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Question 1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Answer

The first circle's radius is 19 cm (given)

The circumference of the first circle is 2π x 19 cm, which equals 38 cm

The second circle's radius is 9 cm (given) 

The circumference of the second circle is 2π x 9 of the circumference of the first circle, which is 18π cm

As a result,

The circumference of the two circles added together is 38π+18π = 56 cm

Let R be the radius of the third circle

The circumference of the third circle is equal to 2πR

The circumference of two circles added together equals the circumference of the third circle

As a result, 56π = 2πR 

Alternatively, R = 28 cm

Question 2. The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Answer

(Given) 1st circle radius = 8 cm

The area of the first circle is π(8)2 = 64

The radius of the second circle is 6 cm (given) 

Area of the second circle = π(6)2 = 36π  

So, 

The sum of the first and second circles is 64π + 36π = 100

Assume that the radius of the third circle is R

The area of the third circle is equal to πR2

It is stated that the area of the third circle = the area of the first circle + the area of the second circle

Alternatively, πR² = 100π cm². 

100 cm² = R² 

As a result, R = 10 cm.

Question 3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Answer

r1 = 21/2 cm is the radius of the first circle (as diameter D is given as 21 cm) 

As a result, the area of the gold region = πr12 = (10.5)2 = 346.5 cm2

Given that each of the remaining bands is 10.5 cm wide, 

As a result, the radius of the second circle, r2, equals 10.5 cm +10.5 cm = 21 cm

Thus, 

The area of the red region is equal to the area of the second circle - The area of the gold region

= ( πr₂²−346.5) cm²

= (π(21)² − 346.5) cm² 

= 1386 − 346.5 

= 1039.5 cm²

Similarly, 

The radius of the third circle, r3, is 21 cm + 10.5 cm = 31.5 cm

The radius of the fourth circle, r4, equals 31.5 cm + 10.5 cm = 42 cm

The radius of the fifth circle, r₅, is 42 cm + 10.5 cm = 52.5 cm

For the nth region's area, 

A = n Circle Area – (n-1) Circle Area

The area of the blue region (n=3) is the area of the third circle − the area of the second circle

= π(31.5)² – 1386 cm² 

= 3118.5 – 1386 cm²  

= 1732.5 cm²  

The area of the black region (n=4) is the area of the fourth circle minus the area of the third circle

= (1.386) – 1386 cm²  

= 5544 – 3118.5 cm²  

= 2425.5 cm²  

The area of the white region (n=5) equals the area of the fifth circle − the area of the fourth circle 

= (52.5)2 – 5544 cm²  

= 8662.5 – 5544 cm² 

= 3118.5 cm²

Question 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer

80/2 = 40 cm (as D = 80 cm) is the radius of a car's wheel

As a result, the circumference of the wheels = 2πr = 80 cm 

The distance reached in one revolution is now equal to the circumference of the wheel, which is 80π cm

The distance travelled by car in one hour is indicated as 66 kilometres 

When we convert kilometres to centimetres, we obtain 

The distance travelled by the car in one hour is (66 x 10⁵) cm

The distance travelled in 10 minutes will be = (66 x 10⁵ x 10)/60 = 1100000 cm/s

The distance travelled by car is 11 x 10⁵ cm

Now, the number of wheel revolutions = (distance travelled by car/circumference of the wheels)

(11 x 10⁵)/80π = 4375

Question 5. Tick the correct Solution: in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

Answer

Because the perimeter of the circle equals the area of the circle, 

2πr = πr² 

Alternatively, r = 2 

As a result, option (A) is correct, indicating that the circle's radius is 2 units.

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

Answer

The sector's angle is assumed to be 60°

We know that the area of sector = (θ/360°)×πr² 

∴ Area of the sector with angle 60° = (60°/360°)×πr² cm² 

= (36/6)π cm²  

= 6×22/7 cm² 

= 132/7 cm²

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

Answer

Circle circumference, C = 22 cm (given) 

A quadrant of a circle is a sector that forms a 90° angle. 

Let's say the circle's radius is r. 

Since, 

C = 2πr = 22, 

R = 22/2π cm = 7/2 cm 

∴ Area of the quadrant = (θ/360°) × πr² 

Now, 

θ = 90° 

Therefore, 

A = (90°/360°) × π r2 cm²  

= (49/16) π cm²  

= 77/8 cm²  

= 9.6 cm² 

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

Answer

Minute hand length = clock radius (circle) 

The circle's radius (r) = 14 cm (given) 

In 60 minutes, the angle swept by the minute hand = 360°. 

Thus, the angle swept by the minute hand in 5 minutes 

= 360° x 5/60 = 30° 

We know, 

Area of a sector = (θ/360°) × πr²

Therefore, 

the area of the sector with a 30° angle = (30°/360°) r² cm² 

= (1/12) × π14² 

= (49/3)×(22/7) cm² 

= 154/3 cm²

Question 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)

Answer

Here, AB is the chord that forms a 90° angle at the centre O. 

The circle's radius (r) is assumed to be 10 cm. 

I Minor sector area = (90/360°) πr². 

= (¼)×(22/7)×10² 

Alternatively, 78.5 cm2 is the minor sector area.

In addition, the area of AOB = 1/2 x OB x OA 

The circle's radii, OB and OA, are equal to 10 cm. 

So, the area of ΔAOB = ½×10×10 

= 50 cm²

Now, the area of the minor segment equals the area of the minor sector minus the area of the AOB. 

= 78.5 – 50 

= 28.5 cm² 

(ii) Major sector area = circle area − minor sector area 

= (3.14×10²)-78.5

= 235.5 cm²

Question 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

Answer

Radius = 21 cm 

θ = 60° 

(i) Length of an arc = θ/360°×Circumference(2πr) 

∴ Length of an arc AB = (60°/360°) ×2×(22/7)×21 

= (1/6)×2×(22/7)×21 

Alternatively, 

Arc AB Length = 22cm 

(ii) The angle subtended by the arc is 60 °. 

So, the area of the sector making an angle of 60° = (60°/360°)×π r² cm² 

= 441/6×22/7 cm²

Alternatively, the sector created by the arc APB has a surface area of 231 cm². 

(iii) APB segment area = OAPB sector area – OAB segment area 

Δ OAB is an equilateral triangle since its two arms are the radii of the circle and hence equal, and one angle is 60 °. So, its area will be √3/4×a2 sq. Units.

Area of segment APB = 231- (3/4) (OA)² 

= 231- (3/4) (212)² 

Or, Area of segment APB 

= 231- (3/4) (441)²cm²

• Students can use NCERT Solutions to reinforce their learning of circle concepts. 

• Answering questions with diagrams makes learning more dynamic and complete. 

• The language used in NCERT Solutions is straightforward and easy to understand. 

• A step-by-step approach to problem solving is beneficial to students. 

• It helps students solve challenging problems at their own pace.