Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 8 Circle Measures Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 8 Solutions Circle Measures
Circle Measures Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 8 Circle Measures Solutions Questions and Answers
Class 9 Maths Chapter 8 Kerala Syllabus – Diameter And Perimeter
Textual Questions And Answers
Question 1.
The circumference of a circle of diameter 2 metres is measured and found to be 6.28 metres.
i) How do we compute the circumference of a circle of diameter 4 metres, without actually measuring it?
ii) What about the circumference of a circle of diameter 1 metre?
iii) And the circumference of a circle of diameter 3 metre?
Answer:
i) If the diameter is 2 meters, perimeter is 6.28 meters.
If the diameter is 4 meters, perimeter = 6.28 × 2 = 12.56 meters
ii) If the diameter is 1 metre, perimeter = \(\frac{6.28}{2}\) = 3.14 meters
iii) If the diameter is 3 metre, perimeter = \(\frac{6.28 \times 3}{2}\) = 9.42 meters
Question 2.
A piece of wire is bent into a circle of diameter 4 centimetres. If a wire of half the length is bent into a circle, what would be its diameter?
Answer:
The circumferences of circles change in proportion to their diameters. Therefore, if the length of the string is halved, the diameter of the circle formed by the string will also be reduced by half.
This means the diameter of the circle will be 2 centimeters.
Class 9 Maths Kerala Syllabus Chapter 8 Solutions – A New Number
Question 1.
Calculate the circumferences of the circles shown below:
Answer:
Figure 1
AB = 2 cm
In the figure triangle are equilateral triangles, therefore radius OA = 2 cm
Circumference of circle = 2πr = 2 × π × 2 = 4π cm
Figure 2
ABCD is a square AB = BC = 2 cm
AC = \(\sqrt{2^2+2^2}\) = √8 = 2√2 cm
Radius of circle = \(\frac{2 \sqrt{2}}{2}\)= √2 cm
Circumference of circle = 2πr = 2 × π × √2 = 2√2π cm
Figure 3
PR = \(\sqrt{2^2+(1.5)^2}=\sqrt{6.25}\) = 2.5 cm
Radius of circle = \(\frac{2.5}{2}\) = 1.25 cm
Circumference of circle = 2πr = 2 × π × 1.25 = 2.5π cm
Question 2.
In a circle, a chord 4 centimetres away from the centre is 6 centimetres long. What is the circumference of the circle?
Answer:
Length of chord (AB) = 6cm
In the figure, P is the midpoint of AB
AB = 2AP
AP = 3cm
OP = 4cm
r = \(\sqrt{4^2+3^2}=\sqrt{25}\) = 5 cm
Circumference of the circle = 2πr = 2 × π × 5 = 10πcm
Question 3.
The figure below shows an isosceles triangle of base and height 4 centimetres drawn with vertices on a circle.
Calculate the circumference of the circle.
Answer:
Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r, OD = 4 – r, AD = 2 cm and AO = r
In triangle AOD
(AO)² = (OD)² + (AD)²
r² = 2² + (4 – r)²
r² = 4 + 16 – 8r + r²
8r = 20
Circumference of the circle = 2πr = 2 × π × 2.5 = 5n cm
Question 4.
In each of the pictures below, the centres of the large and small circles are on the same line. In the first and the second pictures, all the small circles have the same diameter. In each of these figures, show that the circumference of the large circle is sum of the circumferences of the small circles.
Answer:
Figure 1
Smaller circles have same diameters.
Consider the diameter as d, circumference of smaller circle = π × diameter = πd
Circumference of two small circles = 2πd
Diameter of the large circle = d + d = 2d
Circumference of the large circle = π × 2d = 2πd
Therefore, Circumference of the large circle is the sum of the circumference of the small circles.
Figure 2
Let d be the diameter of the small circle. .
Sum of circumference of three small circles =3πd
Diameter of the large circle = d + d + d = 3d
Circumference of the large circle = π × 3d
Therefore Circumference of the large circle is the sum of the circumference of the small circles.
Figure 3
In figure diameter of three circles are different, let consider the diameters of small circles are p, q and r.
Circumference of first small circle = πp
Circumference of second small circle = πq
Circumference of third small circle = πr
Sum of circumference of three small circles = πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Circumference of large circle = π (p + q + r)
Therefore, Circumference of the large circle is the sum of the circumference of the small circles.
Question 5.
In the picture below, the two circles have the same centre. How much more is the circumference < of the larger circle than that of the smaller circle?
Answer:
If r be the radius of the small circle Radius of the large circle = r + 1
Circumference of the small circle = 2πr
Circumference of the large circle = 2π (r + 1) = 2πr + 2π
Therefore, Circumference of the large circle is 2n units more than the circumference of the small circle.
SCERT Class 9 Maths Chapter 8 Solutions – Area
Intext Questions And Answers
Question 1.
What is the relation between the areas of the shaded regions in the two figures?
Answer:
Figure 1
Area = π(2.5)² – π(2)²
= 6.25π – 4π
= 2.25π cm²
Figure 2
r² = (2.5)² – (2)² = 2.25
Area = πr² = 2.25π cm²
Both the area are equal
Textual Questions And Answers
Question 1.
The length of a chord of a circle, 3 centimetres from the centre, is 4 centimetres. What is the area of the circle?
Answer:
Radius = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\) cm
Area = πr = π\((\sqrt{13})^2\) = 13π cm²
Question 2.
In each of the pictures below, compute the difference between the area of the circle and the area of the regular polygon, correct up to two decimal places.
Answer:
Figure 1
Radius of the circle = 2cm
Area = πr² = π(2)² = 4π = 3.14 × 4 = 12.56 cm²
Diagonal of the square = 4 cm
One side of a square = \(\frac{4}{\sqrt{2}}\) cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}\) = 8 cm²
Differences between the areas = 12.56 – 8 = 4.56 cm²
Figure 2
Radius of the circle = 2cm
Area = πr² = π(2)² = 4π = 3.14 × 4 = 12.56 cm²
A regular hexagon is made up of six equilateral triangles.
The side of the equilateral triangle is equal to the radius of the circle.
Side of the equilateral triangle = 2cm
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × (2)² = √3 = 1.73 cm²
Area of a regular hexagon = 6 × Area of an equilateral triangle
= 6 × √3 = 6 × 1.73 = 10.38 cm²
Differences between the areas = 12.56 – 10.38 = 2.18 cm²
Question 3.
In the pictures below, circles are drawn through the vertices of a square and a rectangle.
Calculate the areas of the circles.
Answer:
Figure 1
Diameter of the circle = diagonal of the square
Diameter = \(\sqrt{(3)^2+(3)^2}=\sqrt{2 \times 9}\) = 3√2 cm
Area of the circle = πr² = π × \(\left(\frac{3 \sqrt{2}}{2}\right)^2\) = 4.5 π cm²
Figure 2
Diameter = \(\sqrt{(4)^2+(2)^2}=\sqrt{20}\) cm
Radius = \(\frac{\text { Diameter }}{2}=\frac{\sqrt{20}}{2}\) cm
Area of the circle = π × \(\left(\frac{\sqrt{20}}{2}\right)^2\) = 5 π cm²
Question 4.
Draw a square and draw circles with its vertices as centres and radius as half the side. Draw another square composed of four smaller squares of the same size as the first square, and draw a circle that just fits inside it. Prove that the area of the large circle is the sum of the areas.of the four small circles.
Answer
Figure 1
Radius of one small circle = r cm
Area of one small circle = πr² cm²
Area of the four small circles – 4πr² cm²
Figure 2
Radius of large circle = sum of the radius of two small circles = 2r
Area of the large circle = π(2r)² = 4πr² cm²
The sum of the areas of four small circles is equal to the area of the large circle.
Question 5.
In the pictures below, the squares are of the same size. Prove that the areas of the green regions (shaded regions) in the pictures are equal.
Answer:
Figure 1
Side of the square = 2a
Area of the square = (2a)² = 4a² cm²
A circle is divided into four equal parts and placed at the four comers of a square.
So, radius of the circular segment = a cm
Area of the circular segment = \(\frac{\pi a^2}{4}\)cm²
Area of the circular segment placed at the four comers of the square = 4 × \(\frac{\pi a^2}{4}\) = πa² cm²
Area of the green region (shaded region) = (4 a² – πa²) cm²
Figure 2
Side of the square = 2a
Area of the square = (2 a)² = 4a² cm²
Radius of the circle = a cm
Area of the circle = πr² cm²
Area of the green region (shaded region) = (4 a² – πa²) cm²
Question 6.
Parts of circles are drawn with the vertices of a regular hexagon as centres and the figure below is cut out. Calculate the area of the figure cut out.
Answer:
In the Figure, circular segments are placed at all the six comers of the regular hexagon. The angle at each comer of the regular hexagon isl20°. Combining three circular segments forms a complete circle. The figure contains six such circular segments, which means they can form two complete circles. In other words, the six circular segments are equivalent to two complete circles.
Radius of the circular segment = 1 cm
Area of the circle = πr² = π cm²
Area of two complete circle = 2 × πr² = 2π cm²
Area of the regular hexagon = \(\frac{1}{2}\) × 2 × √3 × 6 = 6√3 cm²
The area of the cut-down portion = Area of the regular hexagon – Area of two complete circle
= (6√3 — 2π) cm²
Question 7.
Parts of a circle are drawn within a square like this. Prove that the area of blue region (shaded region) is half the area of the square.
Answer:
Side of the square = a cm
Area of the square = a² cm²
Area of the semicircle = \(\frac{πa^2}{2}\) cm²
Area of the region found between two quarter circles = [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)] cm²
Area of the shaded region = area of the semicircle + area of the region found between two quarter circles.
= \(\frac{πa^2}{2}\) + [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)]
= \(\frac{a^2}{2}\) cm²
Circle Measures Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
A wire is bent to form a circle with a diameter of 6 centimeters. What will be the diameter of the circle formed by bending a wire that is twice the length of the first one ?
Answer:
The circumferences of circles change in proportion to their diameters. Therefore, if the length of the wire is doubled, the diameter of the circle formed will also get double. This means the diameter of the new circle will be 12 centimeters.
Question 2.
The circumference of a circle with a diameter of 3 meters has been measured to be approximately 9.42 meters.
i) How can we calculate the circumference of a circle with a diameter of 6 meters without measuring it?
ii) Also, what is the circumference of a circle with a diameter of 1.5 meters?
Answer:
i) Since the circumference of a circle with a diameter of 2 meters is 9.42 meters, the circumference of a circle with a diameter of 6 meters will be twice that amount.
Therefore, it will be 2 × 9.42 = 18.84 meters.
ii) Circumference of a circle with diameter 1.5 meters = \(\frac{9.42}{2}\) = 4.7 meters
Question 3.
A wire of 20 centimeters long has been bent to form a circle. If half of that wire is cut and bent into another circle, what will be the diameter of the new circle?
Answer:
Length of the wire = 20 cm
Half the length of the wire = 10 cm
The length of the wire will be equal to the circumference of the circle.
10 = π × diameter
Diameter = \(\frac{10}{\pi}\) cm
Question 4.
Find the length of a chord which is at a distance of 2 cm from the centre of a circle with a circumference of 8ir centimetres.
Answer:
Circumference of a circle = 8π
2πr = 8π
r = 4cm
Considering triangle POR, (OP)² = (OR)² + (PR)²
4² = 2² + (PR)²
PR² = 16 – 4 = 12
PR = √12 = 2√3 cm
Question 5.
i) What Is the area of the circle?
ii) What is the circumference of the circle?
Answer:
The perimeter of the regular hexagon is equal to 6 times the length of one of its sides.
Length of one of its sides = \(\frac{24}{6}\) = 4cm
The regular hexagon is divided into six congruent isosceles triangles.
The length of the side is equal to the radius of the circle.
i) Radius of the circle = 4 cm
ii) Perimeter of the circle = 2πr
= 2 × π × 4
= 8π cm
Question 6.
Calculate the circumference and area of a circle with a radius of 8 centimeters.
Answer:
Radius = 8 cm
Circumference = 2πr = 2 × π × 8 = 16π cm
Area = πr² = π × 8² = 64 π cm²
Question 7.
The diameter of a circle is 10 centimeters. Calculate the area of the circle.
Answer:
Diameter of a circle = 10cm
Radius = \(\frac{10}{2}\) = 5cm
Area of the circle = πr² = π × 5²
= 25π cm²
Question 8.
i) Calculate the perimeter of a wheel of radius 15 cm. What is the distance covered by the wheel in 5 rotations?
ii) What is the distance covered in 5 rotations by another wheel of twice the radius of the circle?
Answer:
i) Radius = 15cm
Perimeter = 2πr = 2 × π × 15 = 30 π cm
Distance travelled when the circle completes 5 rotations = 5 × 30π = 60π cm
ii) Radius = 2 × 15 = 30 cm
Perimeter = 2πr = 2 × π × 30 = 60 π cm
Distance travelled when the circle completes 5 rotations = 5 × 60π = 300π cm
Question 9.
In the diagram, AB is the diameter of the larger semicircle. The segments AC, CD, and DB are the diameters of the smaller semicircles, where AC = 4 centimetres, CD = \(\frac{A C}{2}\), DB = \(\frac{C D}{2}\).What is the area of the shaded region?
Answer:
Area of a circle with diameter AB = πr² = π(3.5)² = 12.25π cm²
Area of a circle with diameter AC = πr² = π(2)² = 4π cm²
Area of a circle with diameter CD = πr² = π(1) = π cm²
Area of a circle with diameter BD = πr² – π(0.5)² = 0.25π cm²
Area of the shaded region = Area of a circle with diameter AB – (Area of a circle with diameter AC + Area of a circle with diameter CD+ Area of a circle with diameter BD)
= 12.25π – (4π + π + 0.25π) = 7π cm²