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1.

In Figure, two circular flower beds have been shown on two sides of asquare lawn ABCD of side 56m. If the centre of each circular flower bed isthe point of intersection of the diagonals of the square lawn, find the sumof the areas of the lawns and the flower beds.

Answer» We know that the diagonals of a square bisect each other perpendicularly
`:. Angle DOC = 90^(@)`
Also, diagonal BD = side `sqrt2 = 56 sqrt2 m`
`:. OD = (1)/(2) xx BD = 28 sqrt2 m`
`:.` Area of `Delta DOC = (1)/(2) xx 28 sqrt2 xx 28 sqrt2 = (28)^(2) m^(2)`
Area of sector ODCO `= pi (OD)^(2) xx (theta)/(360^(@))`
`= pi (28sqrt2)^(2) xx (90^(@))/(360^(@)) = (11 xx 28 xx 28)/(7) m^(2)`
`= 1232 m^(2)`
`:.` Area of 1 flower bed = Area of sector - Area of `Delta`
`:.` Area of 2 flower beds = `2 xx 448 = 896 m^(2)`
Area of square lawn `= (56)^(2) = 3136 m^(2)`
`:.` Required area `= 896 + 3136 = 4032 m^(2)`
Discussion
2.

The perimeter of a semi circular protractor is 32.4 cm. Calculate : (i) the radius of protractor in cm, (ii) the arc of protractor in `cm^(2)`

Answer» (i) Let the radius of protractor be r cm.
Perimeter of semicircle protractor `= (pi r + 2r) cm`
`:. R (pi + 2) = 32.4`
`rArr r ((22)/(7) + 2) = 32.4`
`rArr r xx (36)/(7) = 32.4 " " rArr r = (32.4 xx 7)/(36) rArr r = 6.3 cm`
Hence, radius of protractor = 6.3 cm
(ii) Area of semi circular protractor `= (1)/(2) pi r^(2)`
`= (1)/(2) xx (22)/(7) xx 6.3 xx 6.3 = 62.37 cm^(2)`
Hence, area of protractor `= 62.37 cm^(2)`
Discussion
3.

The inner perimeter of a running track (shown in figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semicircular. If the track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

Answer» Length of the interior curved protion `= (400 - 2 xx 90) m = 220 m`
Each curved part = 110 m
`pi r = 110`
`r = (110)/(pi) = (110 xx 7)/(22) m = 35 m`
`:.` Inner radius = 35 m
`:.` Outer radius `= (35 + 14) m = 49 m`
`:.` Area of the track `= 2 xx` Area of rectangles each of `90 m xx 14 m` + Area of (circular ring with `r_(1) = 49 m and r_(2) = 35 m`)
`= 2(90 xx 14) + (22)/(7) xx [(49)^(2) - (35)^(2)]`
`= 2520 m^(2) + (22)/(7) (49 + 35) (49 - 35) m^(2)`
`= 2520 m^(2) + (22 xx 84 xx 14)/(7) m^(2)`
`= 2520 m^(2) + 3694 m^(2) = 6216 m^(2)`
Length of outer track `= (2 xx 90 + 2 xx (22)/(7) xx 49) m`
`180 m + 308 m = 488 m`
Discussion
4.

In the adjoining figure, two concentric circle with centre O have radii 21 cm and 42 cm. If `angleAOB = 60^(@)`, find the area of the shaded region. (Use `pi = (22)/(7)`)

Answer» Required area = area of large circle with radius 42 cm `-` area of smaller circle with radius 21 cm `-` (area of unshaded portion CDBA)
`= pi (42)^(2) - pi (21)^(2) -` [(area of larger sector with radius 42 cm) `-` (area of smaller sector with radius 21 cm)]
`= pi (42)^(2) - pi(21)^(2) - [pi (42)^(2) xx (60)/(360) - pi (21)^(2) xx (60)/(360)]`
`= pi (42)^(2) - pi (21)^(2) - (1)/(6) pi (42)^(2) + (1)/(6) pi (21)^(2) = (5)/(6) pi (42)^(2) - (5)/(6) pi (21)^(2)`
`= (5)/(6) pi xx (21)^(2) [4 - 1] = (5)/(6) xx (22)/(7) xx 21 xx 21 xx 3 = 3465 cm^(2)`
Discussion
5.

Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the adjoining figure. Find the area of the shaded region.

Answer» Required area = (area of large semicircle with radius 4.5 cm)
`-` (area of two smaller semicircle with radius of each `(3)/(2) cm` and a circle with radius `(4.5)/(2) cm`) `+` (area of smaller semicircle with radius `(3)/(2) cm`)
`= (1)/(2) pi (4.5)^(2) - [2 xx (1)/(2) pi ((3)/(2))^(2) + pi ((4.5)/(2))^(2)] + [(1)/(2)pi ((3)/(2))^(2)]`
`= (1)/(2) pi (4.5)^(2) - pi ((3)/(2))^(2) - pi ((4.5)/(2))^(2) + (1)/(2) pi ((3)/(2))^(2)`
`= (1)/(4) pi [2 xx (4.5)^(2) - 9 - (4.5)^(2) + (9)/(2)] = (1)/(4) pi xx 4.5 [2 xx 4.5 - 2 - 4.5 + 1]`
`= (1)/(4) pi xx 4.5 (3.5) = (1)/(4) xx (22)/(7) xx 4.5 xx 3.5 = 12.375 cm^(2)`
Discussion
6.

The circumference of a field is 220m. Find (i) its radius (ii) its area

Answer» (i) Circumference of circle = 220m
`2 pi r = 220`
`r = (220 xx 7)/(2 xx 22) = 35 m`
(ii) Area of circle `= pi r^(2) (22)/(7) xx 35 xx 35 = 3850 m^(2)`
Discussion
7.

Find the area of a circular park whose radius is 4.5 m

Answer» Area of circular park `= pi r^(2)`
`rArr " Area" = (22)/(7) xx 4.5 xx 4.5`
Area `= 63.63 m^(2)`
Discussion
8.

The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle

Answer» Let the radius of the circle be r.
Diameter = 2r
Circumference of circle `= 2 pi r`
Using the given information, we have
`2 pi r= 2r + 16.8`
`rArr 2 xx (22)/(7) xx r = 2r + 16.8 " " rArr 44r = 14 r + 16.8 xx 7`
`rArr 30 r = 117.6 " " rArr r = (117.6)/(30) = 3.92 cm`
Discussion
9.

The area of a circular plot is `346.5 m^(2)`. Calculate the cost of fencing the plot at the rate of Rs 6 per metre

Answer» Area of plot `= 346.5 m^(2)`
`rArr pi r^(2) = 346.5 m^(2) " " rArr r^(2) = (346.5 xx 7)/(22)`
`rArr r^(2) = 110.25 " " rArr r = 10.5 m`
Circumference of plot `= 2 pi r = 2 xx (22)/(7) xx 10.5 = 66 m`
Cost of fencing of = Circumference `xx` Cost of fencing per metre
`= Rs 66 xx 6 = Rs 396`
Discussion
10.

The diameter of a cycle wheel is 28 cm. How many revolutions will it make in moving 13.2 km?

Answer» Distance travelled by the wheel in one revolution `= 2 pi r = (22)/(7) xx 28 = 88 cm`
Total distance travelled by the wheel `= 13.2 xx 1000 xx 100 cm`
Number, of revolutions made by the wheel `= ("Total distance")/("circumference")`
`= (13.2 xx 1000 xx 100)/(88) = 15000` revolutions
Discussion
11.

The radius of a wheel of a bus is 45 cm. Determine its speed in kilometres per hour, when its wheel makes 315 revolutions per minute

Answer» Radius of wheel of bus = 45 cm
`:.` Circumference of the wheel `= 2 pi r`
`= 2 xx (22)/(7) xx 54 = (1980)/(7) cm`
`:.` Distance covered by the wheel in one revolution `= (1980)/(7) cm`
Distance covered by the wheel in 315 revolutions `= 315 xx (1980)/(7)`
`45 xx 1980 = 89100 cm`
`= (89100)/(1000 xx 100) km = (891)/(1000) km`
`:.` Distance covered in 60 minutes or 1 hr `= (891)/(1000) xx 60 = (5346)/(100) = 53.46 km`
Hence, speed of bus = 53.46 km/hr
Discussion