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

Let r be the radius of the circle whose circumference is equal to the sum of the circumference of two circles of radii 19 cm and 9 cm.

∴ \( 2 \pi r \ = \ 2 \pi (19) \ + \ 2 \pi (9) \)

\( => \ r \ = \ 19 \ + \ 9 \ = \ 28 \)

Hence, the radius of the new circles is 28 cm.

Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the area of the two circles.

Let r be the radius of the circle whose area is equal to the sum of the area of the circles of radii 8 cm and 6 cm.

\( \pi r^2 \ = \ \pi (8)^2 \ + \ \pi(6)^2 \ => \ r^2 \ = \ 8^2 \ + \ 6^2 \)

\( => \ r^2 \ = \ 64 \ + \ 36 \ => \ r \ = \ 10 \)

Hence, the radius of the new circle is 10 cm.

Q3. Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue , Black and White. The diameter of the region representing Gold score is 21 cm and reach of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

The area of each of five scoring regions are as under :

For Gold : \(\pi (10.5)^2 \) \( = \ \frac{22}{7} × 110.25 \)

\( = \ \frac{2425.5}{7} \ = \ 346.5 \ cm^2\)

For Red : \( \pi [ (21)^2 \ - \ (10.5)^2 ] \ = \ \frac{22}{7}(441 \ - \ 110.25) \ = \ \frac{22}{7} × 330.75 \)

\(= \ \frac{7276.5}{7} \ = \ 1039.5 \ cm^2 \)

For Blue : \( \pi [(31.5)^2 \ - \ (21)^2] \ = \ \frac{22}{7}(992.25 \ - \ 441) \)

\( = \ \frac{22}{7} × 551.25 \ = \ \frac{12127.5}{7} \ = \ 1732.5 \ cm^2 \)

For Black : \( \pi [(42)^2 \ - \ (31.5)^2] \ = \ \frac{22}{7} (1764 \ - \ 992.25) \)

\( = \ \frac{22}{7} × 771.75 \ = \ \frac{16978.5}{7} \ = \ 2425.5 \ cm^2 \)

For White : \( \pi [(52.5)^2 \ - \ (42)^2] \ = \ \frac{22}{7} (2756.25 \ - \ 1764) \)

\(= \ \frac{22}{7} × 992.25 \ = \ \frac{21829.5}{7} \ = \ 3118.5 \ cm^2 \)

Q4. 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?

Distance covered by the car in 10 minutes

\( = \ \frac{66 \ × \ 1000 \ × \ 100 \ × \ 10}{60} \) \( = \ \frac{6600000}{6} \) cm

Circumference of the wheel of the car

\( = \ (2 \ × \ \frac{22}{7} \ × \ 40) \)cm

∴ Number of revolutions in 10 minutes when the car is travelling at a speed of 66 km/hr.

\( = \ \frac{6600000}{6} \ × \ \frac{7}{2 \ × \ 22 \ × \ 40} \ = \ 4375 \)

Hence, the wheel of car makes 4375 revolutions in 10 minutes.

Q5. Tick the correct answer 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) \( \pi \) units

(c) \( 4 \) units (d) \( 7 \) units

(a) \( 2 \) units (b) \( \pi \) units

(c) \( 4 \) units (d) \( 7 \) units

Here, \( 2 \pi r \ = \ \pi r^2 \), where r is the radius

\( => \ r^2 \ - \ 2r \ = \ 0 \ => \ r(r \ - \ 2) \ = \ 0 \)

\(=> \ r \ = \ 0 \ or \ r \ = \ 2 \)

But, \( r \ \ne \ 0 \) ∴ \(r \ = \ 2 \) units

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°

We know that the area A of a sector of angle \( \theta \) in a circle of radius r is given by \( A \ = \ \frac{ \theta}{360} \ × \ \pi r^2 \).

Here \( r \ = \ 6 \ cm \), \( \theta \ = \ 60° \)

∴ \( A \ = \ \frac{60}{360} \ × \ \frac{22}{7} \ × \ 36 \ = \ ( \frac{132}{7} ) \ cm^2 \)

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

Let r be the radius of the circle. Then,

Circumference \( = \ 22 \) cm

\( => \ 2 \pi r \ = \ 22 \ => \ 2 \ × \ \frac{22}{7} \ × \ r \ = \ 22 \) \( => \ r \ = \ \frac{7}{2} \)cm

Area of the quadrant of a circle

\( = \ \frac{1}{4} \pi r^2 \ = \ ( \frac{1}{4} \ × \ \frac{22}{7} \ × \ \frac{49}{4}) \)

\(= \ \frac{539}{56} \ = \ \frac{77}{8} \ cm^2 \)

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm. Since the minutes hand rotates through 6° in one minute, therefore, area swept by the minute hand in one minute is the area of sector of angle 6° in a circle of radius 14 cm. Hence, the required area i.e., the area swept in 5 minutes.

\(= \ ( \frac{ \theta}{360} \ × \ \pi r^2 \ × \ 5) \ = \ \frac{6}{360} \ × \ \frac{22}{7} \ × \ (14)^2 \ × \ 5 \)

\(= \ \frac{1}{60} \ × \ \frac{22}{7} \ × \ 196 \ × \ 5 \ = \ \frac{154}{3} \ cm^2 \)

Q4. 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 \( \pi \ = \ 3.14 \))

(i) minor segment

(ii) major sector (Use \( \pi \ = \ 3.14 \))

Here, \( r \ = \ 10 \) cm, \( \theta \ = \ 90° \)

(i) Area of the minor segment

\( = \ r^2[\frac{ \pi \theta }{360} \ - \ \frac{1}{2} sin \theta] \)

\( = \ (10)^2[ \frac{3.14 \ × \ 90}{360} \ - \ \frac{1}{2} sin90°] \)

\(= \ 100(0.785 \ - \ 0.5) \ = \ 100 \ × \ 0.285 \ = \ 28.5 \ cm^2 \)

(ii) Area of the major sector

\(= \ \frac{ \theta }{360} \ × \ \pi r^2 \), where \(r \ = \ 10 \) and \( \theta \ = \ 270° \)

\(= \ \frac{270}{360} \ × \ 3.14 \ × \ 10^2) \) \( = \ (\frac{3}{4} \ × \ 3.14 \ × \ 100) \ = \ 235.5 \ cm^2 \)

Q5. In a circle of a radius 21 cm, an arc substends an angle of 60º 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.

(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.

Here \( r \ = \ 21 \)cm \( \theta \ = \ 60° \)

(i) Length of the arc, \(l \ = \ \frac{ \theta }{180} \ × \ \pi r \)

\( = \ \frac{60}{180} \ × \ \frac{22}{7}\ × \ 21 \ = \ 22 \)cm

(ii) Area of the sector, \( A \ = \ \frac{ \theta }{360} \ × \ \pi r^2 \)

\(= \ \frac{ \theta }{360} \ × \ \frac{22}{7} \ × \ 21 \ × \ 21 \ = \ 231 \) cm^{2}

(iii) Area of the segment

\( = \ r^2[ \frac{ \pi \theta }{360} \ - \ \frac{1}{2}sin \theta ] \)

\(= \ (21)^2[ \frac{22}{7} \ × \ \frac{60}{360} \ - \ \frac{1}{2}sin60°] \ = \ 441( \frac{11}{21} \ - \ \frac{1}{2} \ × \ \frac{ \sqrt{3}}{2}) \)

\( = \ 21 \ × \ 11 \ - \ \frac{441 \sqrt{3}}{4} \ = \ (231 \ - \ \frac{441 \sqrt{3}}{4}) \)cm^{2}

Q6. A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of the corresponding minor and major segment of the circle. (Use \( \pi \ = \ 3.14 \)and\( \sqrt{3} \ = \ 1.73 \))

Here \(r \ = \ 15 \) cm and \( \theta \ = \ 60° \)

∴ Area of the minor segment

\( = \ r^2[ \frac{ \pi \theta}{360} \ - \ \frac{1}{2}sin\theta] \)

\(= \ (15)^2[ \frac{3.14 \ × \ 60}{360} \ - \ \frac{1}{2} \ × \ sin60] \)

\(= \ 225[ \frac{3.14}{6} \ - \ \frac{1}{2} \ × \ \frac{ \sqrt{3}}{2}] \ = \ 225( \frac{3.14 \ × \ 2 \ - \ 3 \sqrt{3}}{12}) \)

\( = \ 225( \frac{6.28 \ - \ 3 \ × \ 1.73}{12} ) \ = \ \frac{225 \ × \ (6.28 \ - \ 5.19)}{12} \)

\( = \ \frac{225 \ × \ 1.09}{12} \ = \ 20.4375 \)cm^{2}

Area of the major segment

= Area of the circle - Area of the minor segment

\( = \ 3.14 \ × \ 225 \ - \ 20.4375 \ = \ 706.5 \ - \ 20.4375 \)

\( = \ 686.0625 \) cm^{2}

Q.7 A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle. (Use \( \pi \ = \ 3.14 \)and\( \sqrt{3} \ = \ 1.73 \))

Here\(r \ = \ 12 \) cm and \( \theta \ = \ 120° \)

Area of the corresponding segment of the circle

= Area of the minor segment

\( = \ r^2[ \frac{ \pi \theta}{360} \ - \ \frac{1}{2}sin\theta] \)

\(= \ (12)^2[ \frac{3.14 \ × \ 120}{360} \ - \ \frac{1}{2} \ × \ sin120] \)

\(= \ 144[ \frac{3.14}{3} \ - \ \frac{1}{2} \ × \ \frac{ \sqrt{3}}{2}] \ = \ 225( \frac{3.14 \ × \ 2 \ - \ 3 \sqrt{3}}{12}) \) [∵ \(sin120° \ = \ sin(180° \ - \ 60°) \ = \ sin60° \ = \ \frac{\sqrt{3}}{2} \)]

\( = \ 144 \ × \ \frac{3.14 \ × \ 4 \ - \ 3 \sqrt{3}}{12} \ = \ 12 \ × \ (12.56 \ - \ 3 \ × \ 1.73) \)

\(= \ 12 \ × \ (12.56 \ - \ 5.19) \ = \ 12 \ × \ 7.37 \ = \ 88.44 \)cm^{2}

Q8. 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 figure). 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 \( \pi \ = \ 3.14 \) )

(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 \( \pi \ = \ 3.14 \) )

(i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.

Then the area of the quadrant of this circle

\( = \ \frac{ \pi \ × \ 5^2}{4} \ = \ \frac{3.14 \ × \ 25}{4} \ = \ 19.625 \)m^{2}

(ii) In the 2nd case, radius = 10 m.

The area of the quadrant of this circle

\(= \ \frac{ \pi \ × \ 10^2}{4} \ = \ \frac{3.14 \ × \ 100}{4} \ = \ 78.5 \) m^{2}

∴ Increase in the grazing area

\( = \ 78.5 \ - \ 19.625 \ = \ 58.875 \)m^{2}

Q.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 sector as shown in figure. Find :

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

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

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

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

(i) Silver wire used to make a brooch

\( = \ 2 \pi r \),where \( r \ = \ \frac{35}{2} \)mm

\( = \ 2 \ × \ \frac{22}{7} \ × \ \frac{35}{2} \ = \ 110 \) mm

Wire used in 5 diameters \( = 5 \ × \ 35 \ = \ 175 \) mm

∴ Total wire used \(= (110 \ + \ 175) \ = \ 285 \) mm

(ii) The area of each sector of the brooch

\( = \ \frac{1}{10} \ × \) Area of the circle

\( = \ \frac{1}{10} \ × \ \pi r^2 \),where \(r \ = \ \frac{35}{2} \)mm

\(= \ \frac{1}{10} \ × \ \frac{22}{7} \ × \ ( \frac{35}{2} )^2 \ = \ \frac{385}{4} \) mm^{2}

Q.10 An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Area between two consecutive ribs

\( = \ \frac{1}{8} \ × \ r^2 \) ,where \( r \ = \ 45 \)cm

\( = \ \frac{1}{8} \ × \ \frac{22}{7} \ × \ (45)^2 \ = \ \frac{22275}{28} \) cm^{2}

Q.11 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades.

Here \(r \ = \ 25 \) cm, \( \theta \ = \ 115º \)

Total area cleaned at each sweep of the blades

\( = \ 2 \ × \) Area of the sector (having \( r \ = \ 25 \) and \( \theta \ = \ 115º \) )

\( = \ 2 \ × \ \frac{ \theta }{360} \ × \ \pi r^2 \)

\(= 2 \ × \ \frac{115}{360} \ × \ \frac{22}{7} \ × \ (25)^2 \)

\( = \ \frac{23 \ × \ 11 \ × \ 625}{18 \ × \ 7} \ = \ \frac{158125}{126} \) cm^{2}

Q.12 To warn ships for underwater rocks a lighthouse spreads a red coloured light over a sector of angle 80º to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use \( \pi \ = \ 3.14 \) )

Area of the sea over which the ships are warned

= Area of the sector (having \(r \ = \ 16.5 \) km and \(\theta \ = \ 80° \))

\( = \ \frac{ \theta }{360} \ × \ \pi r^2 \)

\( = \ \frac{80}{360}\ × \ 3.14 \ × \ (16.5)^2 \)

\( = \ \frac{68389.2}{360} \)

\(= \ 189.97 \) km^{2}

Q.13 A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm2.
(Use \( \sqrt{3} \ = \ 1.7 \))

Clearly from the figure,

Area of one desgin

= Area of the sector AOB - Area ( \(∆ \ AOB \) )

\( = \ \frac{ \theta }{360} \ × \ \pi r^2 \ - \ 12 \ × \ (r)^2 \ × \ sin60° \), where \(r \ = \ 28 \) cm

\( = \ \frac{60}{360} \ × \) \( \frac{22}{7} \ × \ (28)^2 \) \( - \ \frac{1}{2} \ × \ (28)^2 \ × \ \frac{ \sqrt{3}}{2} \)

\( = \ \frac{1232}{3} \ - \ 333.2 \)

\( = \ \frac{1232 \ - \ 999.6}{3} \)

\( = \ \frac{232.4}{3} \)cm^{2}

Area of 6 such design \( = \ 6 \ × \ \frac{232.4}{3} \ = \ 464.8 \)cm^{2}

Cost of making such designs @ Rs. 0.35 per cm^{2}

= Rs \(0.35 \ × \ 464.8 \)

= Rs 162.68

Q.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) \( \frac{p}{180} \ × \ 2\pi R \) (b) \( \frac{p}{180} \ × \ \pi R^2 \)

(c) \( \frac{p}{360} \ × \ 2\pi R \) (d) \( \frac{p}{720} \ × \ 2\pi R^2 \)

(a) \( \frac{p}{180} \ × \ 2\pi R \) (b) \( \frac{p}{180} \ × \ \pi R^2 \)

(c) \( \frac{p}{360} \ × \ 2\pi R \) (d) \( \frac{p}{720} \ × \ 2\pi R^2 \)

We know that area A of a sector of angle ? in a circle of radius r is given by

\( A \ = \ \frac{ \theta }{360} \ × \ \pi r^2 \)

But here , \( r \ = \ R \) and \( \theta \ = \ p \)

∴ \( A \ = \ \frac{p}{360} \ × \ \pi R^2 \ = \ \frac{p}{720} \ × \ 2\pi R^2 \)

∴ (D) is the correct answer.

Q.1 Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

∵ ROQ is a diameter, ∴ \( ∠ \ RPQ \ = \ 90° \)

In rt \( ∆ \ PRQ \) , \( RQ^2 \ = \ RP^2 \ + \ PQ^2 \)

\( => \ RQ^2 \ = \ 7^2 \ + \ 24^2 \ = \ 49 \ + \ 576 \ = \ 625 \)

\( => \ RQ \ = \ 25 \) cm

∴ Radius \( r \ = \ \frac{1}{2} RQ \ = \ \frac{25}{2} \)cm

Area of the semi circle \( = \ \frac{1}{2} \pi r^2 \ = \ \frac{1}{2} \ × \ \frac{22}{7} \ × \

( \frac{25}{2})^2 \ = \ \frac{6875}{28} \)

and area of \( ∆ \ RPQ \ = \ \frac{1}{2} \ × \ RP \ × \ PQ \)

\( = \ 12 \ × \ 7 \ × \ 24 \ = \ 84 \) cm^{2}

Area of the shaded region

= Area of the semi circle - Area ( \(∆ \ RPQ \) )

\( = \ \frac{6875}{28} \ - \ 84 \ = \ = \ \frac{4523}{28} \) cm^{2}

Q.2 Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \( ∠ \ AOC \ = \ 40° \)

Area of the shaded region

= Area of sector AOC - Area of sector OBD

\( = \ \frac{40}{360} \ × \ \pi \ × \ 142 \ - \ \frac{40}{360} \ × \ \pi \ × \ 7^2 \)

\( = \ \frac{1}{9} \ × \ \frac{22}{7}(196 \ - \ 49) \) \(= \ \frac{22}{63} \ × \ 147 \ = \ \frac{154}{3} \)cm^{2}

Q.3 Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Area of the square ABCD \(= \ (14)^2 \ = \ 196 \) cm^{2}

Diameter of the semicircles = AD or BC = 14 cm

∴ Radius of each semicircle \( = \ 7 \) cm

Area of the semicircular regions \( = \ 2 \ × \ \frac{1}{2} \pi \ r^2 \ = \ \pi r^2 \)

\( = \ \frac{22}{7} \ × \ 49 \ = \ 154 \)cm^{2}

∴ Area of the shaded portion

= Area of the square ABCD - Area of the semicircular regions

\(= \ 196 \ - \ 154 \ = \ 42 \)cm^{2}

Q.4 Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Area of the circular portion

= Area of the circle - Area of the sector

\(= \ \pi r^2 \ \ \frac{60}{360} \ \pi \ r^2 \ = \ \pi r^2(1 \ - \ \frac{1}{6}) \)

\(= \ \frac{5}{6} \pi r^2 \), where r = 6

\(= \ \frac{5}{6} \ × \ \frac{22}{7} \ × \ 36 \ = \ \frac{660}{7} \) cm^{2}

Area of the equilateral \( ∆ \ OAB \)

\( = \ \frac{ \sqrt{3}}{4}(side)^2 \ = \ ( \frac{ \sqrt{3}}{4} \ × \ 144) \)

\( = \ 36 \sqrt{3} \)cm^{2}

∴ Area of the shaded region

\(= \ ( \frac{660}{7} \ + \ 36 \sqrt{3}) \) cm^{2}

Q.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 figure. Find the area of the remaining portion of the square.

The area of the whole square ABCD \( = \ 4^2 \ = \ 16 \)cm^{2}

The sum of the area of the four quadrants at the four corners of the square

= The area of a circle of radius 1 cm

\(= \ \frac{22}{7} \ × \ 1^2 \ = \ \frac{22}{7} \)cm^{2}

The area of the circle of diameter 2 cm, i.e., radius 1cm

\( = \ \frac{22}{7} \ × \ 1^2 \ = \ \frac{22}{7} \) cm^{2}

∴ Area of the remaining portion

= The area of the square ABCD - The sum of the area of 4 quadrants at the four corners of the square - The area of the circle of diameter 2 cm

\(= \ 16 \ - \ \frac{22}{7} \ - \ \frac{22}{7} \ = \ \frac{112 \ - \ 22 \ - \ 22}{7} \)

\( = \ \frac{68}{7} \)cm^{2}

Q.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 figure. Find the area of the design (shaded region).

Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.

Area of the circle \( = \ \pi r^2 \)

\( = \ \frac{22}{7} \ × \ (32)^2 \ = \ \frac{22528}{7} \) cm^{2}

Area of \( ∆ \ ABC \) \(= \ 3 × \) Area of \(∆ \ BOC \)

\( = \ 3 \ × \ \frac{1}{2} \ × \ OB \ × \ OC \ × \ sinBOC \) \(= \ \frac{3}{2} \ × \ 32 \ × \ 32 \ × \ sin120° \)

\( = \ 3 \ × \ 16 \ × \ 32 \ × \ \frac{ \sqrt{3}}{2} \ = \ 768 \sqrt{3} \)cm^{2}

∴ Area of the design (i.e., shaded region)

= Area of the circle - Area of ∆ ABC

\(= \ ( \frac{22528}{7} \ - \ 768 \sqrt{3}\)) cm^{2}

Q.7 In figure , ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

The area of the whole square ABCD \( = \ 14^2 \ = \ 196 \)cm^{2}

The sum of the area of the four quadrants at the four corners of the square

= The area of a circle of radius 7cm

\(= \ \pi(7)^2 \ = \ \frac{22}{7} \ × \ 49 \ = \ 154 \) cm^{2}

Area of the shaded portion = The area of the square ABCD - The sum of the area of four quadrants at the four corners of the square

\(= \ 196 \ - \ 154 \ = \ 42 \) cm^{2}

Q.8 Figure depicts a racking track whose left and right ends are semicircular. The distance between the two inner parallel line segment 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.

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

(ii) The area of the track.

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

= BC + EH + 2 × circumference of the semicrircle of radius 30 m

\(= \ 106 \ + \ 106 \ + \ 2 \ × \ \frac{1}{2} \ × \ 2 \pi 30 \ = \ 212 \ + \ 2 \ × \ \frac{22}{7} \ × \ 30 \)

\(= \ 212 \ + \ \frac{1320}{7} \ = \ \frac{2804}{7} \)m

(ii) Area of the track

= Area of rectangle ABCD + Area of rectangle EFGH + 2 (Area of the semicircle of radius 40 m - Area of the semicircle with radius 30 m)

\(= \ (10 \ × \ 106) \ + \ (10 \ + \ 106) \ + \ 2[ \frac{1}{2} \ × \ \frac{22}{7} \ × \ (40)^2 \ - \ \frac{1}{2} \ × \ \frac{22}{7} \ × \ (30)^2] \)

\( = \ 1060 \ + \ 1060 \ + \ \frac{22}{7} (40^2 \ - \ 30^2) \ = \ 2120 \ + \ \frac{22}{7} \ × \ 70 \ × \ 10 \)

\(= \ 2120 \ + \ 2200 \ = \ 4320 \)m^{2}

Q.9 In figure AB and CD are two diameters of a circle (with centre 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.

Area of the sector \( = \ \frac{90}{360} \ × \ \frac{22}{7} \ × \ 7 \ × \ 7 \ =\ \frac{77}{2} \) cm^{2}

Area of \(∆ \ OCB \ = \ \frac{1}{2} \ × \ OC \ × \ OB \ = \ \frac{1}{2} \ ×\ 7 \ × \ 7 \

= \ \frac{49}{2} \)cm^{2}

∴ The area of the segment BPC

\(= \ \frac{77}{2} \ - \ \frac{49}{2} \ = \ \frac{28}{2} \ = \ 14 \)cm^{2}

Similarly the area of the segment AQC = 14 cm^{2}

Also, the area of the circle with DO as diameter

\(= \ \frac{22}{7} \ × \ \frac{7}{2} \ × \ \frac{7}{2} \ = \ \frac{77}{2} \) cm^{2}

∴ The total area of the shaded region

\(= \ 14 \ + \ 14 \ + \ \frac{77}{2} \)

\(= \ \frac{28 \ + \ 28 \ + \ 77}{2} \ = \ \frac{133}{2} \ = \ 66.5 \) cm^{2}

Q.10 The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use \( \pi \ = \ 3.14 \) and \( \sqrt{3} \ = \ 1.73205 \))

Let each side of the triangle be a cm. Then,

Area = 17320.5 cm^{2}

\( => \ \frac{ \sqrt{3}}{4}a^2 \ = \ 17320.5 \) [∵ Area \(= \ \frac{ \sqrt{3}}{4}(side)^2 \)]

\( => \ a^2 \ = \ \frac{17320.5 \ × \ 4}{ \sqrt{3}} \) \( = \ 40000 \ a \ = \ 200 \)

∴ Radius of each circle is 100 cm.

Now, required area

= Area of ∆ ABC - 3 ×(Area of a sector of angle 60º in a circle of 100 cm)

\( = \ 17320.5 \ - \ 3( \frac{60}{360} \ × \ 3.14 \ × \ 100 \ × \ 100) \)

\( = \ 17320.5 \ - \ 15700 \ = \ 1620.5 \) cm^{2}

Q.11 On a square hand kerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Side of the square ABCD = AB

= 3 × diameter of circular design

\( = \ 3 \ × \ (2 \ × \ 7) \ = \ 42 \) cm

∴ Area of the square ABCD

\( = \ 42 \ × \ 42 \ = \ 1764 \) cm^{2}

Area of one circular design

\( = \ \pi r^2 \ = \ \frac{22}{7} \ × \ 7 \ × \ 7 \ = \ 154 \) cm^{2}

∴ Area of 9 such designs

\( = \ 9 \ × \ 154 \ = \ 1386 \) cm^{2}

∴ Area of the remaining portion of the handkerchief

= Area of the square ABCD - Area of 9 circular designs

\(= \ 1764 \ - \ 1386 \ = \ 378 \) cm^{2}

Q.12 In figure OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm , find the area of the

(i) quadrant OACB,

(ii) Shaded region,

(i) quadrant OACB,

(ii) Shaded region,

(i) Area of quadrant \( = \ \frac{1}{4} \ \pi r^2 \)

\(= \ \frac{1}{4} \ × \ \frac{22}{7} \ × \ (3.5)^2 \)

\( = \ \frac{1}{4} \ × \ \frac{22}{7} \ × \ ( \frac{7}{2} )^2 \ = \ \frac{77}{8} \) cm^{2}

(ii) Area of ∆ AOD = \( \frac{1}{2} \) Base × Height

\( = \ \frac{1}{2} \ (OB \ × \ OD) \)

\( = \ \frac{1}{2} (3.5 \ × \ 2) \ = \ \frac{7}{2} \) cm^{2}

So, area of the shaded region

= Area of quadrant - Area of AOD

\(= \ \frac{77}{8} \ \ \frac{7}{2} \ = \ \frac{49}{8} \) cm^{2}

Q.13 In figure, a square OABC is inscribed in a quadrant OPBQ . If OA = 20 cm, find the area of the shaded region. (Use \( \pi \ = \ 3.14 \))

Radius of the quadrant \( = \ OB = \sqrt{OA^2 \ + \ AB^2} \ = \sqrt{20^2 \ + \ 20^2} \ = \ 20 \sqrt{2} \) cm

∴ Area of quadrant OPBQ

\(= \ \frac{1}{4} \ \pi r^2 \ = \ \frac{1}{4} \ × \ 3.14 \ × \ (20 \sqrt{2})^2 \ = \frac{1}{4} \ × \ 3.14 \ × \ 800 \ = \ 628 \) cm^{2}

Area of the square OABC

\(= \ (20)^2 \ = \ 400 \) cm^{2}

So, area of the shaded region

= Area of quadrant - Area of square OABC

\( = \ 628 \ - \ 400 \ = \ 228 \) cm^{2}

Q.14 AB and CD are respectively across of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If \( ∠ \ AOB \ = \ 30º \), find the area of the shaded region.

Let \( A_1 \) and \( A_2 \) be the areas of sectors OAB and OCD respectively. Then,

\( A_1 \) = Area of a sector of angle 30º in a circle of radius 21 cm.

\( = \ \frac{30}{360} \ × \ \frac{22}{7} \ × \ 21 \ × \ 21 \) [Using \(A \ = \ \frac{ \theta }{360} \ × \ \pi r^2 \)]

\( = \frac{231}{2} \) cm^{2}

\( A_2 \) = Area of a sector of angle 30º in a circle of radius 7 cm

\( = \ \frac{30}{360} \ × \ \frac{22}{7} \ × \ 7 \ × \ 7 \)

\(= \ \frac{77}{6} \) cm^{2}

So, area of the shaded region

\( = \ A_1 \ - \ A_2 \ = \ \frac{231}{2} \ - \ \frac{77}{6} \ = \ \frac{69}{3} \ - \ \frac{77}{6} \)

\(= \ \frac{308}{3} \) cm^{2}

Q.15 In figure , ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Let 14 cm be the radius of the quadrant with A as the centre.

Then the area of the quadrant ABMC

\(= \ \frac{1}{4} \ \pi r^2 \ = \ \frac{1}{4} \ × \ \frac{22}{7} \ × \ 196 \)

\(= \ 154 \)cm^{2}

Area of ∆ BAC

\( = \ \frac{1}{2} \ × \ AC \ × \ AB \)

\( = \ \frac{1}{2} \ × \ 14 \ × \ 14 \ = \ 98 \) cm^{2}

∴ Area of the segment of the circle, BMC

= Area of the quadrant ABMC - Area of ∆ BAC

\(= \ 154 \ - \ 98 \ = \ 56 \) cm^{2}

Now, since AC = AB = 14 cm and \( ∠ \ BAC \ = \ 90 \deg \)

∴ By Pyrthagoras Theorem,

\( BC \ = \ \sqrt{AC^2 \ + \ AB^2} \ = \ \sqrt{14^2 \ + \ 14^2} \ = \ 14 sqrt{2} \)

∴ Radius of the semicircle BNC \( = \ 7\sqrt{2} \)cm

Area of the semicircle BNC \( = \ \frac{ \pi}{2} (7 \sqrt{2})^2 \)

\(= \ \frac{1}{2} \ × \ \frac{22}{7} \ × \ 98 \ = \ 154 \) cm^{2}

∴ The area of the region between two arcs BMC and BNC

= The area of the shaded region

= The area of semicircle BNC - The area of the segment of the circle BMC

\(= \ 154 \ - \ 56 \ = \ 98 \) cm^{2}

Q.16 Calculate the area of the the designed region in figure common between the two quadrants of circle of radius 8 cm each.

Here, 8 cm is the radius of the quadrants ABMD and BNDC.

Sum of their areas

\( = \ 2 \ × \ \frac{1}{4} \pi r^2 \ = \ \frac{1}{2} \pi r^2 \)

\( = \frac{1}{2} \ × \ \frac{22}{7} \ × \ 64 \ = \ \frac{704}{7} \) cm^{2}

Area of the square ABCD

\( = \ 8 \ × \ 8 \ = \ 64 \) cm^{2}

Area of the designed region

= Area of the shaded region = Sum of the area of quadrants - Area of the square ABCD

\( = \frac{704}{7} \ - \ 64 \ = \ \frac{704 \ - \ 448}{7} \)

\( = \ \frac{256}{7} \) cm^{2}