# Question-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 circumferences of the two circles.**

**Solution:**

Let r

_{1}and r

_{2}be the radii of the 2 circles then, r

_{1}= 19 cm and r

_{2}= 9 cm

Circumference of 1

^{st}circle = 2Ï€ r

_{1}

= 2 Ã— 19 Ã— Ï€ = 38Ï€

Circumference of 2

^{nd}circle = 2Ï€ r

_{2}

= 2 Ã— 9 Ã— Ï€ = 18Ï€

Sum of circumference of 2 circle = 38Ï€ + 18Ï€ = 56Ï€ â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

Given that the have a circle whose circumference is equal to the sum given by (1)

Let R be the radii of the circle âˆ´ 2Ï€ R = 56Ï€ (given) â‡’ R = 28 cm.

# Question-2

**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 areas of the two circles.**

**Solution:**

# Question-3

**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 each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.**

**Solution:**

Let r

_{1}, r

_{2}, r

_{3}, r

_{4}, r

_{5}be the radii of the 5 scoring areas namely Gold, Red, Blue, Black, white

r

_{1}= = 10.5 cm

r

_{2}= 10.5 cm + 10.5 = 21 cm

r

_{3}= 21 + 10.5 = 31.5 cm

r

_{4}= 31.5 + 10.5 = 42.0 cm

r

_{5}= 42 + 10.5 = 52.5 cm

Area of Gold circle = Ï€ Ã— (10.5)

^{2}

= 346.5 cm

^{2}â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

Area of circle is Ï€ r

_{2}

^{2}with radii = r

_{2}

= Ï€ Ã— (21)

^{2 }= 441 Ã— = 1386 cm

^{2}

Area of red scoring area = Area with radius r

_{2}= 1386 cm

^{2}â€¦â€¦â€¦â€¦. (2)

Hence area of red scoring area = Area given by (2) - Area given by (1)

= 1386 - 346.5

= 1039.5 cm

^{2}

Area of circle with radii r

_{3}= Ï€ r

_{3}

^{2}

=

= 3118.5 sq. cm â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

âˆ´ Area of blue scoring area = Area given by (3) - Area given by (2)

= 3118.5 - 1386

= 1732.5 sq.cm.

Area of circle with radii r

_{4}Ï€ r

^{2}

_{4}= Ï€ Ã— (42)

^{2}

= 5544 sq cm â€¦â€¦â€¦â€¦â€¦â€¦. (4) âˆ´ Area of black scoring area = Area given by (4) - Area given by (3)

= 5544 - 3118.5

= 2425.5 sq.cm

Area of the circle with radii r

_{5}= Ï€ Ã— (52.5)

^{2}

=

= 8662.5 sq.cm â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (5)

âˆ´ Area of white scoring region = Area of circle given by (5) - Area of circle given by (4)

= 8662.5 - 5544

= 3118.5 sq.cm

âˆ´ The areas given by the required region

Gold = 346. 5 cm

^{2}

Red = 1039.5 cm

^{2}

Blue = 1732.5 cm

^{2}

Black = 2425.5 cm

^{2}

White = 3118.5 cm

^{2}.

# 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?**

**Solution:**

Radius of the wheel of the car = = 40 cm

Distance travelled with one revolution = Its circumference

= 2Ï€ r

= 2 Ã— = cm

Let the no. of revolution made by each wheel be n âˆ´ Total distance travelled = Distance travelled with one revolution of the wheel Ã— No. of revolution

= n cm

Time taken = 10 min = = hr

Speed of the car =

= (âˆµ 1 cm = 1/ 100000 Km)

=

66 = (Given speed = 66 km/hr)

n =

= 4375

Hence no. of revolutions n = 4375.

# Question-5

**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) Ï€ units

(C) 4 units

(D) 7 units.

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units

(B) Ï€ units

(C) 4 units

(D) 7 units.

**Solution:**

Perimeter of a circle = 2Ï€ r

Area of a circle = Ï€ r

^{2}

Given 2Ï€ r = Ï€ r

^{2}

Hence radius = 2 units

Answer is (A).

# Question-6

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

**Solution:**

r = 6 cm âˆ Q = 60Â°

Area of the sector of the circle =

=

=

= 18.8 cm^{2}

# Question-7

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

**Solution:**

Circumference of circle is 22 cm â‡’ 2Ï€ r = 22 (If r = radius of circle) â‡’ 2 Ã— = 22

r = 7/2 cm

Area of a quadrant of a circle

=

=

= .

# Question-8

**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 the minute hand = 14 cm

Radius the circle covered = length of minute hand

âˆ´ r = 14 cm.

The minute hand makes an angle of 360Â° in 60 minutes.

so, angle made by it in 5 mins = 30Â°

Area =

where Î¸ is the angle of the sector

Area covered =

= 51.33 cm

^{2}

# Question-9

**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 Ï€ = 3.14)

(i) minor segment

(ii) major sector. (Use Ï€ = 3.14)

**Solution:**

(i) Let AB be the chord of the circle with centre O

âˆ´ Radius AO = OB = 10 cm

Area of minor sector = Ã— Ï€ r

^{2}

=

=

= 78.571 cm

^{2}

In Î”OAB, Sin 90Â° =

âˆ´ AB = 10 cm, Area of Î”OAB = Ã—10 Ã—10 = 50 cm

^{2}

Area of minor segment = 78.5 - 50 = 28.5 cm

^{2}

(ii) For the major sector Î¸ = (360Â° - 90Â°) = 270Â°

Area of major sector =

=

= 78.571 Ã— 3

= 235.713 cm

^{2}.

# Question-10

**In a circle of radius 21 cm, an arc subtends an angle of 60Â° 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

**Solution:**

r = 21 cm Î¸ = 60Â°

(i) length of the arc = =

= 22 cm

(ii) Area of the sector formed by the arc

=

=

= 231 cm

^{2}

(iii) Area of segment formed by corresponding chord.

= Area of sector â€“ Area of triangle

= 231 â€“ Area of Î” OAB â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

Let OM be the âŠ¥ r bisector to the chord AB.

â‡’ âˆ AOM = âˆ BOM (âˆµ AM = BM)

Let OM = x cm

In Î” OMA, = cos 30Â°

= (cos 30Â° = )

x = 18.2 cm

In Î” OMA, = sin 30Â° = (sin 30Â° = )

AM = 21 Â´ = 11.5 cm

AB = 2 AM

= 2 Ã— 11.5 = 21 cm

In Î” OAB, area of Î” OAB =

= Â´ 18.2 Â´ 21

= 191 cm^{2}

from (1), Area of segment formed by corresponding chord (231 -191 ) = 40 cm^{2}.

# Question-11

**A chord of a circle of radius 15 cm subtends an angle of 60Â° at the centre. Find the areas of the corresponding minor and major segments of the circle.**

(Use Ï€ = 3.14 and âˆš3 = 1.73)

(Use Ï€ = 3.14 and âˆš3 = 1.73)

**Solution:**

Area of sector formed by the arc =

=

= = 117.82 cm^{2}

In Î” OAB, let OM be âŠ¥ r bisector of AB

âˆ AOM = âˆ BOM

Let OM be = x cm

In Î” OMA, = cos 30Â°

= (cos 30Â° = )

x = 15 cm=

=

OM = 12.975 cm

In Î” OMA

sin 30Â° =

=

AM = 15/2 cm

AB = 2AM = 15 cm

Area of Î” OAB = OM Ã— AB

= Ã— 12.975 Ã— 15

=

= 97.3125 Sq.cm

Area of minor segment = Area of sector formed by the arc â€“ Area of Î” OAB

= 117.82 - 97.31

= 20.51 sq.cm

Area of major segment = Ï€ r^{2} â€“ 20.51

= 3.14 Ã— 15 Ã— 15

= 706.5 - 20.51

= 686.0 sq.cm.

# Question-12

**A chord of a circle of radius 12 cm subtends an angle of 120Â° at the centre. Find the area of the corresponding segment of the circle. (Use Ï€ = 3.14 and âˆš3 = 1.73)**

**Solution:**

OA = OB = r = 12 cm Î¸ = 120Â°

Area of sector formed by the are =

=

= 150.72 sq.cm.

In Î” OAB âˆ AOB = 120Â°

Let OM = x, be âŠ¥ r bisector to chord AB

In Î” OMA,

= cos 60Â°

OM = OA Â´

OM = 12 Ã— = 6 cm

sin 60Â° = â‡’ AM = 12 Ã— = 6 = 6 Ã— 1.73

AM = 10.38 cm

AB = 2AM = 20.76 cm

Area of Î” OAB = Ã— OM Ã— AB

= Â´ 6 Â´ 20.76

= 62.28 sq. cm

Area of segment of the circle

150.72 â€“ 62.28 = 88.44 sq.cm.

# Question-13

**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

(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

**p**

**= 3.14)**

**Solution:**

(i) Area of the part that the horse can graze = Ã— p r

^{2}

=

= 19.642 sq.m.

(ii) If the rope were 10 m Long instead of 5 m Area =

=

= 3.14 Ã— 25

= 78.57 sq.m

âˆ´ Increase in grazing area = 78.570 - 19.642

= 58.92 sq.m.

# Question-14

**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 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.

**Solution:**

Diameter = 35 mm

The wire is used to form the circumference of the circle + 5 diameter of the circle.

Circumference of circle = 2Ï€ r

= 2 Ã—

= 110 mm

The total length of silver wire required

= 110 + 5 Ã— 35

= 110 + 175

= 285 mm

The brooch is divided into 10 equal sectors.

But angle at the centre = 360Â°

Area made by each sector =

= 36Â°

Area of each sector =

=

= = 96.25 sq.m

# Question-15

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

**Solution:**

The umbrella is divided into 8 ribs

If we consider one sector the angle Î¸ at the centre = = 45Â°

Area between two consecutive ribs = Area of sector =

=

=

= 1591 cm

^{2}.

# Question-16

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

**Solution:**

The Area of the region wiped by both the wipers of the car which do not over lap each other

= 2(Area of the region wiped by one wiper)

Sector angle Î¸ = 115Â°. Radius of the sector = Length of the wiper = 25 cm

Area of the region wiped by both the wiper

= 2 = 1254.96 cm

^{2}.

# Question-17

**To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80Â° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use p = 3.14).**

**Solution:**

Î¸ = 80Â°

r = 16.5 km âˆ´ Area of the sector of the circle =

=

=

= 189.97 sq. km.

Area of the sea over which the ships are warned = Area of the sector of the circle.

# Question-18

**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 cm**

^{2}. (Use**= 1.7)**

**Solution:**

R = 28 cm

The no. of sides = 6

Hence the angle Î¸

Made by an segment of the circle = = 60Â°

Area of the sector of angle Î¸ =

=

=

= 410.29 sq. cm.

__To find the area of Î” OAB__

Set O be the circle

OA = OB = 28 cm

Let OM be âŠ¥ r to AB then In Î” OAM,

sin 30^{0} =

âˆ´ AM = OA sin 30Â°

= 28 Ã— 1/2 = 14 cm

AB = 28 cm

OM = OA cos 30^{0} = 28 Ã—

= 24.248 cm

Hence, Area of Î” OAB =

=

= 339.472 sq.cm.

Area of segment made by AB

410.29

__339.47__

__ 70.82__

= 70.82 Ã— 6

= 424.92 sq.cm.

Cost of making the designs = 424.92 Ã— 0.35

= Rs.148.72

Similarly the area of segment made by 6 chords.

# Question-19

**Tick the correct answer in the following:**

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

(A)

(B)

(C)

(d)

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

(A)

(B)

(C)

(d)

**Solution:**

**Ans.**

=

=

**------**Answer: d

# Question-20

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

**Solution:**

In the figure, PQ = 24 cm, PR = 7 cm âˆ QPR = 90Â° (Angle in a semi circle)

Thus, QR = =

= =

= 25 cm

Radius of the circle (OR) = = 12.5 cm

Area of semicircle =

= = 245.54 cm

^{2}

Area of triangle = =

= = 84 cm

^{2}

Area of shaded region = Area of semicircle - Area of triangle

= 245.54 â€“ 84

= 161.54 cm

^{2}.

# Question-21

**Find the area of the shaded region in Fig. if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and**

âˆ AOC = 40Â°.

âˆ AOC = 40Â°.

**Solution:**

style=""> Radius of bigger circle = 14 cm

Radius of smaller circle = 7 cm

Area of bigger sector =Ã—

= Ã— = 68.44 cm

^{2}

Area of smaller sector = Ã—

**=**Ã— = 17.11 cm

^{2 }

Area of shaded portion = Area of bigger sector - Area of smaller sector

= 68.44 - 17.11 = 51.33 cm

^{2}.

# Question-22

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

**Solution:**

To find the area of the shaded region,

It can be found out by subtracting the two semicircle area from the area of the square.

Area of the square = a

^{2}= 14 Ã— 14 = 196 cm

^{2}

[Diameter of the semicircle = Side of the square = 14 m

and hence the radius of the semicircle = = 7 cm]

Area of a semicircle = = = 77 cm

^{2 }â€¦â€¦â€¦.(1)

As the diameter of both the semicircle is the side of the same square, the combined area of both the semicircle

= 2 Ã— Area of a each semicircle

= 2 Ã— 77 = 154 cm

^{2}(from 1)

Area of the shaded region

196 â€“ 154 = 42 cm

^{2}.

# Question-23

**Find the area of the shaded region in Fig., 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.**

**Solution:**

In the above figure what students should absence is that there is a common area in the form of a sector of a circle shown in the figure.

Therefore in order to get the shaded area, we have to remove the common area in the combined areas of the area of the equilateral and the area of the circle.

Area of the sector =

=

= 6Ï€ = 6 Ã—

= = 18.86 cm

^{2}â€¦â€¦â€¦â€¦â€¦â€¦.(1)

Combined area of the circle + Area of the equilateral triangle

= Ï€ r

^{2}+( ) [The relevant formula for area of a circle]

=

= 113.14 + 62.352

= 175.50 cm

^{2}â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Area of the shaded region

= 175.50 - 18.86 [(2) - (1)]

= 156.64 cm

^{2}.

# Question-24

**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. Find the area of the remaining portion of the square.**

**Solution:**

The area of the square excluding the four sectors and the area of the circle of the centre of square can be obtained by subtracting the combined area of the four sectors and the central and for the area of the square).

Area of 1 sector =

=

= = cm

^{2}

âˆ´ Area of the 4 sectors = 0.785 Ã— 4 = 3.14 cm

^{2}

Area of the circle at the centre = Ï€ Ã— 1 Ã— 1 = = 3.14 cm

^{2}

âˆ´ Combined area of the four sectors and area of the circle

= 3.14 + 3.14 = 6.28 cm

^{2}

Area of the square = a

^{2}= 4

^{2}= 16 cm

^{2}âˆ´ Shaded region area = 16 - 6.28 = 9.72 cm

^{2}.

# Question-25

**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.**

**Solution:**

Side of a square ABCD = 14 cm. âˆ´ Area of square ABCD = 14 Ã— 14 = 196 cm

^{2}

Radius of each circle = = 7 cm

Area of four quadrants at the four corners = Area of circle

= Ï€ r

^{2}

= = 154 cm

^{2}

Required area = Area of square ABCD â€“ 4( of area of each circle)

= Area of square â€“ Area of one circle

= 196 â€“ 154 = 42 cm

^{2.}

# Question-26

**Figure 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.

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.

**Solution:**

(i) The distance around the track along its inner edge = Perimeter of the inner track

= 106 + 106 + 2 Ã— Ã— 30

= 400.60 m.

(ii) Area of the two rectangles = 2 Ã— 106 Ã— 10

= 2120

Area of the 2 semicircular region

= Ï€ (R

^{2}â€“ r

^{2}) = Ï€ (R + r) (R â€“ r)

=

=

= 2200

Total area of the track = 2120 + 2200 = 4320 m

^{2}

# Question-27

**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.**

**Solution:**

Consider semicircle BCA. By subtracting the area of the triangle ABC from this semicircular area, we will be able to get the combined segmented area =

Area of the bigger semicircular with radius 7m â€“ area of the Î” ABC.

=

= 77 â€“ 49 = 28 cm

^{2}

Total Area of the shaded Region = 28 + 38.5 = 66.5 cm

^{2}.

# Question-28

**The area of an equilateral triangle ABC is 17320.5 cm**

^{2}. 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 Fig. 12.28). Find the area of the shaded region.**(Use Ï€**

**= 3.14 and**

**= 1.73205).**

**Solution:**

Area of equilateral ABC with = , where a is the side of a triangle

â‡’ 17320.5 =

â‡’ a

^{2}=

â‡’ a

^{2 }= 40000 cm.

â‡’ a = 200

â‡’ Radius of each circle = = 100 cm

Area of sector =Ã—

= Ã—

= 5233.33 cm

^{2}

Area of three sectors formed = 5233.33 Ã— 3 = 15699.99 cm

^{2}

Area of shaded portion = Area of triangle â€“ Area of three sectors

= 17320.5 - 15699.99

= 1620.51 cm

^{2}.

# Question-29

**On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.**

**Solution:**

Area of remaining portion of the handkerchief = Area of the square kerchief - 9(Area of circle)

Radius of a circle = 7 cm

Length of a side of square = 3(diameter of the circle) = 3 Ã— 14 = 42 cm

Area of a square = Side Ã— Side = (42)

^{2}= 1764

Area of one circle = r

^{2}= (7)

^{ 2}= 154

âˆ´ Area of remaining portion of the handkerchief = 1764 â€“ 9(154)

= 1764 â€“ 1386

= 378 cm

^{2}.

# Question-30

**In Fig., 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.

**Solution:**

(i) Area of the Quadrant OACB =

= = 9.625 cm

^{2}

(ii) Area of OBD =

= = = 3.5 cm

^{2}

Hence the area of shaded portion = Area of the Quadrant - Area of OBD

= 9.625 - 3.5

= 6.125 cm^{2}.

# Question-31

**In Fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use Ï€**

**= 3.14)**

**Solution:**

Area of square = Side Ã— Side = 20 Ã— 20

= 400 cm

^{2}

Diagonal of square = Radius of the quadrant âˆ´ Diagonal of square =

= 20cm

Thus radius of a circle = 20

Area of the Quadrant OPBQ =

=

= 628.58 cm

^{2}

Hence the area of shaded portion = Area of the Quadrant - Area of square

= 628.58 - 400 = 228.58 cm

^{2}.

# Question-32

**AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If**

**âˆ AOB = 30Â°**

**, find the area of the shaded region.**

**Solution:**

Radius of bigger sector = 21 cm

Radius of smaller sector = 7 cm

Area of bigger sector =Ã—

= = 115.5 cm

^{2}

Area of smaller sector =Ã—

**=**= 12.83 cm

^{2}

Area of shaded portion = Area of bigger sector - Area of smaller sector

= 115.5 - 12.83 = 102.67 cm

^{2}.

# Question-33

**In Fig., 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.**

**Solution:**

In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircular is drawn with BC as diameter.

The area of the above, shaded region can be got by first finding the area of the semicircle with BC of diameter and then subtract the segment of the sector BAC.

ABC is a right angled Î”, by applying Pythagoras theorem

BC =

=

=

= 14 cm

Radius = = 7

Area of the semicircle with BC as diameter

=

=

= 154 cm

^{2}..............(i)

Area of the segment of sector BAC = Area of quadrant of a circle - Area of ABC

= -

= 154 â€“ 98 = 56 cm

^{2}...........(ii) âˆ´ Area of the shaded sector = Area of the semi circle with BC as diameter - Area of the segment of sector BAC

= (i) â€“ (ii)

= 154 â€“ 56 = 98 cm

^{2}.

# Question-34

**Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each.**

**Solution:**

Side of a square = 8 cm

Area of a square = Side Ã— Side = 8 Ã— 8 = 64 cm

^{2}

If we consider a diagonal, two sectors are formed

Area of two sectors = 2 Ã— = 2 Ã— Ã—

= cm

^{2}

Thus area of design = Area of both sectors â€“ Area of square

= - 64

= cm

^{2}= 36.57 cm

^{2}

Hence the area of design = 36.57 cm

^{2}.