96.25 mm²)

10. 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.

Solution:
Radius (r) = 45 cm.
Total ribs = 8, so the circle is divided into 8 equal sectors.
Area between two consecutive ribs = Area of one sector.
Area = (1/8) × πr²
= (1/8) × (22/7) × 45 × 45
= (11 × 45 × 45) / (4 × 7)
= 22275 / 28 cm²

Answer: The area between two consecutive ribs is 22275/28 cm² (or approx 795.53 cm²).

11. 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.

Solution:
Radius (r) = 25 cm.
Angle (θ) = 115°.
Since there are two identical wipers, we multiply the area of one sector by 2.
Total Area = 2 × (θ/360) × πr²
= 2 × (115/360) × (22/7) × 25 × 25
= 2 × (23/72) × (22/7) × 625
= (23 × 11 × 625) / (18 × 7)
= 158125 / 126 cm²

Answer: The total area cleaned is 158125/126 cm² (or approx 1254.96 cm²).

12. 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 π = 3.14)

Solution:
Radius (distance) r = 16.5 km.
Angle (θ) = 80°.
Area = (θ/360) × πr²
= (80/360) × 3.14 × 16.5 × 16.5
= (2/9) × 3.14 × 272.25
= 2 × 3.14 × 30.25
= 189.97 km²

Answer: The area of the sea is 189.97 km².

13. 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. (Use √3 = 1.7)

Solution:
Radius (r) = 28 cm.
The designs are in the form of 6 segments.
Angle of each sector = 360° / 6 = 60°.
Area of one design (segment) = Area of sector - Area of triangle.
Area of sector = (60/360) × (22/7) × 28 × 28
= (1/6) × 22 × 4 × 28
= 1232 / 3 = 410.67 cm²
Since the angle is 60°, the triangle is equilateral.
Area of triangle = (√3 / 4) × 28 × 28
= 1.7 × 7 × 28 = 333.2 cm²
Area of one design = 410.67 - 333.2 = 77.47 cm²
Total area of 6 designs = 6 × 77.47 = 464.82 cm²
Cost = Area × Rate = 464.82 × 0.35 = ₹ 162.68 (approx).
(Using exact fractions: Total Area = 6 × (1232/3 - 196√3) = 2464 - 1176(1.7) = 2464 - 1999.2 = 464.8. Cost = 464.8 × 0.35 = 162.68)

Answer: The cost of making the designs is ₹ 162.68.

14. Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) (p/180) × 2πR
(B) (p/180) × πR²
(C) (p/360) × 2πR
(D) (p/720) × 2πR²

Answer:
Formula for area of sector = (Angle/360) × πR²
Here Angle = p.
So, Area = (p/360) × πR².
Looking at option (D): (p/720) × 2πR² = (p/360) × πR².
Correct Option: (D)

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Quick Review Flashcards - Click to flip and test your knowledge!

Question

How is a 'sector' of a circle defined?

Answer

The portion of the circular region enclosed by two radii and the corresponding arc.

Question

What defines a 'segment' of a circle?

Answer

The portion of the circular region enclosed between a chord and the corresponding arc.

Question

In Figure 11.1, what is the term for the angle $\angle AOB$ in the shaded region OAPB?

Answer

The angle of the sector.

Question

What is the name of the smaller region of a circular region divided by two radii?

Answer

The minor sector.

Question

What is the name of the larger region of a circular region divided by two radii?

Answer

The major sector.

Question

How is the angle of a major sector calculated if the angle of the minor sector is $\angle AOB$?

Answer

$360^\circ - \angle AOB$.

Question

Which specific segment is meant when the term 'segment' is used without any qualification?

Answer

The minor segment.

Question

Which specific sector is meant when the term 'sector' is used without any qualification?

Answer

The minor sector.

Question

What is the formula for the area of a circle with radius $r$?

Answer

$\pi r^2$

Question

In the unitary method derivation, what is the area of a sector with a degree measure of $360^\circ$?

Answer

$\pi r^2$

Question

Using the unitary method, what is the area of a sector when the degree measure of the angle at the centre is $1^\circ$?

Answer

$\frac{\pi r^2}{360}$

Question

What is the formula for the area of a sector with radius $r$ and angle $\theta$ (in degrees)?

Answer

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

Question

What is the formula for the length of an arc of a sector with radius $r$ and angle $\theta$?

Answer

$\frac{\theta}{360} \times 2 \pi r$

Question

How is the area of a segment calculated relative to its corresponding sector?

Answer

Area of the corresponding sector minus the area of the corresponding triangle.

Question

What is the formula for the area of a major sector in terms of the minor sector?

Answer

$\pi r^2 - \text{Area of the minor sector}$

Question

What is the formula for the area of a major segment in terms of the minor segment?

Answer

$\pi r^2 - \text{Area of the minor segment}$

Question

Example 1: What is the area of a sector with radius $4$ cm and angle $30^\circ$ (use $\pi = 3.14$)?

Answer

$4.19$ $cm^2$ (approx.)

Question

Example 1: How can the area of the major sector be calculated directly using the angle for $r=4$ and $\theta=30^\circ$?

Answer

$\frac{360 - 30}{360} \times 3.14 \times 16$

Question

Example 2: What is the area of the sector $OAYB$ if the radius is $21$ cm and $\angle AOB = 120^\circ$?

Answer

$462$ $cm^2$

Question

Example 2: When drawing $OM \perp AB$ for triangle $OAB$, what is the relationship between $\triangle AMO$ and $\triangle BMO$?

Answer

They are congruent by RHS congruence.

Question

Example 2: If $\angle AOB = 120^\circ$, what is the measure of $\angle AOM$ after drawing the perpendicular $OM$ to chord $AB$?

Answer

$60^\circ$

Question

Example 2: In $\triangle OMA$, which trigonometric ratio is used to find the length of $OM$ relative to $OA$?

Answer

$\cos 60^\circ$

Question

Example 2: In $\triangle OMA$, which trigonometric ratio is used to find the length of $AM$ relative to $OA$?

Answer

$\sin 60^\circ$

Question

Example 2: How is the length of chord $AB$ related to $AM$ in an isosceles triangle with altitude $OM$?

Answer

$AB = 2 \times AM$

Question

Example 2: What is the area of $\triangle OAB$ when $r=21$ cm and $\angle AOB = 120^\circ$?

Answer

$\frac{441 \sqrt{3}}{4}$ $cm^2$

Question

Exercise 11.1, Q1: What is the area of a sector with radius $6$ cm and angle $60^\circ$?

Answer

$\frac{132}{7}$ $cm^2$

Question

Exercise 11.1, Q2: What is the degree measure of the angle of a quadrant of a circle?

Answer

$90^\circ$

Question

Exercise 11.1, Q2: What is the radius of a circle whose circumference is $22$ cm?

Answer

$\frac{7}{2}$ cm

Question

Exercise 11.1, Q3: What is the degree measure of the angle swept by a minute hand in $5$ minutes?

Answer

$30^\circ$

Question

Exercise 11.1, Q4: What is the area of a major sector if the chord subtends a right angle and $r=10$ cm?

Answer

$235.5$ $cm^2$

Question

Exercise 11.1, Q5: What is the length of an arc that subtends an angle of $60^\circ$ at the centre of a circle with $r=21$ cm?

Answer

$22$ cm

Question

Exercise 11.1, Q8: If a horse is tied to the corner of a square field, what shape is the area it can graze?

Answer

A sector of a circle with a $90^\circ$ angle.

Question

Exercise 11.1, Q9: How many equal sectors are formed in the brooch by $5$ diameters?

Answer

$10$

Question

Exercise 11.1, Q9: What is the angle of each sector in a brooch divided into $10$ equal sectors?

Answer

$36^\circ$

Question

Exercise 11.1, Q10: What is the angle between two consecutive ribs of an $8$-rib umbrella assumed to be a flat circle?

Answer

$45^\circ$

Question

Exercise 11.1, Q11: What angle does each car wiper sweep through?

Answer

$115^\circ$

Question

Exercise 11.1, Q12: What part of a circle represents the area of sea over which a lighthouse spreads light?

Answer

A sector.

Question

Exercise 11.1, Q13: How many equal designs are on the round table cover?

Answer

$6$

Question

Exercise 11.1, Q14: What is the correct formula for the area of a sector with angle $p$ and radius $R$ among the given choices?

Answer

$\frac{p}{720} \times 2 \pi R^2$

Question

What is the constant value of $\pi$ to be used if not stated otherwise in the text?

Answer

$\frac{22}{7}$

Question

Summary: What is the area of a sector with radius $r$ and degree measure $\theta$?

Answer

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

Question

Summary: How do you find the area of a segment of a circle?

Answer

Subtract the area of the corresponding triangle from the area of the corresponding sector.

Question

Summary: What is the formula for the length of an arc with radius $r$ and angle $\theta$?

Answer

$\frac{\theta}{360} \times 2 \pi r$

Question

In Example 2, what value is used for $\cos 60^\circ$?

Answer

$\frac{1}{2}$

Question

In Example 2, what value is used for $\sin 60^\circ$?

Answer

$\frac{\sqrt{3}}{2}$

Question

The portion of a circular region enclosed by a chord and the corresponding arc is a _____.

Answer

Segment

Question

If the angle of a minor sector is $60^\circ$, what is the angle of the corresponding major sector?

Answer

$300^\circ$

Question

Formula: Area of a quadrant of a circle with radius $r$.

Answer

$\frac{1}{4} \pi r^2$

Question

For Exercise 11.1, Q6 and Q7, what numerical value is used for $\sqrt{3}$?

Answer

$1.73$

Question

For Exercise 11.1, Q13, what numerical value is used for $\sqrt{3}$?

Answer

$1.7$

Question

How does the grazing area change if a horse's rope is lengthened from $5$ m to $10$ m?

Answer

The area increases.

Question

Exercise 11.1, Q9: What is the total length of silver wire required for a brooch with diameter $35$ mm and $5$ diameters?

Answer

Circumference plus $5 \times$ diameter.

Question

Exercise 11.1, Q11: How many wipers are on the car?

Answer

$2$

Question

In Figure 11.7, $M$ is the _____ of the chord $AB$.

Answer

Mid-point

Question

In Example 2, if $OA = 21$ cm and $OM = x$, find $x$ using $\cos 60^\circ$.

Answer

$10.5$ cm

Question

What is the degree measure of the angle at the centre for a whole circular region?

Answer

$360^\circ$

Question

Cloze: The area of the _____ sector is calculated as $\pi r^2 - \text{Area of the minor sector}$.

Answer

Major

Question

If a chord subtends an angle of $60^\circ$ at the centre, what type of triangle is formed by the chord and the two radii?

Answer

Equilateral triangle.

Question

What is the radius of the circle in Exercise 11.1, Q7?

Answer

$12$ cm

Question

What is the radius of the circle in Exercise 11.1, Q6?

Answer

$15$ cm