## RD Sharma Class 10 Solutions Areas related to Circles Ex 15.1

### RD Sharma Class 10 Solutions Areas related to Circles Exercise 15.1

Question 1.

Find the circumference and area of a circle of radius 4.2 cm.

Solution:

Radius of a circle = 4.2 cm

Question 2.

Find the circumference of a circle whose area is 301.84 cm^{2}.

Solution:

Area of a circle = 301.84 cm^{2
}Let r be the radius, then πr^{2} = 301.84

Question 3.

Find the area of a circle whose circumference is 44 cm.

Solution:

Circumference of a circle = 44 cm

Let r be the radius,

then 2πr = circumference

Question 4.

The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle. (C.B.S.E. 1996)

Solution:

Let r be the radius of the circle

∴ Circumference = 2r + 16.8 cm

⇒ 2πr = 2r + 16.8

⇒ 2πr – 2r = 16.8

Question 5.

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7)

Solution:

Radius of the circle (r) = Length of the rope = 28 m .

Area of the place where the horse can graze

Question 6.

A steel wire when bent in the form of a square encloses an area of 121 cm^{2}. If the same wire is bent in the form of a circle, find the area of the circle. (C.B.S.E. 1997)

Solution:

Area of square formed by a wire =121 cm^{2
}∴ Side of square (a) = \(\sqrt { Area } \) = \(\sqrt { 121 } \) = 11 cm Perimeter of the square = 4 x side = 4 x 11 = 44 cm

∴Circumference of the circle formed by the wire = 44cm

Let r be the radius

Question 7.

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

Solution:

Let R and r be the radii of two circles and their ratio between them circumference = 2 : 3

Question 8.

The sum of radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles.

Solution:

Let R and r be the radii of two circles Then R + r = 140 cm …….(i)

and difference of their circumferences

Question 9.

Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm. [NCERT Exemplar]

Solution:

Let the radius of a circle be r.

Circumference of a circle = 2πr

Let the radii of two circles are r_{1} and r_{2} whose

values are 15 cm and 18 cm respectively.

i.e., r_{1} = 15 cm and r_{2} = 18 cm

Now, by given condition,

Circumference of circle = Circumference of first circle + Circumference of second circle

⇒ 2πr = 2π r_{1} + 2πr_{2 }=

⇒ r = r_{1} + r_{2
}⇒ r = 15 + 18

∴ r = 33 cm

Hence, required radius of a circle is 33 cm.

Question 10.

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

Solution:

Radius of first circle (r_{1}) = 8 cm

and radius of second circle (r_{2}) = 6 cm

Question 11.

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

Solution:

Radius of the first circle (r_{1}) = 19 cm

and radius of the second circle (r_{2}) = 9 cm S

um of their circumferences = 2πr_{1} + 2πr_{2
}= 2π (r_{1 }+ r_{2}) = 2π (19 + 9) cm

= 2π x 28 = 56π cm

Let R be the radius of the circle whose circumference is the sum of the circumferences of given two circles, then

Question 12.

The area of a circular playground is 22176 m^{2}. Find the cost of fencing this ground at the rate of ₹50 per metre. [NCERT Exempiar]

Solution:

Given, area of a circular playground = 22176 m^{2}

Question 13.

The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.

Solution:

ABCD is a square whose each side is 10 cm

∴ AB = BC = CD = DA = 10 cm

AC and BD are its diagonals

Question 14.

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Solution:

Let r be the radius of the circle a be the side of the square

Question 15.

The area of a circle inscribed in an equilateral triangle is 154 cm^{2}. Find the perimeter of the triangle. (Use π = 22/7 and \(\sqrt { 3 } \) = 1.73)

Solution:

Area of the inscribed circle of ΔABC = 154 cm^{2}

Question 16.

A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be ₹2640 at the rate of ₹12 per metre. Then, the field is to be thoroughly ploughed at the cost of ₹0.50 per m^{2}. What is the amount required to plough the field ? (Take π = 22/7)

Solution:

Cost of the fencing the circular field = ₹2640

Rate = ₹12 per metre 2640

∴ Circumference = \((\frac { 2640 }{ 12 } )\) = 220 m

Let r be the radius of the field, then = 2πr = 220

Question 17.

A park is in the form of a rectangle 120 m x 100 m. At the centre of the park there is a circular lawn. The area of park excluding lawn is 8700 m^{2}. Find the radius of the circular lawn. (Use π = 22/7).

Solution:

Area of the park excluding lawn = 8700 m^{2
}Length of rectangular park = 120 m

and width = 100 m

∴ Area of lawn = l x b

= 120 x 100 m^{2} = 12000 m^{2
}Let r be the radius of the circular lawn, then area of lawn = πr^{2}

Question 18.

A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.

Solution:

Distance covered by the car in 450 revolutions = 1 km = 1000 m

∴ Distance covered in 1 revolution = \((\frac { 1000 }{ 450 } )\)

= \((\frac { 20 }{ 9 } )\) m

Question 19.

The area of enclosed between the concentric circles is 770 cm^{2}. If the radius of the outer circle is 21 cm, find the radius of the inner circle.

Solution:

Area of enclosed between two concentric circles = 770 cm^{2
}Radius of the outer circle (R) = 21 cm

Question 20.

An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.[NCERT Exemplar]

Solution:

Let the diameters of concentric circles be k, 2k , 3k

Question 21.

The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/hr? [NCERT Exemplar]

Solution:

Given, radius of wheel, r = 35 cm

Circumference of the wheel = 2πr

= 2 x \((\frac { 22 }{ 7 } )\) x 35 = 220 cm

But speed of the wheel = 66 kmh^{-1
}= \((\frac { 66 x 1000 }{ 60 } )\) m/ mm

= 1100 x 100 cm min^{-1
}= 110000 cm min^{-1}^{
}∴ Number of revolutions in 1 min

= \((\frac { 110000 }{ 220 } )\)= 500 revolution

Hence, required number of revolutions per minute is 500.

Question 22.

A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per m^{2}. [NCERT Exemplar]

Solution:

Given that, a circular pond is surrounded by a wide path.

The diameter of circular pond = 17.5 m

Question 23.

A circular park is surrounded by a rod 21 m wide. If the radius of the park is 105 m, find the area of the road. [NCERT Exemplar]

Solution:

Given that, a circular park is surrounded by a road.

Width of the road = 21 m

Radius of the park (r_{1}) = 105 m

.’. Radius of whole circular portion (park + road),

r_{e} = 105 + 21 = 126 m

Now, area of road = Area of whole circular portion – Area of circular park

= πr^{2} – πr^{2} [∵ area of circle = πr^{2}]

Question 24.

A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying outside the circle and inside the square. [NCERT Exemplar]

Solution:

Let the side of a square be a and the radius of circle be r.

Given that, length of diagonal of square = 8 cm

Question 25.

A path of 4 m width runs round a semicircular grassy plot whose circumference is 81 \((\frac { 5 }{ 7 } )\)m. Find:

(i) the area of the path

(ii) the cost of gravelling the path at the rate of ₹1.50 per square metre

(iii) the cost of turfing the plot at the rate of 45 paise per m^{2}.

Solution:

Width of path around the semicircular grassy plot = 4 m

Circumference of the plot = 81 \((\frac { 5 }{ 7 } )\)m

= \((\frac { 572 }{ 7 } )\) m

Let r be the radius of the plot, then

Question 26.

Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.

Solution:

Radius of first circle (r_{1}) = 3.5 cm

Radius of second circle (r_{2}) = 7 cm

Question 27.

A path of width 3.5 m runs around a semicircular grassy plot whose perimeter is 72 m. Find the area of the path. (Use π = 22/7) [CBSE 2015]

Solution:

Perimeter of semicircle grassy plot = 72 m

Let r be the radius of the plot

Question 28.

A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14) [CBSE 2014]

Solution:

Diameter of circular pond (d) = 17.5 m

Radius (r) =\((\frac { 1725 }{ 2 } )\) = 8.75 m

Width of path = 2m

∴ Radius of outer cirlce (R) = 8.75 + 2 = 10.75 m

Area of path = (R^{2} – r^{2})π

= [(10.75)^{2} – (8.75)^{2}](3.14)

= 3.14(10.75 + 8.75) (10.75 – 8.75)

= 3.14 x 19.5 x 2 = 122.46 m^{2
}Cost of 1 m^{2} for constructing the path ₹25 m^{2
}∴ Total cost = ₹ 122.46 x 25 = ₹3061.50

Question 29.

The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use π= 22/7).

Solution:

Let R and r be the radii of the outer and inner of track.

Outer circumference of the race track = 528 m

Question 30.

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

Solution:

Width of the road = 7 m

Circumference of the park = 352 m

Let r be the radius, then 2πr = 352

Question 31.

Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

Solution:

Radius of inner circle = r

Width of path = h

∴ Outer radius (R) = (r + h)

∴ Area of path = πR^{2} – πr^{2
}= π {(r + h)^{2} – r^{2}}

= π {r^{2} + h^{2} + 2rh – r^{2}}

= π {2rh + h^{2}}

= πh (2r + h) Hence proved.