CBSE Solutions For Class 10 Mathematics Chapter 12 Area Related To Circles

CBSE Solutions For Class 10 Mathematics Chapter 12 Area Related To Circles

Question 1. Find the area of a circle whose Circumference is 440m. 
Solution: 

Circumference of a circle = 440m

2πr = 440

⇒ \(r=\frac{440 \times 7}{2 \times 22}\)

r = 70m

Area of a circle = πr² = \(\frac{22}{77} \times 70 \times 70\)

= 22 ×10×70

= 15400 m²

Question 2. Find the radius of a Circular sheet whose area is 55442″. 
Solution: 

Area of Circular sheet

= 5544 m²

πr² = 5544

πr²= \(\frac{5544 \times 7}{22}\)

r² = \(\frac{38808}{22}\)

r² = 1764

r = 42m

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Question 3. The area of a Circular plot is 346.5m. Calculate the Cost of fencing the plot at the rate of ± 6 Per metre. 
Solution:

Area of plot = 346.5m²

⇒ πr² = 346.5m²

⇒ r² = \(\frac{3465 \times 7}{22}\)

⇒ r²= 110.25

⇒ r = 10.5m

Circumference of plot = 2πY = 2x \(\frac{22}{7}\) x 10.5 =  66m

Cost of fencing = Circumference X Cost of fencing per metre.

= 66 × 6 =7396

Question 4. The radii of the two Circles are 19 cm and 9cm respectively. Find the radius of the Circle which has a Circumference equal to the Sum of the circumferences of two circles. 
Solution: 

Here, r₁ =19 cm and r₂ = 9 cm

Let the radius of the new circle =R Cm

Given that,

Circumference of given two new circles = Sum of the Circumferences of given two Circles

⇒ 2πR = 2πr₁ + 2πr₂

⇒ R = r₁+r₂ = 19+9 = 28cm

Question 5. The distance of a Cycle wheel is 28cm How many revolutions will it make in moving 13.2km? 
Solution: 

Distance travelled by the wheel in one revolution = 271 = 22 = x 28 = 88m Total distance travelled by the wheel = 13.2 x 1000 x 100 cm

Number of revolution made by the wheel = \(\frac{\text { total distance }}{\text { Circumference }}\)

⇒ \(\frac{13.2 \times 1000 \times 100}{88}\)

= 15000 revolutions.

Question 6. The Circumference of a Circle exceeds the diameter by 16-8 cm. Find the radius of the Circle. 

Solution: 

let the radius of the Circle be r.

Diameter = 2r

Circumference of Circle =2πr

using the given Information, we have

2πr = 2r+ 16.8

⇒ \(2 \times \frac{22}{7} \times r\) = 2r+16·8

⇒ 44r = l4r+ 16.8×7

⇒ 30r = 117.6

⇒ \(r=\frac{117.6}{30}=3.92 \mathrm{~cm}\)

Question 7. Find the area of 15cm. ring whose outer and Inner radii are respectively 20cm and 

Solution: 

Outer radius R = 200m

Immer radius r = 15 Cm

Area of ring = πT (Rr² – r²)

Area of ring = \(\frac{22}{7}\) [(20)² = (15)²]

⇒ \(\frac{22}{7}\) (400-225)

⇒ \(\frac{22}{7}\)  x 175

⇒ 22×25

⇒ 550cm²

Question 8. A race track is in the form of a · ring with an inner circumference of 352m and an outer Circumference of 396m. Find the width of the track. 

Solution: 

Let R and r be the outer and inner radii of the Circle.

The width of the track = (R-r) (m

Now,  2πr = 352

⇒ 2 x  \(\frac{22}{7}\) x r  = 352

⇒ r = \(\frac{352 \times 7}{2 \times 22}\)

r = 7 x 8= 56m

Again, 2πR=396

⇒ \(2 x \frac{352 \times 7}{2 \times 22}\) x R = 396

⇒ R= \(\frac{396 \times 7}{2 \times 22}\) = 7×9= 63m

R= 63m, r=56m

width of the track = (R-r)m = (63-56)m = 7m

Question 9. Two Circles touch internally. The Sum of their areas is 116π (m² and the distance between their Centres is 6cm. Find the radii of the circles. 

Solution: 

let two circles with Centers ‘o’ and o  having radial R and r respectively touch each other at P.

It is given that θ = 6

⇒ R-r=6

⇒ R= 6+r²

Also, πR²+ πr² = 116π

⇒ π(R² + r²) = 116π

⇒ R² + r² = 116 → 2

From equations (1) and (2), we get (6+r)² + r² = 116

⇒ 36+r²+12r+r²=116

⇒ 27 2 +(20-80=0

⇒ r² 768-40=0

⇒ (1+(0)(x-4)=0

So r=-10 and r=4

But the radius Cannot be negative. So, we reject r=-10

r = 4cm

R=6+4=10cm

Hence, the radii of the two circles are 4cm and 10 Cm.

Question 10. The radius of a wheel of a bus is 5 cm. Determine its Speed in kilometers per hour, when its wheel makes 315 revolutions per minute. 

Solution: 

The radius of the wheel of the bus = 45 Cm

Circumference of the wheel = 2πr

⇒ \(=2 \times \frac{22}{7} \times 45=\frac{1980}{7} \mathrm{~cm}\)

Distance Covered by the wheel in One revolution = \(=\frac{1980}{7} \mathrm{~cm}\)

Distance Covered by the wheel in 315 revolution = \(\frac{1980}{7}\)

= 45 x 1980 = 89100 cm

⇒ \(\frac{89100}{1000 \times 100} \mathrm{~km}=\frac{891}{1000} \mathrm{~km}\)

Distance Covered in 60 minutes to Ihr  = \(\frac{891}{1000} \times 60=\frac{5346}{100}=\) 53.46km

Hence, Speed of bus  = 53.46 km/hr.

Question 11. A Square of the largest circle is lost as trimmings? area is cut out of a circle. What /% of the area of 

Solution: 

Let the radius of the Circle be v units,

Area of Circle = πr² Sq. units

Let ABCD be the largest Square.

Length of diagonal = 2r= Side√2

Side = \(\frac{2 r}{\sqrt{2}}\) =√2r

Area (Square ABCD) = (Side)2 = (2x)2 = 2r²

Area of Circle lost by cutting out of a Square of largest area = 2r²

Required percentage of the area of c\(=\frac{2 r^2}{\pi r^2} \times 100=\frac{200}{\pi} \%\)

Question 12. The perimeter of a Semi Circular protractor is 32.4cm Calculate: 

1. The radius of the protractor in Cm, 
2.  The are of protractors in m². 

Solution:

1 . let the radius of the protractor be 5cm.

Perimeter of semicircle protractor = (πr+21) Cm

r(π+2)=324

⇒ r \(\left(\frac{22}{7}+2\right)\) = 32.4

⇒ \(r \times \frac{36}{7}=32.4\)

⇒ \(\frac{324 \times 7}{36}\)

⇒ r=6.3 cm

Hence, the radius of the protractor = 6.3 cm

2) Area of Semi Circular protractor =\(\frac{1}{2} \pi r^2\)

⇒ \(\frac{1}{2} \times \frac{22}{7} \times 6.3 \times 6.3\)

⇒ 62.37 Cm²

Hence, the area of the protractor = 62.37 cm²

Question 13. The minute hand of a clock Is √21 cm long. Find the area described as the minute hand on the face of the clock between 6 am. and 605 am. 

Solution: 

In 60 minutes, the minute hand of a clock move through an angle of 360°:

In 5 minutes hand will move through an angle = 360° x 5 = 30°,

Now, r=√21 Cm and 0=30°

Area of Sector described by the minute hand between 6 am and 6.05 am.

⇒ \(\frac{\pi r^2 \theta}{360^{\circ}} \)

⇒ \(\frac{22}{7} \times(\sqrt{21})^2 \times 30^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times 21 \times \frac{1}{12}\)

Question 14.  In the adjoining figure, Calculate: 

1. The length of minor arc ACB 
2. Area of shaded Sector.

Solution: 

Here, θ =150°, r=14cm

1) Length of minor are = \(\frac{\pi r \theta}{180^{\circ}}\)

\(\frac{22}{7} \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 36.67 Cm

2) Area of shaded sector= \(\frac{\pi r^2 \theta}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times(14)^2 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times 4^2 \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(44 \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 256.67 cm²

Question 15. Find the area of a sector of a circle with a radius of 6 cm if the angle of the Sector is 60°. 

Solution: 

Here, the radius of the circle, r=6cm

The angle of Sector, θ = 60°

Area of Sector =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)

⇒ \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

⇒ \(\frac{1}{6} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

⇒ \(=\frac{132}{7} \mathrm{~cm}^2\)

= 18.86cm²

Question 16. Find the area of a quadrant of a circle whose Circumference is 22cm.

Solution: 

Circumference of Circle, 2πr=22

⇒ \(2 \times \frac{22}{7} \times r=22\)

⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

Now, the area of the quadrant of the Circle,

⇒ \(\frac{1}{4} \pi r^2\)

⇒ \(\frac{1}{4} \times \frac{2 \pi}{7} \times \frac{7}{2} \times \frac{7}{2}\)

⇒ \(\frac{1}{2} \times 11 \times \frac{1}{2} \times \frac{7}{2}\)

⇒ \(\frac{11}{4} \times \frac{7}{2}\)

⇒ \(\frac{77}{8} \mathrm{~cm}^2\)

Question 17. The length of the minute hand of a clock is 14cm. Find the area an Sq ft take Swept by the Minute hand in 5 minutes. 

Solution: 

Length of a minute hand of clock = 14cm

Radius of Circle = 14cm

Angle Subtended by minute hand in 60 min  = 360°

Angle subtended by minute hand in Iminute =\(\frac{360^{\circ}}{60^{\circ}}=6^{\circ}\)

The angle subtended by minute hand in 5 minutes = 30°

From the formula,

Area of Sector of Circle = \(\frac{\theta \pi r^2}{360^{\circ}}\)

⇒ \(=30^{\circ} \times \frac{22 \times(14)^2}{7 \times 360^{\circ}}\)

⇒ \(\frac{22 \times 14 \times 2}{12}\)

⇒ \(\frac{616}{12}\)

⇒ \(\frac{154}{3} \mathrm{~cm}^2\)

Question 18. A chord of a circle of radius 10 cm Subtends a right angle at the  Centre. Find the area of the Corresponding: 

1. minor Segment 
2. major segment 

Solution: 

Given the radius of the Circle, A0-10cm.

The perpendicular is drawn from the Centre of the circle to the chord of the Circle to the chord of the circle which bisects this chord.

AD=DC

and LAOD = <COD  = 45°

LAOC = LAOD + LCOD

= 45° +45° =90°

In the right AAOD,

⇒ Sin45\(=\frac{A D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{A D}{10}\)

⇒ AD=5√2 Cm

and cos 45° = \(\frac{O D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{O D}{10}\)

⇒ OD = 5√2 Cm

Now, AC=2AD

= 2X5√2 = 10√2 cm

Now, the area of AAOC

⇒ \(\frac{1}{2}\) X AC X OD

⇒ \(\frac{1}{2}\) x 10√2 × 5√52

= 50cm²

Now, area of Sector

⇒ \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2\)

⇒ \(\frac{314}{4}\)

=78.5cm²

(1)Area of minor Segment AEC

= area of sector OAEC – area of Aoc

= 78.5-50 =  28,50m²

(2)Area of major SegSector OAFGCO

= area of a circle – an area of sector OAEC = πr²-78.5

= 3-14x(10) 278.5

= 314-785

= 235.5cm²

Question 19. A chord of a circle of radius 12CM Subtends an angle of 120° at the Centre Find the area of the Corresponding Segment of the Circle. (use πT = 3.14 and √3 = 1-73) 

Solution: 

Here, the radius of the circle, r = 12 Cm

Angle Subfended by the chord at the Centre, 0=120°

Area of Corresponding minor segment

⇒ \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

⇒ \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

⇒ \(12 \times 12 \times\left(\frac{3.14 \times 120^{\circ}}{360^{\circ}}-\frac{1}{2} \times \sin 120^{\circ}\right)\)

⇒ \(144\left(\frac{3.14}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

⇒ \(144\left(\frac{3.14}{3}-\frac{1.73}{4}\right)\)

= 88,44 cm²

Question 20.  A Car has two wipers which do not overlap. Each wiper has a blade of length 25 cm Sweeping through each Sweep of the blades. an angle of 115 find the total area cleaned at 

Solution:

Given, length of wiper blade  = 25 cm = r (Say)

The angle formed by this blade, 0=115°

Area cleaned by a blade = area of Sector formed  by blade

⇒ \(\frac{\theta \pi r^2}{360^{\circ}}\)

⇒ \(115^{\circ} \times \frac{22}{7 \times 360^{\circ}} \times(25)^2\)

⇒ \(\frac{23 \times 22}{7 \times 72} \times 625\)

⇒ \(\frac{23 \times 11 \times 625}{7 \times 36}\)

Total area cleaned by two blades =  2x area cleaned by a blade

⇒ \(\frac{2 \times 158125}{252}\)

⇒ \(\frac{158125}{126} \mathrm{~cm}^2\)