Circle Measures
Class 9th Mathematics Part 2 Kerala Board Solution
- Prove that the circumcentre of an equilateral triangle is the same as its centroid. i)…
- Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.…
- Calculate perimeter of a regular hexagon with vertices on a circle of diameter 1…
- The perimeter of a regular hexagon with vertices on a circle is 24 centimetres. i) What…
- A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of…
- The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28…
- In the picture below, a regular hexagon, square and a rectangle are drawn with their…
- An isosceles triangle with its vertices on a circle is shown in this picture. (1) What…
- In all the pictures, below, centres of the circles are on the same line. In the first…
- In this picture, the circles have the same centre and the line drawn is a diameter of…
- In the pictures, below, find the difference between the areas of the circle and the…
- The pictures below show circles through the vertices of a square and a rectangle:…
Questions Pg-157
Question 1.Prove that the circumcentre of an equilateral triangle is the same as its centroid.
i) Calculate the length of side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.
ii) Calculate the perimeter of such a triangle.
Answer:
The figure is given below:
Let ABC be a triangle
AD, BE and CF are medians intersecting at point O.
In triangles ABD and ACD,
BD = CD (AD is the median)
∠ABD = ∠ACD = 60°
AB = AC (Equilateral Triangle)
∴ triangles ABD and ACD are congruent.
∴ ∠BAD = ∠CAD = 
30°
In triangle ABD
Sum of all angles of a triangle = 180°
∴ ∠ABD + ∠DAB + ∠ADB = 180°
∴ 60° + 30° + ∠ADB = 180°
90° + ∠ADB = 180°
∴ ∠ADB = 180° – 90° = 90°
∴ ∠ADC = 180 – ∠ADB (Supplementary angles)
= 180 – 90°
= 90°
Similarly,
∠OBD = ∠OCD = 30° …(eq)1
In triangles OBD and OCD,
∠OBD = ∠OCD = 30° (from eq 1)
∠ADB = ∠ADC = 90°
OD = OC is a common side
By RHS criterion,
Triangles OBD and OCD are congruent.
∴ OB = OC
Similarly, OC = OA and OA = OB
∴ OA = OB = OC
Hence, pt. O is a circumcentre and also a intersection of medians.
Hence, the circumcentre of an equilateral triangle is the same
as its centroid.
i) 
Diameter of circle = 1 centimetre
∴ Radius of circle = 
cm = 0.5 cm
As proved above, the circumcentre of an equilateral triangle is the same as its centroid.
Also, centroid divides equilateral in the ratio 2:1
∴ radius 
∴ 
∴ 
= 
× 0.5 = 
But, for an equilateral triangle,
Height of triangle = 
=
∴ side of triangle = 
ii) Perimeter of triangle = Sum of all sides of triangle
As for an equilateral triangle, all sides are equal
Let a = side of triangle = 
Perimeter of triangle = a + a + a = 3a = 
cm
Question 2.
Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.
Answer:
The figure is shown below:
The square ABCD is inscribed within the circle.
Diameter of circle, DB = 1 cm
∴ Radius of circle = cm = 0.5 cm
As seen in figure,
Diameter of circle = side of square
∴ the diagonal of the square = 1 cm
For a square,
If side = “s” cm, then,
Diagonal of square, AC2 = s2 + s2
⇒ AC2 = 2s2
⇒ AC = √2 
= 1
∴ a = 
Perimeter of a square = 4 × Side of square
= 
= 
= 
Question 3.
Calculate perimeter of a regular hexagon with vertices on a circle of diameter 1 centimetre.
Answer:
The figure is shown below:
Diameter of circle, CF = 1 centimetre
∴ Radius of circle = cm = 0.5 cm
As can be seen from figure,
Radius of circle = Side of Hexagon
∴ Side of Hexagon = 0.5 cm
Perimeter of Hexagon = 6 × Side of Hexagon
= 6 × 0.5
= 3 cm
Questions Pg-160
Question 1.The perimeter of a regular hexagon with vertices on a circle is 24 centimetres.
i) What is the perimeter of a square with vertices on this circle?
ii) What is the perimeter of a square with vertices on a circle of double the diameter?
iii) What is the perimeter of an equilateral triangle with vertices on a circle of half the diameter of the first circle?
Answer:
Given:
Perimeter of Hexagon = 24 cm
Let a = side of Hexagon
But, Perimeter of Hexagon = 6 × side of Hexagon
24 = 6 × a
∴ a = 24/6=4
As can be seen from figure,
Radius of circle = Side of Hexagon
∴ Radius of circle = a = 4 cm
i) Diameter of circle = 2 × radius
= 2 × 4
= 8 cm
As seen in figure,
Diameter of circle = Diagonal of square
Diagonal of square = 8 cm
For a square,
If side = a cm
Diagonal of square = = 8
∴ 
Perimeter of square = 4 × side of square
= 
= 
= 
cm
ii) Diameter of circle =2 × Initial diameter
=2 × 8=16 cm
As seen in figure,
Diameter of circle = Diagonal of square
Diagonal of square = 16 cm
For a square,
If side = a cm
Diagonal of square = √2a = 16
∴ 
Perimeter of square = 4 × side of square
= 
= 
= 
cm
iii) 
Diameter of circle =1/2 × Initial diameter
=
∴ Radius of circle = 
cm = 2 cm
As proved earlier, the circumcentre of an equilateral triangle is the same as its centroid.
Also, centroid divides equilateral in the ratio 2:1
∴ radius = 
∴ 2 =
∴ height of triangle = × 2 = 3
But, for an equilateral triangle,
Height of triangle =
3 =
∴ side of triangle = 3
Perimeter of equilateral triangle = 
=
=
=
cm
Question 2.
A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of a circle made by bending a wire of half the length?
Answer:
Diameter of circle = 4 cm
Length of wire = Circumference of circle
If r = radius of circle
Circumference of circle = 2 × π × r=2 × π × 4
=8 × π
If length of wire is halved,
Circumference is also halved
∴ New Circumference = 1/2 × (8 × π)= 4 × π
Let r’be new radius
∴ 2 × π × r'= New Circumference = 4 × π
∴ r’ = 
Hence, New diameter = 2 × r'= 2 × 2=4 cm
Question 3.
The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28 metres. How do we compute the perimeter of a circle of diameter 3 metres, without measuring?
Answer:
If r = radius of circle
Perimeter = Circumference of circle = 2 × π × r
When, diameter=2 m
Radius = 
Perimeter = 2 × π × 1=6.28 …(eq)1
When, diameter=3 m
Radius = 
Perimeter = 2 × π × 1.5=1.5 × (2 × π × 1)
= 1.5 × 6.28 …from (eq)1
= 9.42 m
Questions Pg-163
Question 1.In the picture below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle. Calculate the perimeter of each circle.
Answer:
Case (i) for regular hexagon:
In the figure given below:
Side of regular hexagon = 2 cm
In this figure,
Radius of circle = Side of hexagon = 2cm
In triangle OAB,
∠OAB = ∠OBA = 60° (OA = OB = radius = 2 cm)
If r = radius of circle
Circumference = 2 × π × r
Hence, perimeter of each circle = 2 × π × 2≈4 × 3.14 = 12.56 cm
Case (ii) for square:
The figure is shown below:
In triangle OAB,
OA = OB = radius
In triangle OAB,
∠OAB = ∠OBA = 45°
Sum of all angles of a triangle = 180°
∠OAB + ∠OBA + ∠AOB = 180°
45° + 45° + ∠AOB = 180°
90° + ∠AOB = 180°
∴ ∠AOB = 180-90 = 90°
By Pythagoras theorem,
(Hypotenuse)2 = (One side)2 + (Other side)2
(AB)2 = (OA)2 + (OB)2
22 = 2 × (OA)2 (OA = OB)
∴ (OA)2 = 
OA = √2 = radius
If r = radius of circle
Circumference = 2 × π × r
Hence, perimeter of each circle = 2 × π × √2
≈4 × 3.14 = 8.87 cm
Case (iii) for rectangle drawn within the circle. The figure is displayed below:
In triangle ABC,
∠ABC = 90° (angle subtended in the semicircle is right angle).
By pythagoras theorem,
(Hypotenuse)2 = (One side)2 + (Other side)2
∴ (AC)2 = (AB)2 + (AC)2
∴ (AC)2 = 22 + 1.52
∴ (AC)2 = 4 + 2.25 = 6.25
∴ AC = √6.25 = 2.5 = Diameter of circle.
∴ Radius = 
If r = radius of circle
Circumference = 2 × π × r
∴ Perimeter = 2 × π × r≈2 × 3.14 × 1.25 = 7.85 cm
Question 2.
An isosceles triangle with its vertices on a circle is shown in this picture.
What is the perimeter of the circle?
Answer:
The figure is shown below:
Let OA = radius of circle.
AD = 1 cm
OA = r
∴ OD = 1-r
OA = OB = r
In triangle OBD,
∠ODB = 90°
AB = AC and AD is perpendicular on BC from point A.
∴ BD = DC = 
∴ By pythagoras theorem,
(Hypotenuse)2 = (One side)2 + (Other side)2
∴ (OB)2 = (OD)2 + (BD)2
∴ r2 = (1-r)2 + (
)2
∴ r2- (1-r)2 = 
∴ (r + {1-r})(r-{1-r}) = {a2-b2 = (a + b) × (a-b)}
∴ 1 × (2r-1) =
∴ 2r = 
∴ r = 
If r = radius of circle
Circumference = 
∴ Perimeter = 
Question 3.
In all the pictures, below, centres of the circles are on the same line. In the first two pictures, the small circles are of the same diameter.
Prove that in all pictures, the perimeters of the large circle is the sum of the perimeters of the small circles.
Answer:
Let R = radius of larger circle in all figures.
If r = radius of circle
Circumference =
∴ Perimeter of large circle = 
…(eq)1
In first figure,
Diameter of larger circle = 2 × Diameter of smaller circle(as seen from figure)
Hence,
Diameter of smaller circle = 
∴ Radius of smaller circle = 
(Radius = 
)
Let r = Radius of smaller circle
∴ r = 
Perimeter of a smaller circle = 2 × π × r
= 2 × π × = π × R
Perimeter of 2 smaller circles = 2 × (π × R) = …(eq)1
Hence, proved.
In second figure,
Diameter of larger circle = 3 × Diameter of smaller circle(as seen from figure)
Hence,
Diameter of smaller circle = 
∴ Radius of smaller circle = 
(Radius = )
Let r = Radius of smaller circle
∴ r = 
Perimeter of a smaller circle = 2 × π × r
= 2 × π × R
Perimeter of 3 smaller circles = 3 × (
× π × R) = …(eq)1
Hence, proved.
In 3rd figure,
There are 3 small circles.
Right half has radius =
Perimeter of right half = 2 × π × = π × R …(eq)2
The left half’s radius is divided in the ratio 1:2
Hence, radius of smallest circle = 
Hence, radius of middle circle = 
Perimeter of smallest circle = 2 × π × 
…(eq)3
Perimeter of middle circle = 
…(eq)4
Sum of perimeter of all circles = (eq)2 + (eq)3 + (eq)4
= (π × R) + (π × ) + (2 × π × )
= π × R …(same as (eq)1)
Hence, proved.
Question 4.
In this picture, the circles have the same centre and the line drawn is a diameter of the large circle. How much more is the perimeter of the large circle than the perimeter of the small circle?
Answer:
The figure is shown below:
Diameter of larger circle = 1 cm
Hence, diameter of smaller circle = cm
Let r = Radius of smaller circle = 
If r = radius of circle
Circumference =
As seen from figure,
Radius of Larger circle = 2r
Perimeter of smaller circle = 2 × π × r …(eq)2
Perimeter of larger circle = 2 × π × (2r) = 4 × π × r …(eq)1
Perimeter of larger circle is = (eq)1 – (eq)2
More then smaller circle by
= (4 × π × r) – (2 × π × r)
= 2 × π × r
Questions Pg-167
Question 1.In the pictures, below, find the difference between the areas of the circle and the polygon, up to two decimal places:
Answer:
As seen in figure 1,
Diameter of circle = Diagonal of square
∴ Diameter of circle = 4 cm
∴ Radius of circle = 
If r = radius of circle
Area of circle = 
r2 cm2
= 22
= 
cm2
As seen in figure 2,
Diameter of circle = 4 cm
∴ Radius of circle =
As can be seen from figure 2,
Radius of circle = Side of Hexagon
∴ Side of Hexagon = 2 cm
The regular hexagon is made of 6 equilateral triangles.
Area of polygon (here, regular hexagon) = 6 × Area of triangle
Side of Hexagon = Side of triangle = 2 cm
Area of equilateral triangle = 
(side)2
= 22
= 
cm2
Area of polygon (here, regular hexagon) = 6 × Area of triangle
= 
cm2
∴ Difference between = Area of circle – Area of polygon
Areas
= (4 × π)-(6 × √3)
= 
(3.14159)-
(1.73205)
= 12.56636-10.3923
= 2.17406
≈2.17(upto 2 decimal places).
Question 2.
The pictures below show circles through the vertices of a square and a rectangle:
Calculate the areas of the circles.
Answer:
For figure 1,
Side of square = 3 cm
If, Side of square = a cm
Diagonal of square = 
= 
As seen in figure,
Diameter of circle = Diagonal of square
∴ Diameter of circle = √2 × 3 cm
∴ Radius of circle = 
If r = radius of circle
Area of circle = π × r2 cm2
= (
)2
= 
cm2
≈3.14
4.5
≈14.13 cm2
For figure 2,
From Pythagoras theorem,
(Hypotenuse)2 = (one side)2 + (other side)2
Diagonal of rectangle = (42 + 22)1/2
= (16 + 4)1/2
= √20 cm
As seen from figure,
Diagonal of rectangle = Diameter of circle = √20 cm
∴ Radius of circle = 
If r = radius of circle
Area of circle = π × r2 cm2
= (
)2
= 
cm2
≈3.1415200
≈628.3 cm2
Questions Pg-173
Question 1.In a circle, the length of an arc of central angle 40° is 3π centimetres. What is the perimeter of the circle? What is its radius?
Answer:
X = 40°
Length of arc = 3cm
In a circle of radius r,
If central angle = x°
Length of arc = 
Putting in above equation,
3 = 
∴ 3 = 
∴ 
= r
∴ r≈3 cm
If r = radius of circle,
Perimeter = 2 × π × r
Hence, Perimeter = 2 × π × 3 ≈18.84 cm
Question 2.
In the same circle, what is the length of an arc of central angle 25° is 4 centimetres.
i) In the same circle, what is the length of an arc of central angle 75°?
ii) In a circle of radius one and a half times the radius of this circle, what is the length of an arc of central angle 75°?
Answer:
In a circle of radius r,
If central angle = x°
Length of arc =
Here,
x = 25°
r = 4 cm
Length of arc = 
i) Here,
x = 75°
r = 4 cm
Length of arc = 
ii) Here,
x = 25°
r = 1.5 × 4 = 6 cm
Length of arc =
Question 3.
From a bangle of radius 3 centimetres, a piece is to be cut out to make a ring of radius 
centimetres.
i) What should be the central angle of the piece to be cut out?
ii) The remaining part of the bangle was bent to make a smaller bangle. What is its radius?
Answer:
i) Radius of bangle = 3 cm
Radius of ring =
If r = radius of circle
Circumference of circle = 2 × π × r cm
∴ Length of ring = 2 × π × = π cm
Hence, length of arc = π cm
In a circle of radius r,
If central angle = x°
Length of arc =
Here r = 3 cm
π = 
∴ 
= x
∴ x = 
ii) Length of remaining part = Length of bangle-Length of smaller part
= (2 × π × 3)-(π)
= 5 × π ≈15.7 cm
Let a be the new radius
Since,
Circumference = 2 × π × a
15.7 = 2 × π × a
∴ a = 
Questions Pg-176
Question 1.What is the area of a sector of central angle 120° in a circle of radius 3 centimetres? What is the area of a sector of the same central angle in a circle of radius 6 centimetres?
Answer:
Radius = r = 3 cm
If x° is the angle at subtended at the sector,
Area of sector = 
r2
Here, x = 120°
∴ Area = 
π× 32
cm2
If r2 = 6 cm
∴ Area = π× 62 = 
cm2
Question 2.
Calculate the area of the shaded part of this picture.
Answer:
Inner radius = r = 2 cm
Outer radius = R = 4 cm
Area of shaded part = Area of larger sector –Area of smaller sector
If x° is the angle at subtended at the sector,
Area of sector = r2
Here, x = 120°
Area of larger sector = 42× π = 
Area of smaller sector = 22× π = 
Area of shaded part = Area of larger sector –Area of smaller sector
= –
cm2
Question 3.
Centred at each corner of a regular hexagon, a part of a circle is drawn and a figure is cut out as shown below:
What is the area of this figure?
Answer:
As seen from the figure,
Radius of each arc = 
For a regular polygon with n sides,
Sum of angles = (n-2)× 180°
Here, n = 6
∴ Sum of angles = (6-2)× 180
= 4× 180
= 720°
Hence, each angle = 
Thus for each sector,
X = 120°
r = 1 cm
If x° is the angle at subtended at the sector,
Area of sector = r2
∴ Area of sector = 
12 = 
cm2
Area of 6 sectors = 
cm2
As the hexagon is made up of 6 equilateral triangles,
Area of polygon = 6× Area of equilateral triangle with side as 2 cm
If a = side
Then area of triangle = 
(side)2
Here, side = 2 cm
∴ area of a triangle = (2)2 = 
cm2
∴ area of polygon = 6× √3 = 6√3 cm2
∴ area of shaded region = area of polygon-area of 6 sectors
= 6√3 – 2π = 4.112 cm2