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 9 Maths are prepared by experts and are 100% accurate.
Page No 168:
Question 1:
Compute the perimeter of the circles shown below, correct to a millimetre.
(i)
(ii)
(iii)
(iv)
Answer:
(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.
In ΔADC, ∠ADC = 90°.
Using Pythagoras theorem:
AC2 = AD2 + DC2
= (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 =
Using Pythagoras theorem in ΔADC:
AC2 = AD2 + DC2
= (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
Page No 169:
Question 1:
In the figure below, how much more is the perimeter of the larger circle than that of the smaller circle?
Answer:
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.
Page No 169:
Question 2:
A wheel of radius 20 centimetres rolls along. How much would it move ahead after 10 rotations?
Answer:
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
Page No 175:
Question 1:
In a circle of radius 3 centimetres, what is the length of an arc of central angle 30°?
Answer:
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
Page No 175:
Question 2:
In a circle of radius 2 centimetres, what is the length of an arc of central angle 300°?
Answer:
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
Page No 175:
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?
Answer:
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
Page No 175:
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?
Answer:
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
Page No 175:
Question 5:
In the figure below, each curved line is an arc of a circle centered at a vertex of the triangle.
Answer:
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
Page No 179:
Question 1:
In each of the figures below, find the area of the shaded part
Answer:
(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.
In ΔADC, ∠ADC = 90°
Using Pythagoras theorem:
AC2 = AD2 + DC2
= (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
Page No 181:
Question 1:
In each of the figures below, find the area of the shaded part
Answer:
(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°
⇒ 2∠OBA + 60° = 180°
⇒ 2∠OBA = 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
Page No 182:
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?
Answer:
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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