**S Chand ICSE Mathematics Class 9 Solution Seventeenth Chapter Circle – Circumference and Area Exercise 17B**

**(1) Find the area of the circles whose**

**(i) diameter = 7 cm**

**(ii) radious = 14 cm**

**(iii) diameter = 2.8 cm**

solution:

**(2) A horse is tied to a pole with 28 m long string. Find out the area which the horse can graze.**

solution:

**(2) Find the radious of a circular field whose area is –**

**(i) 154 cm ^{2}**

**(ii) 1386 cm ^{2}**

solution:

**(4) The area of a circle is 24.64 cm ^{2}. Find its circumference.**

solution:

**(5) A copper wire, when bent in the form of square, encloses an area of 121 cm ^{2}, If the same wire is bent into the form of a circle, find the area of the circle.**

solution: Let ‘a’ be the side of the square.

∴ a^{2} = 121

a = 11

**(6) The circumference of a circle is equal to the perimeter of a square. The area of the square is 484 sq m. Find the area of the circle.**

solution:

**(7) A wire is in the form of circle of radius 42 cm. It is bent into a square. Determine the area of the square and compare the areas of the regions enclosed in the two cases (Use π = 22/7). **

**Solution: **

**(8) From a copper plate, which is square of side 12.5 cm, a circular disc of diameter 7 cm is cut off. Find the weight of the remaining part, if 1 sq cm of the plate weighs 0.8 g (Assume π = 22/7). **

**Solution: **Area of the remaining part = {(12.5)^{2} – π (7/2)^{2}} cm^{2}

={156.25 – 38.5} cm^{2}

= 117.75 cm^{2}

∴ Weight of the remaining part = (117.75 x 0.8) gm

= 94.2 gm

**(9) The circumference of two circles are in the ratio 2:3. Find the ratio of their areas.**

**Solution: **

**(10) A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m as shown in the figure. Find the area OF THE REMAINING PART OF THE PARK. (Take π = 22/7).**

**Solution: **

**(11) The radius of a circular field is 20 m. Inside it runs a path 5 m wide all around. Find the area of the path.**

**Solution: **

Area of the circular field = π(20)^{2} m^{2}

Area of the inside area of the path = π (15)^{2} m^{2}

∴ Area of the path = { π(20)^{2} – π(15)^{2}}m^{2}

= 22/7 x 35 x 5 m^{2}

= 22 x 25 m^{2}

= 550 m^{2}

**(12)**** A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at Rs. 10 per m ^{2}**

**Solution:**

Let, r be the radius of surroundings,

2πr = 44

r = (44/2 x 22) x 7

= 7

∴ Area of circular plot = π(7)^{2} m^{2}

Area of portion with road is = π(10.5)^{2} m^{2}

∴ Area of the road = π{(10.5)^{2} – 7^{2}} m^{2}

Cost of paving = π{(10.5)^{2} – 7^{2}} x 10 ₨

= 1925 ₨.

**(13) A lawn is in the shape of a semi-circle of diameter 35 dm. The lawn is surrounded by a flower bed of width 3.5 dm all round. Find the area of the flower bed in dm ^{2}**

**Solution: **

**(14) A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the surface area of the road.**

**Solution: **

Let, r be the radius of the road.

**(15) The shaded area Fig. 17.19 between the circumference of two concentric circles is 346.5 cm ^{2}. The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle. (Take π = 22/7)**

**Solution: **

**(16) Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circles is drawn outside the 7 cm circle so that the area enclosed between it and the 7 cm circle is the same as that between the two inner circles, Find the radius of the third circle, correct to one decimal place. (Take π = 22/7)**

**Solution: **

Let, r be the radius of the third circle,

∴ According to the condition,

Πr^{2} – π(7)^{2} = π(7)^{2} – π(3.5)^{2}

Or, r^{2} = 98 – (3.5)^{2}

Or, r = 9.3 cm

**(17) A circular field has a perimeter of 650 m. A plot, in the shape of a square having its vertices on the circumference of the field, is marked in the field. Calculate the area of the square plot. (Take π = 22/7)**

**Solution: **

Let, r be the radius of the field,

2πr = 650

Or, r = 650/44 x 7

**(18) The inside perimeter of a practice running track with semicircle ends and straight parallel sides is 312 m. The length of the straight portions of the track is 90 m. If the track has a uniform width of 2 m throughout, Find its area.**

**Solution: **

Inside the perimeter of the track = 312 m

Length of the two st portions = 90 x 2 = 180 m

The length of the remaining = (312 – 180) m

= 132 m

Circumference of the remaining semi-circular portions = πr + πr = 2 πr, where r be the radius,

**(19) The boundary of the shaded region in Fig. 17.21 consists of three semi-circular arcs, the smaller ones being equal. If the diameter of the larger arc is 10 cm, calculate:**

**(i) the length of the boundary**

**(ii) the area of the shaded region. (Take π to be 3.14)**

**Solution: **

**(20) A bed of roses is like as shown in fig. 17.22. In the centre of a square and on each side there is semi circle. Side of the square is 21 m. If each rose-plant needs 6 m ^{2} of space, find out the number of plants which can be planted in the whole figure.**

**Solution: **

So, one plant needs 6m^{2} space,

So total space required by = 1134 / 6number of plants

= 189 plants.

**(21) Find the area of the region between two concentric circles given in Fig. 17.23, if the length of the chord of the outer circle touching the inner circle is 14 cm.**

**Solution: **

**(22) The Fig. 17.24 shows a circle with centre at O and <AOB = 90 ^{0}. If the radius of the circle is 40 cm, calculate the area of the shaded portion of the circle. (Take π = 3.14)**

**Solution: **

By the Pythagoras theorem,

AB^{2} = OA^{2} + OB^{2}

AB^{2} = (40)^{2} + (40)^{2}

= 1600 + 1600

= 3200

AB = √1600×2

= 40√2

**(23) ABCD is a flower bed. If OA = 21 m and OC = 14 m, find the area of the bed. (Use π = 22/7)**

**Solution: **

**(24) In fig. 17.26, A, B. C and D are centers of equal circles which touch externally in pairs and ABCD is a square of side 14 cm. Find the area of the shaded region.**

**Solution: **

Radius of the circle is = 7 cm

∴ Area of each circle is = π(7)^{2} cm^{2}

= 22/7 x 7^{2 }cm^{2}

= (22 x 7) cm^{2}

= 154 cm^{2}

∴ Area of the shaded region = Area of Square – 4 x (1/4 of the circle)

= (196 – 154) cm^{2}

= 42 cm^{2}** **