##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 13.2**- An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at…
- The radius of a circular ground is 63 m. There is 7 m wide road inside the…
- A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10…
- The length of a minute hand of a circular dial is 10 cm. Find the area of the…
- The radius of a field in the form of a sector is 21 m. The cost of constructing…
- The length of a side of a square field is 20 m. A cow is tied at the corner by…
- A chord of a circle of radius 42 cm subtends an angle of measure 60 at the…
- A chord of a circle, of length 10 cm, subtends a right angle at the centre.…

**Exercise 13.3**- A rectangle whose length and breadth are 12 cm and 5 cm respectively is…
- ABCD, square park, has each side of length 80 m. There is a flower bed at each…
- What will be the cost of covering the white portion in figure 13.21 with a…
- ABCD is a square plate of 1 m length. As shown in figure circles are drawn with…
- bar {oa} and bar {ob} are two mutually perpendicular radii of a circle of…

**Exercise 13**- The area of a circular park is 616 m2. There is a 3.5 m wide track around the…
- A man is cycling in such a way that the wheels of the cycle are making 140…
- If a chord bar {ab} of (O, 20) subtends right angle at O, find the area of…
- There are two arcs {apb} of 6 (O, OA) and {aqb} of 6 (M, MA) as shown…
- The length of a side of a square garden ABCD is 70 m. A minor segment of 6…
- ABCD is a square of side 20 cm. Find the area of blue coloured region formed by…
- On a circular table top of radius 30 cm a design is formed leaving an…
- Figure 13.27 shows a kite formed by a square PQRS and an isosceles right…
- In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area…
- A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find…
- If an arc of a circle subtends an angle of measure 0 at the centre, then the…
- The area of a sector is given by the formula ….(r is the radius and 1 is the…
- bar {oa} and bar {ob} are the two mutually perpendicular radii of a circle…
- A sector subtends an angle of measure 120 at the centre of a circle having…
- If the area and the circumference of a circle are numerically equal, then r =…
- The length of an arc subtending an angle of measure 60 at the centre of a…
- The area of a minor sector of (O, 15) is 150. The length of the corresponding…
- If the radius of a circle is increased by 10 %, then corresponding increase…
- If the ratio of the area of two circles is 1: 4, then the ratio of their…
- The area of the largest triangle inscribed in a semi-circle of radius 8 is ………
- If the circumference of a circle is 44 then the length of a side of a square…
- The length of minute hand of a clock is 14 cm. If the minute hand moves from…
- The length of minute hand of a clock is 15 cm. If the minute hand moves for…

**Exercise 13.2**

- An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at…
- The radius of a circular ground is 63 m. There is 7 m wide road inside the…
- A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10…
- The length of a minute hand of a circular dial is 10 cm. Find the area of the…
- The radius of a field in the form of a sector is 21 m. The cost of constructing…
- The length of a side of a square field is 20 m. A cow is tied at the corner by…
- A chord of a circle of radius 42 cm subtends an angle of measure 60 at the…
- A chord of a circle, of length 10 cm, subtends a right angle at the centre.…

**Exercise 13.3**

- A rectangle whose length and breadth are 12 cm and 5 cm respectively is…
- ABCD, square park, has each side of length 80 m. There is a flower bed at each…
- What will be the cost of covering the white portion in figure 13.21 with a…
- ABCD is a square plate of 1 m length. As shown in figure circles are drawn with…
- bar {oa} and bar {ob} are two mutually perpendicular radii of a circle of…

**Exercise 13**

- The area of a circular park is 616 m2. There is a 3.5 m wide track around the…
- A man is cycling in such a way that the wheels of the cycle are making 140…
- If a chord bar {ab} of (O, 20) subtends right angle at O, find the area of…
- There are two arcs {apb} of 6 (O, OA) and {aqb} of 6 (M, MA) as shown…
- The length of a side of a square garden ABCD is 70 m. A minor segment of 6…
- ABCD is a square of side 20 cm. Find the area of blue coloured region formed by…
- On a circular table top of radius 30 cm a design is formed leaving an…
- Figure 13.27 shows a kite formed by a square PQRS and an isosceles right…
- In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area…
- A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find…
- If an arc of a circle subtends an angle of measure 0 at the centre, then the…
- The area of a sector is given by the formula ….(r is the radius and 1 is the…
- bar {oa} and bar {ob} are the two mutually perpendicular radii of a circle…
- A sector subtends an angle of measure 120 at the centre of a circle having…
- If the area and the circumference of a circle are numerically equal, then r =…
- The length of an arc subtending an angle of measure 60 at the centre of a…
- The area of a minor sector of (O, 15) is 150. The length of the corresponding…
- If the radius of a circle is increased by 10 %, then corresponding increase…
- If the ratio of the area of two circles is 1: 4, then the ratio of their…
- The area of the largest triangle inscribed in a semi-circle of radius 8 is ………
- If the circumference of a circle is 44 then the length of a side of a square…
- The length of minute hand of a clock is 14 cm. If the minute hand moves from…
- The length of minute hand of a clock is 15 cm. If the minute hand moves for…

###### Exercise 13.2

**Question 1.**An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at the centre. Find the length of the arc and area of the sector.

**Answer:**Given radius of the circle (r)= 21 cm

Measure of angle (θ)= 120^{o}

Length of the arc =

=

= 44 cm

Area of the sector formed by the arc =

=

= 462 cm^{2}

**Question 2.**The radius of a circular ground is 63 m. There is 7 m wide road inside the ground as shown in figure 13.10. The blue coloured portion of the road, shown in figure 13.10 is to be repaired. If the rate of repair work of the road costs Rs. 25 per m2, find the total cost of repair.

**Answer:**

Given the rate of repair work per m^{2} = Rs. 25

The radius of the outer circle (R_{1}) = 63 cm

radius of the inner circle (R_{2}) = 63 – 7

= 56 cm

(∵ the distance between the inner circle and outer circle is 7 cm)

Let the measured angle is 60^{o}

area of the repair road = () – ()

= () – (

= 436.15 m^{2}

cost required to repair the road = 436.15 × 25

= Rs. 10,903.9445 (approximately)

**Question 3.**A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10 cm as shown in the figure 13.11. Find the area of the remaining part of the sheet. ( = 1.73) (= 3.14)

**Answer:**Given radius of the circle (r) = 10 cm

Side of the hexagon (a) = 10 cm

we can divide the hexagon into 6 equilateral triangles as follows.

the side of an equilateral triangle will be 10 cm

we have to find the area of the shaded region as shown in fig. 13.11

area of the shaded region = area of circle - area of hexagon

= area of circle - area of 6 right equilateral triangles

= π r^{2} – 6 ()

= 3.14 × 10^{2} – 6 ( × 10^{2})

= 54.5 cm^{2}

**Question 4.**The length of a minute hand of a circular dial is 10 cm. Find the area of the sector formed by the present position and the position after five minute of the minute hand. (π = 3.14)

**Answer:**length of the minute hand (r) = 10 cm

(In clocks the length of minute hand will be radius of that clock dial)

The sector formed by the minute hand after 5 minutes will be as follows.

θ = = 30^{o}

Area of the sector formed by minute hand =

=

= 26.166 cm^{2}

**Question 5.**The radius of a field in the form of a sector is 21 m. The cost of constructing a wall around the field is Rs. 1875 at the rate of Rs. 25 per meter. If it costs Rs. 10 per m2 to till the field, what will be the cost of tilling the whole field?

**Answer:**Radius of the sector (r) = 21 m

cost for constructing a meter wall = Rs. 25

cost of constructing a wall around the field = Rs. 1875

∴ Total length of wall = = 75 m

perimeter of the sector = 75 m

l + r + r = 75 m

l + 2(21) = 75 m

l = 33 m

∴ Arc length of CDB (L) = 33m

Area of the field =

=

= 346.5 m^{2}

Cost of tiling 1 m^{2} = Rs. 10

∴ Cost of tiling 346.45 m^{2} = 10 × 346.5

= Rs. 3465

**Question 6.**The length of a side of a square field is 20 m. A cow is tied at the corner by means of a 6 m long rope. Find the area of the field which the cow can graze. Also find the increase in the grazing area, if length of the rope is increased by 2 m. (π = 3.14)

**Answer:**length of side of a square field (a) = 20 m

length of the rope (r_{1}) = 6 m

since it is square the center angle (θ) = 90^{o}

area of the gazing field =

=

= 28.26 m^{2}

If the length of rope is increased by 2m

The radius of the gazing field (r_{2}) = 6m + 2m = 8m

area of the gazing field =

=

= 50.24 m^{2}

increase in gazing field = 50.24 – 28.26

= 27.98 m^{2}

**Question 7.**A chord of a circle of radius 42 cm subtends an angle of measure 60 at the centre. Find the area of the minor segment of the circle. ( = 1.73)

**Answer:**radius of the circle (r) = 42 cm

angle of measure (θ) = 60^{o}

area of the segment =

=

= 923.16 cm^{2}

since the center angle of measure is 60^{o} that is a equilateral triangle.

side of the triangle (a) = 42 cm

area of the equilateral triangle =

=

= 762.93 cm^{2}

area of minor segment (BDC) = area of segment – area of triangle

= 923.16 – 762.93

= 160.23 cm^{2}

**Question 8.**A chord of a circle, of length 10 cm, subtends a right angle at the centre. Find the areas of the minor segment and the major segment formed by the chord. (π = 3.14)

**Answer:**Length of the chord = 10 cm

Angle of measure = 90^{o}

since triangle ABC is a right-angled triangle

AB, AC is same as the radius of the circle

BC^{2} = AB^{2} + AC^{2}

10^{2} = r^{2} + r^{2}

100 = 2 r^{2}

50 = r^{2}

r = √50

Area of segment =

=

= 39.25 cm^{2}

Area of the triangle = AB× AC

= × (√50) × (√50)

= 25 cm^{2}

Area of minor segment = Area of segment - Area of the triangle

= 39.25 – 25

= 14.25 cm^{2}

Area of major segment = Area of circle - Area of minor segment

= π r^{2} – 14.25

= 3.14 × (√50)^{2} – 14.25

= 142.82 cm^{2}

**Question 1.**

An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at the centre. Find the length of the arc and area of the sector.

**Answer:**

Given radius of the circle (r)= 21 cm

Measure of angle (θ)= 120^{o}

Length of the arc =

=

= 44 cm

Area of the sector formed by the arc =

=

= 462 cm^{2}

**Question 2.**

The radius of a circular ground is 63 m. There is 7 m wide road inside the ground as shown in figure 13.10. The blue coloured portion of the road, shown in figure 13.10 is to be repaired. If the rate of repair work of the road costs Rs. 25 per m2, find the total cost of repair.

**Answer:**

Given the rate of repair work per m^{2} = Rs. 25

The radius of the outer circle (R_{1}) = 63 cm

radius of the inner circle (R_{2}) = 63 – 7

= 56 cm

(∵ the distance between the inner circle and outer circle is 7 cm)

Let the measured angle is 60^{o}

area of the repair road = () – ()

= () – (

= 436.15 m^{2}

cost required to repair the road = 436.15 × 25

= Rs. 10,903.9445 (approximately)

**Question 3.**

A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10 cm as shown in the figure 13.11. Find the area of the remaining part of the sheet. ( = 1.73) (= 3.14)

**Answer:**

Given radius of the circle (r) = 10 cm

Side of the hexagon (a) = 10 cm

we can divide the hexagon into 6 equilateral triangles as follows.

the side of an equilateral triangle will be 10 cm

we have to find the area of the shaded region as shown in fig. 13.11

area of the shaded region = area of circle - area of hexagon

= area of circle - area of 6 right equilateral triangles

= π r^{2} – 6 ()

= 3.14 × 10^{2} – 6 ( × 10^{2})

= 54.5 cm^{2}

**Question 4.**

The length of a minute hand of a circular dial is 10 cm. Find the area of the sector formed by the present position and the position after five minute of the minute hand. (π = 3.14)

**Answer:**

length of the minute hand (r) = 10 cm

(In clocks the length of minute hand will be radius of that clock dial)

The sector formed by the minute hand after 5 minutes will be as follows.

θ = = 30^{o}

Area of the sector formed by minute hand =

=

= 26.166 cm^{2}

**Question 5.**

The radius of a field in the form of a sector is 21 m. The cost of constructing a wall around the field is Rs. 1875 at the rate of Rs. 25 per meter. If it costs Rs. 10 per m2 to till the field, what will be the cost of tilling the whole field?

**Answer:**

Radius of the sector (r) = 21 m

cost for constructing a meter wall = Rs. 25

cost of constructing a wall around the field = Rs. 1875

∴ Total length of wall = = 75 m

perimeter of the sector = 75 m

l + r + r = 75 m

l + 2(21) = 75 m

l = 33 m

∴ Arc length of CDB (L) = 33m

Area of the field =

=

= 346.5 m^{2}

Cost of tiling 1 m^{2} = Rs. 10

∴ Cost of tiling 346.45 m^{2} = 10 × 346.5

= Rs. 3465

**Question 6.**

The length of a side of a square field is 20 m. A cow is tied at the corner by means of a 6 m long rope. Find the area of the field which the cow can graze. Also find the increase in the grazing area, if length of the rope is increased by 2 m. (π = 3.14)

**Answer:**

length of side of a square field (a) = 20 m

length of the rope (r_{1}) = 6 m

since it is square the center angle (θ) = 90^{o}

area of the gazing field =

=

= 28.26 m^{2}

If the length of rope is increased by 2m

The radius of the gazing field (r_{2}) = 6m + 2m = 8m

area of the gazing field =

=

= 50.24 m^{2}

increase in gazing field = 50.24 – 28.26

= 27.98 m^{2}

**Question 7.**

A chord of a circle of radius 42 cm subtends an angle of measure 60 at the centre. Find the area of the minor segment of the circle. ( = 1.73)

**Answer:**

radius of the circle (r) = 42 cm

angle of measure (θ) = 60^{o}

area of the segment =

=

= 923.16 cm^{2}

since the center angle of measure is 60^{o} that is a equilateral triangle.

side of the triangle (a) = 42 cm

area of the equilateral triangle =

=

= 762.93 cm^{2}

area of minor segment (BDC) = area of segment – area of triangle

= 923.16 – 762.93

= 160.23 cm^{2}

**Question 8.**

A chord of a circle, of length 10 cm, subtends a right angle at the centre. Find the areas of the minor segment and the major segment formed by the chord. (π = 3.14)

**Answer:**

Length of the chord = 10 cm

Angle of measure = 90^{o}

since triangle ABC is a right-angled triangle

AB, AC is same as the radius of the circle

BC^{2} = AB^{2} + AC^{2}

10^{2} = r^{2} + r^{2}

100 = 2 r^{2}

50 = r^{2}

r = √50

Area of segment =

=

= 39.25 cm^{2}

Area of the triangle = AB× AC

= × (√50) × (√50)

= 25 cm^{2}

Area of minor segment = Area of segment - Area of the triangle

= 39.25 – 25

= 14.25 cm^{2}

Area of major segment = Area of circle - Area of minor segment

= π r^{2} – 14.25

= 3.14 × (√50)^{2} – 14.25

= 142.82 cm^{2}

###### Exercise 13.3

**Question 1.**A rectangle whose length and breadth are 12 cm and 5 cm respectively is inscribed in a circle. Find the area of the blue coloured region, as shown in the figure 13.19.

Figure 13.19

**Answer:**given length (l) = 12 cm

breadth (b) = 5 cm

we can notice from the fig. that the Hypotenuse of ADB is same as diameter of the circle.

from the triangle ABD

BD^{2} = AD^{2} + AB^{2}

BD = √(5^{2} + 12^{2})

(∵ AD is breadth, AB is length of the rectangle)

BD = √(25 + 144)

= √169

= 13 cm

∴ diameter of the circle = 13 cm

Radius of the circle (r) = 6.5 cm

Area of the shade region = Area of circle – area of rectangle

= π r^{2} – b × h

= π × 6.5^{2} – 5 × 12

= 72.732 cm^{2}

**Question 2.**ABCD, square park, has each side of length 80 m. There is a flower bed at each corner in the form of a sector of radius 7 m, as shown in figure 13.20. Find the area of the remaining part of the park.

Figure 13.20

**Answer:**side of a square park (a) = 80 cm

radius of the sectors (r) = 7 cm

we can notice that all the sectors can form a circle of radius 7 cm

so, the area of remaining part of park =

area of square – area of circle

= a^{2} - π × r^{2}

= 80^{2} - π × 7^{2}

= 6,240.06 cm^{2}

**Question 3.**What will be the cost of covering the white portion in figure 13.21 with a silver foil if the rate is Rs. 100 per m^{2}? (π = 3.14)

Figure 13.21

**Answer:**Given side of a square (a) = 12 cm

Radius of each sector (r) = 12 cm

since it is square the angle (θ) = 90^{o}

let divide the square into 2 parts. (BAPC, DAQC)

Area of part – 1

The area of the sector BAPC =

= × 12^{2} × 90

= 113.04 cm^{2}

Area of square = a^{2}

= 12^{2}

= 144 cm^{2}

Area of part 1 (APC) = Area of square - The area of the sector BAPC

= 144 – 113.04

= 30.96 cm^{2}

∵ The shapes are identical

The area of part 2 (AQC) = 30.96 cm^{2}

Area of the colored region = area of PART 1+ area of part 2

= 30.96 + 30.96

= 61.92 cm^{2}

Cost for covering 1 m^{2} silver foil = 100

Cost for covering 61.92 m^{2} silver foil = 61.92 × 100

= 6192

**Question 4.**ABCD is a square plate of 1 m length. As shown in figure circles are drawn with their center at A, B, C, D respectively, each with radius equal to 42 cm. The blue coloured part at each corner, as shown in the C figure 13.22 is cut. What is the area of the remaining portion of the plate?

Figure 13.22

**Answer:**Side of square plate (a) = 1 m = 100cm

radius of the segment (r) = 42 cm

we can notice that all the sectors can form a circle of radius 42 cm

the area of the remaining portion of the plate =

Area of square plate - area of the circle

= a^{2} - π r^{2}

= 100^{2} - π × 42^{2}

= 4,458.23 cm^{2}

=0.4458 m^{2}

**Question 5.** and are two mutually perpendicular radii of a circle of radius 10.5 cm. D and OD = 6 cm. Find the area of blue coloured region shown in figure 13.23.

Figure 13.23

**Answer:**given OA = 10.5 cm = radius (r)

OD = 6 CM

we can see that ODA is a right-angled triangle

area of triangle = OA× OD

= 10.5 × 6

= 31.5 cm^{2}

we can see that segment OACB is of the circle.

so, the area of OACB = × r^{2}

= × 10.5^{2}

= 86.54625 cm^{2}

area of blue colored region = area of OACB - area of triangle

= 86.54625 – 31.5

= 55.04625 cm^{2}

**Question 1.**

A rectangle whose length and breadth are 12 cm and 5 cm respectively is inscribed in a circle. Find the area of the blue coloured region, as shown in the figure 13.19.

Figure 13.19

**Answer:**

given length (l) = 12 cm

breadth (b) = 5 cm

we can notice from the fig. that the Hypotenuse of ADB is same as diameter of the circle.

from the triangle ABD

BD^{2} = AD^{2} + AB^{2}

BD = √(5^{2} + 12^{2})

(∵ AD is breadth, AB is length of the rectangle)

BD = √(25 + 144)

= √169

= 13 cm

∴ diameter of the circle = 13 cm

Radius of the circle (r) = 6.5 cm

Area of the shade region = Area of circle – area of rectangle

= π r^{2} – b × h

= π × 6.5^{2} – 5 × 12

= 72.732 cm^{2}

**Question 2.**

ABCD, square park, has each side of length 80 m. There is a flower bed at each corner in the form of a sector of radius 7 m, as shown in figure 13.20. Find the area of the remaining part of the park.

Figure 13.20

**Answer:**

side of a square park (a) = 80 cm

radius of the sectors (r) = 7 cm

we can notice that all the sectors can form a circle of radius 7 cm

so, the area of remaining part of park =

area of square – area of circle

= a^{2} - π × r^{2}

= 80^{2} - π × 7^{2}

= 6,240.06 cm^{2}

**Question 3.**

What will be the cost of covering the white portion in figure 13.21 with a silver foil if the rate is Rs. 100 per m^{2}? (π = 3.14)

Figure 13.21

**Answer:**

Given side of a square (a) = 12 cm

Radius of each sector (r) = 12 cm

since it is square the angle (θ) = 90^{o}

let divide the square into 2 parts. (BAPC, DAQC)

Area of part – 1

The area of the sector BAPC =

= × 12^{2} × 90

= 113.04 cm^{2}

Area of square = a^{2}

= 12^{2}

= 144 cm^{2}

Area of part 1 (APC) = Area of square - The area of the sector BAPC

= 144 – 113.04

= 30.96 cm^{2}

∵ The shapes are identical

The area of part 2 (AQC) = 30.96 cm^{2}

Area of the colored region = area of PART 1+ area of part 2

= 30.96 + 30.96

= 61.92 cm^{2}

Cost for covering 1 m^{2} silver foil = 100

Cost for covering 61.92 m^{2} silver foil = 61.92 × 100

= 6192

**Question 4.**

ABCD is a square plate of 1 m length. As shown in figure circles are drawn with their center at A, B, C, D respectively, each with radius equal to 42 cm. The blue coloured part at each corner, as shown in the C figure 13.22 is cut. What is the area of the remaining portion of the plate?

Figure 13.22

**Answer:**

Side of square plate (a) = 1 m = 100cm

radius of the segment (r) = 42 cm

we can notice that all the sectors can form a circle of radius 42 cm

the area of the remaining portion of the plate =

Area of square plate - area of the circle

= a^{2} - π r^{2}

= 100^{2} - π × 42^{2}

= 4,458.23 cm^{2}

=0.4458 m^{2}

**Question 5.**

and are two mutually perpendicular radii of a circle of radius 10.5 cm. D and OD = 6 cm. Find the area of blue coloured region shown in figure 13.23.

Figure 13.23

**Answer:**

given OA = 10.5 cm = radius (r)

OD = 6 CM

we can see that ODA is a right-angled triangle

area of triangle = OA× OD

= 10.5 × 6

= 31.5 cm^{2}

we can see that segment OACB is of the circle.

so, the area of OACB = × r^{2}

= × 10.5^{2}

= 86.54625 cm^{2}

area of blue colored region = area of OACB - area of triangle

= 86.54625 – 31.5

= 55.04625 cm^{2}

###### Exercise 13

**Question 1.**The area of a circular park is 616 m^{2}. There is a 3.5 m wide track around the park running parallel to the boundary. Calculate the cost of fencing on the outer circle at the rate of Rs. 5 per meter.

**Answer:**Area of the circular park = 616 m^{2}

width of the track = 3.5 m

Area of the circular park = 616 m^{2}

π × r^{2} = 616

r^{2} = 196

r = √196

r = 14 m

∴ the radius of the circular park (r_{1}) = 14 m

the radius of the park with track (r_{2}) = 14 + 3.5

= 17.5 m

Circumference of the track = 2π (r_{2})

= 2× (17.5^{)}

= 110 m

Cost of fencing per meter = Rs. 5

Cost of fencing 110 m = 110 × 5

= 550

**Question 2.**A man is cycling in such a way that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, then how much distance will he cover in 2 hours?

**Answer:**Speed of the cycle = 140RPM

Diameter of the wheel = 60 cm

No. of revolution in 2 hours = 120 × 140

= 16,800

Distance covered in 2 hours = Revolutions × Diameter

= 16,800 × 60

= 1008000 cm

= 10080 m

= 10.8 km

**Question 3.**If a chord of (O, 20) subtends right angle at O, find the area of the minor segment.

**Answer:**Given radius = 20cm

if we connect the chord to the center point that forms a segment.

the centre measure angle (θ) =90^{o}

we can observe a right-angled triangle OAB

Area of segment =

=

= 314 cm^{2}

Area of the triangle = OA× OB

= × (20) × (20)

= 200 cm^{2}

Area of minor segment = Area of segment - Area of the triangle

= 314 – 200

= 114 cm^{2}

**Question 4.**There are two arcs of (O, OA) and of (M, MA) as shown in figure 13.24. Find the area enclosed by two arcs.

Figure 13.24

**Answer:**from the figure we can notice all these

1. OAB is an equilateral triangle

(∵ All the sides are equal)

2. OAPB is a minor segment with Radius 14 cm, the angle (θ) between them is 60^{o}

3. ABQ is a semi-circle with centre as M radius 7 cm

Area of the segment =

=

= 102.62 CM^{2}

Area of the triangle = × OA^{2}

= × 14^{2}

= 84.870 CM^{2}

Area of the semi-circle =

Area of the segment - Area of the triangle

= 102.62 - 84.870

= 17.75 cm^{2}

Area of the semi-circle = × r^{2}

= × 7^{2}

= 76.96 cm^{2}

area enclosed by two arcs =

Area of the semi-circle - Area of the semi-circle

= 76.96 – 17.75

= 59.20 cm^{2} (approximately)

**Question 5.**The length of a side of a square garden ABCD is 70 m. A minor segment of (O, OA) is drawn on each of two opposite sides for developing lawn, as shown in figure 13.25. Find the area of the lawn.

figure 13.25

**Answer:**side of a square garden ABCD (a) = 70 m

From the fig. in the below we can see that

1. OAXD is a segment

2. OAD is an right-angled triangle.

3. AXD is a minor segment

OA, OD are the radii of the segment

from the triangle OAD

AD^{2} = OA^{2} + OD^{2}

70^{2} = r^{2} + r^{2}

2r^{2} = 4900

r^{2} = 2450

r = √2450

Area of minor segment = Area of segment - Area of triangle

= - × OA^{2}

= - ×

= 1924.22 – 1225

=699.22 cm^{2}

= 700 cm^{2}

since AXD, BYC are same

∴ Area of lawn = 2 × area of minor segment

= 2 × 700

= 1400 cm^{2}

**Question 6.**ABCD is a square of side 20 cm. Find the area of blue coloured region formed by the semi-circles drawn on each side as shown in figure 13.26. ( = 3.14)

figure 13.26.

**Answer:**side of a square (a) = 20 cm

we can see the semi-circle in the square having 10 cm as radius (r)

area of the colored region =

area of 4 semi-circles - area of the square

= 4 × ( × r^{2}) - a^{2}

= 2 (π × 10^{2}) - 20^{2}

= 228 cm^{2}

**Question 7.**On a circular table top of radius 30 cm a design is formed leaving an equilateral triangle inscribed in a circle. Find the area of the design. (π = 3.14)

**Answer:**radius of the circle = 30 cm

Area of the circle = π r^{2}

= 3.14 × 30^{2}

= 2826 cm^{2}

PQR is an equilateral triangle

∠ QPR = 60^{O}

∠ QOR = 2 × ∠ QPR = 120^{o}

In OQR triangle,

let OS is perpendicular to QR

In triangle OQR as OQ = OR

∠ QOS = ∠ QPR = 60^{O}

In triangle OSQ ∠S = 90^{o}

sin 60^{o} =

=

QS = 15 √3 cm

Now,

QR = 2 QS = 2 (15 √3) = 30√3 cm

Area of equilateral triangle = × QR^{2}

= × (30√3)^{2}

= 1167.75 CM^{2}

Area of design = Area of circle - area of triangle PQR

= 2826 – 1167.75

= 1658.25 CM^{2}

**Question 8.**Figure 13.27 shows a kite formed by a square PQRS and an isosceles right triangle ARB whose congruent sides are 5 cm long. is an arc of a (R, 42). Find the area of the blue coloured region.

Figure 13.27

**Answer:**Side of square (r) = 42 cm

In triangle RAB,

RA = RB = 5 cm

In minor sector RQCS angle is 90^{o}

Area of the shaded region = Area of minor sector RQCS + Area of RAB

= + × RA × RB

= + × 5 × 5

= 1384.74 + 12.5

= 1372.24 cm^{2}

**Question 9.**In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area of the blue coloured region. (π = 3.14)

Figure 13.28

**Answer:**side of square (a) = 8 cm

from the fig. we can see that of circle is placed at four parts of the square. if we join them a circle is formed with radius (r_{1}) 2 cm

Another circle is placed at centre of the square with radius (r_{2}) 2 cm

Area of colored region = Area of square - Area of circles at 4 corners + Area of circle at centre

= a^{2} – (4 × ( × r_{1}^{2}) + (π r_{2}^{2}))

= a^{2} – (π r_{1}^{2} + π r_{2}^{2})

= 8^{2} – (3.14 (2^{2} + 2^{2}))

= 38.88 cm^{2}

**Question 10.**A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find the area of the blue coloured region shown in figure 13.29. (= 3.14)

Figure 13.29

**Answer:**from the given figure

PQR is a right-angle triangle

PQ = 8 cm

QR = 6 cm

PR^{2} = PQ^{2} + QR^{2}

PR^{2} = 8^{2} + 6^{2}

PR = √100 = 10 cm

Radius of the incircle (r) = = = = 2 CM

area of coloured region = Area of triangle - area of circle

= × PQ × QR - π r^{2}

= × 8 × 6 - π 2^{2}

= 24 – 12.566

= 11.433 cm^{2}

**Question 11.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If an arc of a circle subtends an angle of measure 0 at the centre, then the area of the minor sector is…..

A.

B.

C.

D.

**Answer:**we know that area of minor sector =

So, option(d) is the correct answer.

**Question 12.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a sector is given by the formula ….

(r is the radius and 1 is the length of an arc.)

A. rl

B. r^{2}/ l

C. rl

D. rl

**Answer:**we know that when arc length (l) and radius (r) of sector is given

Area of the sector is = rl

So, option(a) is the correct answer.

**Question 13.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

and are the two mutually perpendicular radii of a circle having radius 9 cm. The area of the minor sector corresponding to AOB is cm^{2}( = 3.14)

A. 63.575

B. 63.585

C. 63.595

D. 63.60

**Answer:**Given radius (r)= 9 cm

∵ They are perpendicular to each other θ = 90^{o}

area of the minor sector =

=

=

= 63.585 cm^{2}

So, option(b) is the correct answer.

**Question 14.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

A sector subtends an angle of measure 120 at the centre of a circle having radius of 21 cm. The area of the sector is ……….. cm^{2}.

A. 462

B. 460

C. 465

D. 470

**Answer:**Given radius (r)= 21 cm

angle(θ) = 120^{o}

area of the minor sector =

=

=

= 461.58 cm^{2}

= 462 cm^{2}

So, option(a) is the correct answer.

**Question 15.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the area and the circumference of a circle are numerically equal, then r = …………

A. π

B.

C. 1

D. 2

**Answer:**Area of the circle = πr^{2}

circumference of the circle = 2πr

given, Area = Circumference

2πr = πr^{2}

2 = r

So, option(d) is the correct answer.

**Question 16.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of an arc subtending an angle of measure 60 at the centre of a circle whose area is 616 is ……….

A.

B. 66

C.

D. 33

**Answer:**Area of circle = 616cm^{2}

πr^{2} = 616

r^{2} = 196

r = 14 cm

Given θ = 60^{o}

area of the minor sector =

=

=

= cm^{2}

So, option(c) is the correct answer.

**Question 17.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a minor sector of (O, 15) is 150. The length of the corresponding arc is (π = 3.14)

A. 30

B. 20

C. 90

D. 15

**Answer:**Radius of the minor sector= 15 cm

Area of minor sector = 150

Area of a minor sector = r l

150 = × 15 × l

l = 20 cm

So, option(b) is the correct answer.

**Question 18.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the radius of a circle is increased by 10 %, then corresponding increase in the area of the circle is ………. (π = 3.14)

A. 19 %

B. 10 %

C. 21 %

D. 20 %

**Answer:**r be the radius of the circle

Area of the circle = πr^{2}

10% increase in radius = 1.1r

Area of circle with new radius = π(1.1r)^{2}

= 1.21πr^{2}

increase in area = 1.21πr^{2} - πr^{2}

= 0.21 πr^{2}

∴ 21% increase in the area of circle

So, option(c) is the correct answer.

**Question 19.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the ratio of the area of two circles is 1: 4, then the ratio of their circumference …..

A. 1: 4

B. 1: 2

C. 4: 1

D. 2: 1

**Answer:**let r_{1}, r_{2} are the radii of two circles

given,

πr_{1}^{2}: πr_{2}^{2} = 1 : 4

r_{1}^{2}: r_{2}^{2} = 1 : 4

r_{1} : r_{2} =1 : 2

r_{2} = 2 r_{1}

circumference of 1^{st} circle = 2πr_{1}

circumference of 2^{nd} circle = 2πr_{2} = 2π(2r_{1}) = 4πr_{1}

ratio of circumference = 2πr_{1} : 4πr_{1} = 1 : 2

So, option(b) is the correct answer.

**Question 20.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of the largest triangle inscribed in a semi-circle of radius 8 is ……

A. 8

B. 16

C. 64

D. 256

**Answer:**Given radius of semi-circle = 8 cm

For a largest triangle in semi-circle

base will be its diameter of semi-circle

height will be radius of semi-circle

so, its base = 2 × radius = 2 × 8 = 16 cm

height = 8 cm

Area of the triangle = × base × height

= × 16 × 8

= 64 cm^{2}

So, option(c) is the correct answer.

**Question 21.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the circumference of a circle is 44 then the length of a side of a square inscribed in the circle is ….

A.

B.

C.

D.

**Answer:**Given circumference = 44 cm

2πr = 44

πr = 22

r = 7 cm

Given that square inscribed in a circle.

so, diagonal of square will be equal to diameter of circle.

diagonal = 2 × radius = 2 × 7 = 14 cm

diagonal = √2 × side of square

14 = √2 × side

side = 7√2 cm

**Question 22.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 14 cm. If the minute hand moves from 2 to 11 on the circular dial, then area covered by it is cm^{2}.

A. 154

B. 308

C. 462

D. 616

**Answer:**The angle moved by minute hand = 90^{o}

length of minute hand (r) = 14 cm

Area of covered by minute hand =

=

=

= 153.93 cm^{2}

= 154 cm^{2}

So, option(a) is the correct answer.

**Question 23.**Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 15 cm. If the minute hand moves for 20 minutes on a circular dial of a clock, the area covered by it is …….. cm^{2}. (= 3.14)

A. 235.5

B. 471

C. 141.3

D. 706.5

**Answer:**The angle moved by minute hand = 120^{o}

length of minute hand (r) = 15 cm

Area of covered by minute hand =

=

=

= 235.619 cm^{2}

= 235.5 cm^{2}

So, option(a) is the correct answer.

**Question 1.**

The area of a circular park is 616 m^{2}. There is a 3.5 m wide track around the park running parallel to the boundary. Calculate the cost of fencing on the outer circle at the rate of Rs. 5 per meter.

**Answer:**

Area of the circular park = 616 m^{2}

width of the track = 3.5 m

Area of the circular park = 616 m^{2}

π × r^{2} = 616

r^{2} = 196

r = √196

r = 14 m

∴ the radius of the circular park (r_{1}) = 14 m

the radius of the park with track (r_{2}) = 14 + 3.5

= 17.5 m

Circumference of the track = 2π (r_{2})

= 2× (17.5^{)}

= 110 m

Cost of fencing per meter = Rs. 5

Cost of fencing 110 m = 110 × 5

= 550

**Question 2.**

A man is cycling in such a way that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, then how much distance will he cover in 2 hours?

**Answer:**

Speed of the cycle = 140RPM

Diameter of the wheel = 60 cm

No. of revolution in 2 hours = 120 × 140

= 16,800

Distance covered in 2 hours = Revolutions × Diameter

= 16,800 × 60

= 1008000 cm

= 10080 m

= 10.8 km

**Question 3.**

If a chord of (O, 20) subtends right angle at O, find the area of the minor segment.

**Answer:**

Given radius = 20cm

if we connect the chord to the center point that forms a segment.

the centre measure angle (θ) =90^{o}

we can observe a right-angled triangle OAB

Area of segment =

=

= 314 cm^{2}

Area of the triangle = OA× OB

= × (20) × (20)

= 200 cm^{2}

Area of minor segment = Area of segment - Area of the triangle

= 314 – 200

= 114 cm^{2}

**Question 4.**

There are two arcs of (O, OA) and of (M, MA) as shown in figure 13.24. Find the area enclosed by two arcs.

Figure 13.24

**Answer:**

from the figure we can notice all these

1. OAB is an equilateral triangle

(∵ All the sides are equal)

2. OAPB is a minor segment with Radius 14 cm, the angle (θ) between them is 60^{o}

3. ABQ is a semi-circle with centre as M radius 7 cm

Area of the segment =

=

= 102.62 CM^{2}

Area of the triangle = × OA^{2}

= × 14^{2}

= 84.870 CM^{2}

Area of the semi-circle =

Area of the segment - Area of the triangle

= 102.62 - 84.870

= 17.75 cm^{2}

Area of the semi-circle = × r^{2}

= × 7^{2}

= 76.96 cm^{2}

area enclosed by two arcs =

Area of the semi-circle - Area of the semi-circle

= 76.96 – 17.75

= 59.20 cm^{2} (approximately)

**Question 5.**

The length of a side of a square garden ABCD is 70 m. A minor segment of (O, OA) is drawn on each of two opposite sides for developing lawn, as shown in figure 13.25. Find the area of the lawn.

figure 13.25

**Answer:**

side of a square garden ABCD (a) = 70 m

From the fig. in the below we can see that

1. OAXD is a segment

2. OAD is an right-angled triangle.

3. AXD is a minor segment

OA, OD are the radii of the segment

from the triangle OAD

AD^{2} = OA^{2} + OD^{2}

70^{2} = r^{2} + r^{2}

2r^{2} = 4900

r^{2} = 2450

r = √2450

Area of minor segment = Area of segment - Area of triangle

= - × OA^{2}

= - ×

= 1924.22 – 1225

=699.22 cm^{2}

= 700 cm^{2}

since AXD, BYC are same

∴ Area of lawn = 2 × area of minor segment

= 2 × 700

= 1400 cm^{2}

**Question 6.**

ABCD is a square of side 20 cm. Find the area of blue coloured region formed by the semi-circles drawn on each side as shown in figure 13.26. ( = 3.14)

figure 13.26.

**Answer:**

side of a square (a) = 20 cm

we can see the semi-circle in the square having 10 cm as radius (r)

area of the colored region =

area of 4 semi-circles - area of the square

= 4 × ( × r^{2}) - a^{2}

= 2 (π × 10^{2}) - 20^{2}

= 228 cm^{2}

**Question 7.**

On a circular table top of radius 30 cm a design is formed leaving an equilateral triangle inscribed in a circle. Find the area of the design. (π = 3.14)

**Answer:**

radius of the circle = 30 cm

Area of the circle = π r^{2}

= 3.14 × 30^{2}

= 2826 cm^{2}

PQR is an equilateral triangle

∠ QPR = 60^{O}

∠ QOR = 2 × ∠ QPR = 120^{o}

In OQR triangle,

let OS is perpendicular to QR

In triangle OQR as OQ = OR

∠ QOS = ∠ QPR = 60^{O}

In triangle OSQ ∠S = 90^{o}

sin 60^{o} =

=

QS = 15 √3 cm

Now,

QR = 2 QS = 2 (15 √3) = 30√3 cm

Area of equilateral triangle = × QR^{2}

= × (30√3)^{2}

= 1167.75 CM^{2}

Area of design = Area of circle - area of triangle PQR

= 2826 – 1167.75

= 1658.25 CM^{2}

**Question 8.**

Figure 13.27 shows a kite formed by a square PQRS and an isosceles right triangle ARB whose congruent sides are 5 cm long. is an arc of a (R, 42). Find the area of the blue coloured region.

Figure 13.27

**Answer:**

Side of square (r) = 42 cm

In triangle RAB,

RA = RB = 5 cm

In minor sector RQCS angle is 90^{o}

Area of the shaded region = Area of minor sector RQCS + Area of RAB

= + × RA × RB

= + × 5 × 5

= 1384.74 + 12.5

= 1372.24 cm^{2}

**Question 9.**

In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area of the blue coloured region. (π = 3.14)

Figure 13.28

**Answer:**

side of square (a) = 8 cm

from the fig. we can see that of circle is placed at four parts of the square. if we join them a circle is formed with radius (r_{1}) 2 cm

Another circle is placed at centre of the square with radius (r_{2}) 2 cm

Area of colored region = Area of square - Area of circles at 4 corners + Area of circle at centre

= a^{2} – (4 × ( × r_{1}^{2}) + (π r_{2}^{2}))

= a^{2} – (π r_{1}^{2} + π r_{2}^{2})

= 8^{2} – (3.14 (2^{2} + 2^{2}))

= 38.88 cm^{2}

**Question 10.**

A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find the area of the blue coloured region shown in figure 13.29. (= 3.14)

Figure 13.29

**Answer:**

from the given figure

PQR is a right-angle triangle

PQ = 8 cm

QR = 6 cm

PR^{2} = PQ^{2} + QR^{2}

PR^{2} = 8^{2} + 6^{2}

PR = √100 = 10 cm

Radius of the incircle (r) = = = = 2 CM

area of coloured region = Area of triangle - area of circle

= × PQ × QR - π r^{2}

= × 8 × 6 - π 2^{2}

= 24 – 12.566

= 11.433 cm^{2}

**Question 11.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If an arc of a circle subtends an angle of measure 0 at the centre, then the area of the minor sector is…..

A.

B.

C.

D.

**Answer:**

we know that area of minor sector =

So, option(d) is the correct answer.

**Question 12.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a sector is given by the formula ….

(r is the radius and 1 is the length of an arc.)

A. rl

B. r^{2}/ l

C. rl

D. rl

**Answer:**

we know that when arc length (l) and radius (r) of sector is given

Area of the sector is = rl

So, option(a) is the correct answer.

**Question 13.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

and are the two mutually perpendicular radii of a circle having radius 9 cm. The area of the minor sector corresponding to AOB is cm^{2}( = 3.14)

A. 63.575

B. 63.585

C. 63.595

D. 63.60

**Answer:**

Given radius (r)= 9 cm

∵ They are perpendicular to each other θ = 90^{o}

area of the minor sector =

=

=

= 63.585 cm^{2}

So, option(b) is the correct answer.

**Question 14.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

A sector subtends an angle of measure 120 at the centre of a circle having radius of 21 cm. The area of the sector is ……….. cm^{2}.

A. 462

B. 460

C. 465

D. 470

**Answer:**

Given radius (r)= 21 cm

angle(θ) = 120^{o}

area of the minor sector =

=

=

= 461.58 cm^{2}

= 462 cm^{2}

So, option(a) is the correct answer.

**Question 15.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the area and the circumference of a circle are numerically equal, then r = …………

A. π

B.

C. 1

D. 2

**Answer:**

Area of the circle = πr^{2}

circumference of the circle = 2πr

given, Area = Circumference

2πr = πr^{2}

2 = r

So, option(d) is the correct answer.

**Question 16.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of an arc subtending an angle of measure 60 at the centre of a circle whose area is 616 is ……….

A.

B. 66

C.

D. 33

**Answer:**

Area of circle = 616cm^{2}

πr^{2} = 616

r^{2} = 196

r = 14 cm

Given θ = 60^{o}

area of the minor sector =

=

=

= cm^{2}

So, option(c) is the correct answer.

**Question 17.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a minor sector of (O, 15) is 150. The length of the corresponding arc is (π = 3.14)

A. 30

B. 20

C. 90

D. 15

**Answer:**

Radius of the minor sector= 15 cm

Area of minor sector = 150

Area of a minor sector = r l

150 = × 15 × l

l = 20 cm

So, option(b) is the correct answer.

**Question 18.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the radius of a circle is increased by 10 %, then corresponding increase in the area of the circle is ………. (π = 3.14)

A. 19 %

B. 10 %

C. 21 %

D. 20 %

**Answer:**

r be the radius of the circle

Area of the circle = πr^{2}

10% increase in radius = 1.1r

Area of circle with new radius = π(1.1r)^{2}

= 1.21πr^{2}

increase in area = 1.21πr^{2} - πr^{2}

= 0.21 πr^{2}

∴ 21% increase in the area of circle

So, option(c) is the correct answer.

**Question 19.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the ratio of the area of two circles is 1: 4, then the ratio of their circumference …..

A. 1: 4

B. 1: 2

C. 4: 1

D. 2: 1

**Answer:**

let r_{1}, r_{2} are the radii of two circles

given,

πr_{1}^{2}: πr_{2}^{2} = 1 : 4

r_{1}^{2}: r_{2}^{2} = 1 : 4

r_{1} : r_{2} =1 : 2

r_{2} = 2 r_{1}

circumference of 1^{st} circle = 2πr_{1}

circumference of 2^{nd} circle = 2πr_{2} = 2π(2r_{1}) = 4πr_{1}

ratio of circumference = 2πr_{1} : 4πr_{1} = 1 : 2

So, option(b) is the correct answer.

**Question 20.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of the largest triangle inscribed in a semi-circle of radius 8 is ……

A. 8

B. 16

C. 64

D. 256

**Answer:**

Given radius of semi-circle = 8 cm

For a largest triangle in semi-circle

base will be its diameter of semi-circle

height will be radius of semi-circle

so, its base = 2 × radius = 2 × 8 = 16 cm

height = 8 cm

Area of the triangle = × base × height

= × 16 × 8

= 64 cm^{2}

So, option(c) is the correct answer.

**Question 21.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the circumference of a circle is 44 then the length of a side of a square inscribed in the circle is ….

A.

B.

C.

D.

**Answer:**

Given circumference = 44 cm

2πr = 44

πr = 22

r = 7 cm

Given that square inscribed in a circle.

so, diagonal of square will be equal to diameter of circle.

diagonal = 2 × radius = 2 × 7 = 14 cm

diagonal = √2 × side of square

14 = √2 × side

side = 7√2 cm

**Question 22.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 14 cm. If the minute hand moves from 2 to 11 on the circular dial, then area covered by it is cm^{2}.

A. 154

B. 308

C. 462

D. 616

**Answer:**

The angle moved by minute hand = 90^{o}

length of minute hand (r) = 14 cm

Area of covered by minute hand =

=

=

= 153.93 cm^{2}

= 154 cm^{2}

So, option(a) is the correct answer.

**Question 23.**

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 15 cm. If the minute hand moves for 20 minutes on a circular dial of a clock, the area covered by it is …….. cm^{2}. (= 3.14)

A. 235.5

B. 471

C. 141.3

D. 706.5

**Answer:**

The angle moved by minute hand = 120^{o}

length of minute hand (r) = 15 cm

Area of covered by minute hand =

=

=

= 235.619 cm^{2}

= 235.5 cm^{2}

So, option(a) is the correct answer.