NCERT-CLASS-10-SOLUTION-AREAS-RELATED-TO-CIRCLE

 AREA RELATED TO CIRCLE: CONCEPT AND SOLUTION

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Minor Sector: The blue region in the circle is said to be a minor sector.

Major Sector:   The yellow region in the circle inside the circle is known as the major sector.

Area of the major and the minor sector:

The area of a circle is Πr²

Area of 360 degrees around a point in a circle = Πr²

If the angle made in the minor sector is ө,

Then, using the unitary method

360 degrees form an area = Πr²

^{Then Ө degree =} Πr²ϴ/360 

This is the area of the minor sector

The area of the major sector can be obtained by = Area of the circle - Area of the minor sector

                                                                              

                                                                            = Πr^{2  -} Πr²ϴ/360 

Segment of a circle: 

Minor segment:  The area of a circle between the chord and the arc is said to be the area of the segment.

The area in blue is the area of the minor segment, and the area in red is the area of the major segment. 

Finding the area of minor and major segments 

                                                                              Exercise 11.1

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

 

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

Solution:  Circumference of the circle = 2pi *r = 22cm

 2*22/7 * r = 22

r = 22*7/22*2 = 7/2 cm

Area of the quadrant of the circle = (1/4) pi * r * r

                                                     = 1/4 * (22/7) * (7/2) * (7/2) 

                                                     = 77/8 cm

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.

Solution: The Length of the minute hand of the clock will be the radius of the circle

R = 14cm

The minute hand makes 360 degrees in 60 minutes

So, the angle formed by the minute hand in 5 minutes is (360/60)*5 = 30 degrees

Area of the sector formed by the minute hand of the clock = pi*r*r*θ/360

 = 22/7 * 14 * 14 * (30/360) = 22 * 2 * 14 * 1/12 = 616/12 cm^2 

Question 4: A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding (i) minor segment (ii) major sector. (Use pi = 3.14)

Solution 

Question 5:  In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find:  (i) the length of the arc, (ii) the area of the sector formed by the arc, (iii) the area of the segment formed by the corresponding chord

Solution

(i) Length of arc formula = pi *r *Ө/180

                                         = pi *21* 60/180 = 7pi cm = 7*22/7 = 22 cm

(ii) 

Question 6. A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle.

Solution

Question 7: A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73.)

Question 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. 12.11). 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 radius of the path = 5 m (length of the rope)

The angle for the sector is 90 degrees.

Area of the sector grazed by the horse

= pi*r*r* θ/360 = 3.14* 5*5*90/360 = 3.14*25*1/4 = 19.625 m*m

(ii) When the length of the rope changed by 10 m 

then new radius = 10 m

Area of new sector = pi* 10*10*90/360 = 3.14*100*1/4 = 78.5 m*m

Increased area = 78.5 - 19.625 = 58.875 m^2 

Question 9:  A brooch is made with silver wire in the form of a circle with a diameter of of 35 mm. The wire is also used in making 5 diameters, which divide the circle into 10 equal sectors, as shown in Fig. 12.12. Find:

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Solution:  Angle of each sector = 360/10 = 36

radius of each sector = 35/2 mm

(i) Total length of the brooch = circumference + 5 * diameter

 = 22/7 * 35 +5*35 

= 110 +175 

= 285 mm

(ii) Area of each sector = 22/7 * (35/2) (35/2) * 36/360

 = 96.25 mm^2 

Question 10:  An umbrella has 8 ribs that are equally spaced (see Fig. 12.13). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution 

Angle of each sector = 360/8 = 45

r = 45

Area of each sector = (3.14 *45* 45 * 45)/360 =  794.8125 cm*cm

Question 11:   A car has two wipers that 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:  Wipers wipes out as a sector; since there are two wipers, they will swipe twice

Area swept out =  2*pi*r*r*115/(360)

 = 2* (22/7)*25*25 *115/360

 = 1254.960 cm*cm

Question 12:  To warn ships of underwater rocks, a lighthouse spreads a red-colored 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.

Solution: 

 The light ray will make a sector

radius = 16.5 km

Angle in the sector = 80 degrees

area = (22/7)*(16.5)(16.5)*(80/360)

        = 16.6375 km*km

Question 13: A round table cover has six equal designs, as shown in Fig. 12.14. 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)

Question 14:Tick the correct solution in the following:

The area of a sector of angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR

(B) p/180 × π R²

(C) p/360 × 2πR

(D) p/720 × 2πR²

Solution: Area of sector = (p/360)π R²

Multiplying the numerator and denominator by 2

Area = (p/760)2π R²     (C)

                                                   Exercise 12.3

Question 1: Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Question 2:Find the area of the shaded region in Fig. 12.20, if the radii of the two concentric circles with centre O are 7 cm and 14 cm, respectively and AOC = 40°.

Question 3:. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centrend the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Question 4: Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.

Note: Value of pi = 22/7 is used; further solution can be obtained by putting the value of the square root of 3 = 1.732

Question 5: From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut, and also a circle of diameter 2 cm is cut, as shown in Fig. 12.23. Find the area of the remaining portion of the square.

Question 6:   In a circular table cover of radius 32 cm, a design is formed, leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

Question 7: In Fig. 12.25, ABCD is a square of side 14 cm. With centers A, B, C, and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Question 8:  Fig. 12.26 depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m, and they are each 106 m long. If the track is 10 m wide, find

(i) the distance around the track along its inner edge

(ii) the area of the track.

Question 9: In Fig. 12.27, AB and CD are two diameters of a circle (with center O) perpendicular to each other, and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Question 10:  The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as the center, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region (use π = 3.14 and √3 = 1.73205).

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