Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 11 Parts of Circles Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 11 Solutions Parts of Circles
Parts of Circles Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 11 Parts of Circles Solutions Questions and Answers
Class 9 Maths Chapter 11 Kerala Syllabus – Length And Angle
Textal Questions and Answers
Question 1.
In a circle, the length of an arc of central angle 400 ¡s 3 centimeters. What is the circumference of the circle? And the radius?
Answer:
Central angle = 40°
Length ofan arc = 3 cm
2πr × \(\frac{40}{360}\) = 3π
2r × \(\frac{1}{9}\) = 3
2r = 3 × 9 = 27cm
Circumference = 2πr = 2πr cm
Radius, r = \(\frac{27}{2}\) cm
Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimeters.
i) What is the length of an arc of central angle 75° in this circle?
ii) What is the length of an arc of central angIe 75° in a circle with radius one and a half times this circle?
Answer:
Central angle = 25°
Length of an arc = 4 cm
2πr × \(\frac{25}{360}\) = 4
2πr = 4 × \(\frac{360}{25}\)
=4 × \(\frac{72}{5}=\frac{288}{5}\)
i) Length of an arc of central angle 75° is
2πr × \(\frac{x}{360}=\frac{288}{5} \times \frac{75}{360}\) = 12cm
ii) Length of an arc of central angIe 75° with radius 1\(\frac{1}{2}\)r
2π × \(\frac{3}{2} \mathrm{r} \times \frac{x}{360}=\frac{288}{5} \times \frac{3}{2} \times \frac{75}{360}\) = 18 cm
Question 3.
From a bangle of radius 3 centimeters, a small piece is to be cut off to make a ring of radius centimeter 2
i) What should be the central angle of the piece to be cut off?
ii) The remaining part of the bangle is bent to make a smaller bangle. What is its radius?
Answer:
i) Perimeter of the bangle having radius 3 cm = 2πr = 2 × π × 3 = 6π cm
Perimeter of the ring having radius \(\frac{1}{2}\) cm = 2πr = 2 × π × \(\frac{1}{2}\) = π cm
Since the circumference of the ring and the arc length of the circle are equal Arc length of the bangle = π cm
2πr × \(\frac{x}{360}\) = π
Central angle, x = 360 × \(\frac{1}{6}\) = 60°
Central angle of the piece to be cut off = 60°
ii) Central angle of the remaining part = 360° – 60°= 300°
Arc length of the remaining part = 2πr × \(\frac{300}{360}\) = 5 π cm
Arc length of the remaining part = circumference of the new bangle
Circumference of the new bangle = 5π
2πr = 5π
r = \(\frac{5}{2}\) = 2.5cm
Question 4.
With each vertex of an equilateral triangle as center, an arc of a circle which passes through the other two vertices is drawn to get the given figure. What is its perimeter?
Answer:
Central angle of an equilateral triangle, x = 60°
Arc length = 2πr × \(\frac{x}{360}\)
= 2 × π × 4 × \(\frac{60}{360}\) = \(\frac{4 \pi}{3} cm\) cm
= 3 × \(\frac{4 \pi}{3} cm\) = 4π cm
Question 5.
With each vertex of a regular octagon as center an arc of a circle is drawn and the resulting figure is cut off, as in the pictures below:
What is the perimeter of the figure cut off.’
Answer:
Sum of the interior angle of a regular octagon = (n – 2) × 180°
= (8 – 2) × 180°
= 6 × 180° = 1080°
Angle measure at a vertex = \(\frac{1080^{\circ}}{8}\) = 135°
Centrai angle, x = 135°
Radius of the circular arc = 1 cm
Arc length = 2πr × \(\frac{x}{360}\)
= 2 π × 1 × \(\frac{135}{360}\)
Perimeter of the figure cut off = 8 × Arc length
= 8 × 2 × π × 1 × \(\frac{135}{360}\)
= 6 π cm
Class 9 Maths Kerala Syllabus Chapter 11 Solutions – Angles And Areas
Textual Questions And Answers
Question 1.
i) In a circle of radius 3 centimeters, What is the area of a sector of central angle 120°?
ii) What is the area of a sector of the same central angle in a circle of radius 6 centimeters?
Answer:
i) Radius = 3 cm
Central angle, x = 120°
Area of a sector = πr² × \(\frac{x}{360}\)
= π × 3² × \(\frac{120}{360}\)
= π × 9 × \(\frac{1}{3}\)
= 3 π cm2
ii) Radius = 6 cm
Central angle, x = 120°
Area of a sector = πr² × \(\frac{x}{360}\)
= π × 6² × \(\frac{120}{360}\)
= π × 36 × \(\frac{1}{3}\)
= 12 π cm²
Question 2.
Find the area of green region(shaded region) in the picture below:
Answer:
Central angle, x = 120°
Radius of the larger sector = 3 cm
Area of the larger sector = πr² × \(\frac{x}{360}\)
= π × 9 × \(\frac{120}{360}\)
= 3 π cm²
Radius of the smaller sector = 2 cm
Area of the smaller sector = πr² × \(\frac{x}{360}\)
= π × 4 × \(\frac{120}{360}\)
= 1.33 π cm²
Area of the shaded region = Area of the larger sector – Area of the smaller sector
= 3π – 1.33 π
= 1.67 π cm²
Question 3.
In the picture below, arc of a circle is drawn with each vertex of an equilateral triangle as center and half the side as radius:
Calculate the area of the blue region (shaded region).
Answer:
Area of the shaded region = area of the equilateral triangle – 3 × area of the sector
= \(\frac{\sqrt{3}}{4}\) × 4² – 3 × π × 2² × \(\frac{60}{360}\)
= (4√3 – 2π) cm²
Question 4.
In the picture below, arcs of circles are drawn centered on two opposite vertices of a square and passing through the other two vertices:
What is the area of green region (shaded region)?
Answer:
Area of the shaded region = 2 × area of the sector – area of the square
= 2 × π × 4² × \(\frac{90}{360}\) – 4 × 4
= (8π – 16) cm²
Question 5.
The picture below shows two circles of the same radii, each passing through the center of the other:
Calculate the area of the region common to both.
Answer:
In the figure, triangle ABC and ABD are equilateral triangle.
Area of the region between the two circles = 2 × area of the sector – area of the equilateral triangle
= 2 × π × 2² × \(\frac{60}{360}\) – 2 × \(\frac{\sqrt{3}}{4}\) × 2²
= (\(\frac{8 \pi}{3}\) – 2√3)cm²
Question 6.
The picture below shows semicircles drawn with the sides of a right triangle as diameters:
Prove that the area of the largest semicircle is the sum of the areas of the other two.
Answer:
Area of the semicircle with diameter AB = \(\frac{1}{2}\) × π × \(\left(\frac{A B}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{A B^2}{4}=\frac{\pi}{8}\)AB²
Area of the semicircle with diameter BC = \(\frac{1}{2}\) × π × \(\left(\frac{B C}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{B C^2}{4}=\frac{\pi}{8}\)BC²
Area of the semicircle with diameter AC = \(\frac{1}{2}\) × π × \(\left(\frac{A C}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{A C^2}{4}=\frac{\pi}{8}\)(AC)²
Sum of the areas of smaller semicircles = \(\frac{\pi}{8}\)(AB)² + \(\frac{\pi}{8}\)(BC)²
= \(\frac{\pi}{8}\) x ((AB)² + (BC)²)
= \(\frac{\pi}{8}\)(AC)²
Parts of Circles Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
In a circle, if the arc length corresponding to a central angle of 40° is 8n centimeters, what will be the arc length for a central angle of 100° in the same circle?
Answer:
Central angle = 40°
Arc length = 8n cm
2πr × \(\frac{40}{360}\) = 8n
2πr = 8π × \(\frac{360}{40}\) = 8π × 9 = 72 π cm
When the central angle is 100°, then the arc length is
2πr × \(\frac{100}{360}\) = 12 π × \(\frac{100}{360}\) = 20 π cm
Question 2.
i) Calculate the area of a circular disc of radius 10 centimeter
ii) It is divided into four parts by drawing two perpendicular diameters .What is the area of each of the sectors obtained?
Answer:
i) Radius = 10 cm
Area of circular disc = πr² = 100π cm²
ii) Area of the sector = πr² × \(\frac{x}{360}\)
= 100π × \(\frac{90}{360}\)
= 25π cm²
Question 3.
In the figure, two arcs are drawn from a point. The distance between the two arcs is 3 cm. The radius of the first arc is 6 cm. Find the difference between the lengths of the two arcs.
Answer:
Arc length having radius 6 cm 2πr × \(\frac{60}{360}=\frac{6 \pi}{3}\)= 2π cm
Arc length having radius 9 cm 2πr × \(\frac{60}{360}=\frac{9 \pi}{3}\) = 3π cm
Difference of arc length = 3π – 2π = π cm
Question 4.
A clock’s minute hand has a length of 10 cm. What is the distance travelled by it between 1:05 PM and 1:40 PM?
Answer:
The minute hand makes a full rotation of 360° in one complete cycle.
In 5 minutes, it rotates \(\frac{360}{12}\) =30°
To rotate from 1:05 to 1:40, it takes 7 × 30 = 210°
Central angle formed by the minute hand = 210°
Distance travelled = 2πr × \(\frac{210}{360}\)
= 2 × π × 10 × \(\frac{210}{360}=\frac{70}{6}\)π cm
Question 5.
A regular hexagon with sides measuring 4 cm has a circle drawn with a diameter equal to half the length of its sides from each vertex. Find the area of the shaded region in the figure?
Answer:
The area of a regular hexagon with sides measuring 4 cm = \(\frac{\sqrt{3}}{4}\) × 4² = 4√3 cm²
The angle at one vertex of the regular hexagon is 60°
The area of the circular segment at one vertex = π × r² × \(\frac{60}{360}\)
= π × r² × \(\frac{60}{360}\)
= \(\frac{2 \pi}{3}\)
The area of the circle at all three vertices = 3 × \(\frac{2 \pi}{3}\) = 2π cm²
Area of the shaded region = area of the regular hexagon – area of the circular segments at the vertices.
= (4√-3 – 2π) cm²
Question 6.
In the diagram, a regular hexagon with sides measuring 2 cm has circular segments with a radius of 1 cm at each vertex. Find the area of the shaded region?
Answer:
The area of a regular hexagon with sides measuring 2 cm = 6 × \(\frac{\sqrt{3}}{4}\) × 22 = 6√3 cm²
Interior angle of the regular hexagon =120°
Exterior angle = 240°
The area of the circular segment at one vertex = π × r² × \(\frac{240}{360}\) = π × 12 × \(\frac{240}{360}\)
= \(\frac{2 \pi}{3}\)
The area of 6 circular segments = 6 × \(\frac{2 \pi}{3}\) = 4π
Area of the shaded portion = area of a regular hexagon with sides measuring 2 cm + area of 6 circular segments
= (6√3 + 4π) cm²