Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 3 Circular Measures are provided here with simple step-by-step explanations. These solutions for Circular Measures are extremely popular among Class 9 students for Maths Circular Measures Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Compute the perimeter of the circles shown below, correct to a millimetre.

(i) (ii) (iii) (iv) (i)

Construction: Join the diagonal AC of the given rectangle.

The diagonal AC will pass through the centre of the circle. So, AC is the diametre of the circle. Using Pythagoras theorem:

= (32 + 42) cm2

= (9 + 16) cm2

= 25 cm2 Perimeter of the circle = π × Diametre

= π × 5 cm

= (3.141 × 5) cm

= 15.705 cm

(ii)

Construction: Join the diagonals of the regular hexagon.

The diagonals of the regular hexagon pass through the centre of the circle. It can be observed that the diagonals of the regular hexagon divide it into six equilateral triangles.

∴ ΔOAB is an equilateral triangle with OA = OB = AB = 2 cm

Radius (r) of the circle = 2 cm

Perimeter of the circle = 2πr

= (2 × 3.141 × 2) cm

= 12.564 cm

(iii)

Diagonal of the given circle = Side of the square = 4 cm

Perimeter of the circle = π × diametre

= (3.141 × 4) cm

= 12.564 cm

(iv) The given circle is the circumcircle for ΔABC. Therefore, AD is the perpendicular bisector of side BC.

DC = BD = = (42 + 22) cm2

= (16 + 4) cm2

= 20 cm2 Similarly, AB = ∴ ΔABC is an isosceles triangle.

We know that perpendicular bisector and median of an isosceles triangle coincide with each other.

AD is the median and O is the centroid.

Also, centroid divides the median in the ratio 2:1.

AO = Radius of the circle is Perimeter of the circle = 2πr

= 2 × 3.141 × = 16.752 cm

#### Question 1:

In the figure below, how much more is the perimeter of the larger circle than that of the smaller circle?

Let the radius of the smaller circle be r cm.

Radius of the larger circle = (r + 1) cm

Perimeter of the smaller circle = 2πr cm

Perimeter of the larger circle = 2π(r + 1) cm = (2πr + 2π) cm

Difference between the perimeters of the larger circle and smaller circle

= {(2πr + 2π) 2πr} cm

= 2π cm

Therefore, perimeter of the larger circle is 2π more than the perimeter of the smaller circle.

#### Question 2:

A wheel of radius 20 centimetres rolls along. How much would it move ahead after 10 rotations?

Radius of the wheel = 20 cm

Distance covered by the wheel in one rotation = Perimeter of the wheel

= 2πr

= (2 × 3.141 × 20) cm

= 125.64 cm

Distance covered by the wheel in 10 rotations = (125.64 × 10) cm = 1256.4 cm

#### Question 1:

In a circle of radius 3 centimetres, what is the length of an arc of central angle 30°?

Radius (r) of the circle = 3 cm

Perimeter of the circle = 2πr

= (2 × 3.141 × 3) cm

= 18.846 cm

Central angle made by the given arc is 30°, which is of 360°.

Length of the arc = of the perimeter of the circle

= = 1.5705 cm

#### Question 2:

In a circle of radius 2 centimetres, what is the length of an arc of central angle 300°?

Radius (r) of the circle = 2 cm

Perimeter of the circle = 2πr

= (2 × 3.141 × 2) cm

= 12.564 cm

Central angle made by the given arc is 300°, which is of 360°.

Length of the arc = of the perimeter of the circle

= = 10.47 cm

#### Question 3:

In a circle, the length of an arc of central angle 40° is 3 centimetres. What is the perimeter of the circle? What is its radius?

Central angle made by the given arc is 40°, which is of 360°.

Length of the arc = of the perimeter of the circle

3 cm = × Perimeter of the circle

Perimeter of the circle = (3 × 9) cm

= 27 cm

Let the radius of the given circle be r cm.

2πr = 27 cm

2 × 3.141 × r = 27 cm

6.282 × r = 27

r = 4.3 cm

#### Question 4:

The length of an arc of a circle is 4 centimetres, what is the length of an arc of the same central angle, in a circle of double the radius?

Let the radius of the first circle be r.

Let the central angle made by the arc be of 360°.

Length of the arc = of the perimeter of the circle

4 cm = × 2πr …(1)

Radius of the second circle = 2r

Given: The central angle of both the circles is same.

Length of the arc of the second circle = of the perimeter of the second circle

= × 2π(2r

= 2 × × 2πr

= 2 × 4 cm (From (1))

= 8 cm

#### Question 5:

In the figure below, each curved line is an arc of a circle centered at a vertex of the triangle.  Arc ADC is an arc of the circle whose centre is the vertex B of the equilateral triangle ABC.

Radius of the circle with ADC as an arc = 4 cm

Each angle of an equilateral triangle = 60°

∴ ∠ABC = 60°

Central angle made by the arc is 60°, which is of 360°.

Length of the arc ADC = of the perimeter of the circle

= × 2π × 4 cm

= cm

Similarly, length of the arc AEB = length of the arc BFC = cm

Perimeter of the figure = Length of the arc AEB + Length of the arc BFC + Length of the arc ADC

= = 3 × cm

= 4π cm

#### Question 1:

In each of the figures below, find the area of the shaded part (i)

Construction: Join the diagonal AC of the given rectangle.

The diagonal AC will pass through the centre of the circle. So, AC is the diametre of the circle. It can be observed that the diagonals of the regular hexagon divide it into six equilateral triangles.

Using Pythagoras theorem:

= (32 + 42) cm2

= (9 + 16) cm2

= 25 cm2 Radius (r) of the circle = Area of the circle = π × square of radius

= (3.14 × 2.5 × 2.5) cm2

= 19.625 cm2

Area of the rectangle, ABCD = Length × Breadth

= 4 × 3 cm2

= 12 cm2

Area of the shaded region = Area of the circle Area of the rectangle, ABCD

= 19.625 cm2 − 12 cm2

= 7.625 cm2

(ii)

Construction: Join the diagonals of the regular hexagon.

The diagonals of the regular hexagon pass through the centre of the circle. It can be observed that the diagonals of the regular hexagon divide it into six equilateral triangles.

∴ ΔOAB is an equilateral triangle with OA = OB = AB = 2 cm

Radius (r) of the circle = 2 cm

Area of the circle = π × square of radius

= (3.14 × 2 × 2) cm2

= 12.56 cm2

Area of the hexagon, ABCDEF =  Area of the shaded region = Area of the circle Area of the hexagon, ABCDEF

= (12.56 10.38) cm2

= 2.18 cm2

(iii)

Radius of the larger circle = 2 cm

Radius of the smaller circle = 1 cm

Area of the larger circle = π × square of radius

= (3.14 × 2 × 2) cm2

= 12.56 cm2

Area of the smaller circle = π × square of radius

= (3.14 × 1 × 1) cm2

= 3.14 cm2

Area of the shaded region = Area of the larger circle Area of the smaller circle

= (12.56 − 3.14) cm2

= 9.42 cm2

(iv)

Side of the square = 4 cm

Radius of the circle whose quadrants are shown in the figure = Area of the square = side × side

= 4 × 4 cm2

= 16 cm2

Area of one quadrant = π × square of radius

= = 3.14 cm2

Area of the four quadrants = 4 × 3.14 cm2 = 12.56 cm2

Area of the shaded region = Area of the square Area of the four quadrants

= (16 − 12.56) cm2

= 3.44 cm2

#### Question 1:

In each of the figures below, find the area of the shaded part  (i)

Central angle = 40°

Radius (r) of the circle = 5 cm

Area of the sector = =  Area of the shaded part = Area of the sector (ii)

Central angle = 60°

Radius (r) of the circle = 6 cm

Area of the sector = = = 6π cm2

Area of the circle = π r2

= π (6)2 cm2

= 36π cm2

Area of the shaded part = Area of the circle Area of the sector

= (36π 6π) cm2

= 30π cm2

(iii) Central angle = 60°

Radius (r) of the circle = 7 cm

Area of the sector = =  In ΔAOB, OA = OB = 7 cm

⇒ ∠OBA = OAB (Angles opposite to equal sides are equal in measure.)

Using angle sum property in ΔAOB:

OBA + OAB + AOB = 180°

2OBA + 60° = 180°

2OBA = 120°

⇒ ∠OBA = 60°

∴ ΔOAB is an equilateral triangle.

Area of the equilateral triangle =  Area of the shaded part = Area of the sector Area of the equilateral triangle

= (25.65 − 21.19) cm2

= 4.46 cm2

(iv)

Radius of the bigger circle = 6 cm

Radius of the smaller circle = 5 cm

Central angle made by the arcs = 120°

Area of the bigger sector = = = 12π cm2

Area of the smaller sector =  Area of the shaded part = Area of the bigger sector Area of the smaller sector

= (12π 8.33π) cm2

= 3.67π cm2n

= (3.67 × 3.14) cm2

= 11.52 cm2

#### Question 1:

In the figure below, the curved lines are all arcs of circles centred on the vertices of the triangle. What is the area of this figure? Arc ADC is an arc of the circle whose centre is the vertex B of the equilateral triangle ABC.

Radius of the circle with ADC as an arc = 4 cm

Each angle of an equilateral triangle = 60°

∴ ∠ABC = 60°

Central angle made by the arc = 60°

Area of the sector, ABCD =  Similarly, area of the sector, ABFC = area of the sector, AEBC = Area of figure = Area of the sector, ABCD + Area of the sector, ABFC + Area of the sector, AEBC

=  = (8 × 3.14) cm2

= 25.12 cm2

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