Get free RD Sharma Solutions for Class 9 Maths Chapter 16 Exercise 16.2 – Circles here. In this exercise, students will study about congruence of circles and arcs. Solving the RD Sharma Class 9 Maths questions helps to build the basics and gives in-depth understanding of the basics of circles. Click on the link given below to get your PDF now.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.2
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise 16.2 Page number 16.24
Question 1: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.
Solution:
Radius of circle (OA) = 8 cm (Given)
Chord (AB) = 12cm (Given)
Draw a perpendicular OC on AB.
We know, perpendicular from centre to chord bisects the chord
Which implies, AC = BC = 12/2 = 6 cm
In right ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
64 = 36 + OC2
OC2 = 64 – 36 = 28
or OC = √28 = 5.291 (approx.)
The distance of the chord from the centre is 5.291 cm.
Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.
Solution:
Distance of the chord from the centre = OC = 5 cm (Given)
Radius of the circle = OA = 10 cm (Given)
In ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
100 = AC2 + 25
AC2 = 100 – 25 = 75
AC = √75 = 8.66
As, perpendicular from the centre to chord bisects the chord.
Therefore, AC = BC = 8.66 cm
=> AB = AC + BC = 8.66 + 8.66 = 17.32
Answer: AB = 17.32 cm
Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.
Solution:
Distance of the chord from the centre = OC = 4 cm (Given)
Radius of the circle = OA = 6 cm (Given)
In ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
36 = AC2 + 16
AC2 = 36 – 16 = 20
AC = √20 = 4.47
Or AC = 4.47cm
As, perpendicular from the centre to chord bisects the chord.
Therefore, AC = BC = 4.47 cm
=> AB = AC + BC = 4.47 + 4.47 = 8.94
Answer: AB = 8.94 cm
Question 4: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.
Solution:
Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm
Draw perpendiculars OP on CD and OQ on AB
Let OP = x cm and OC = OA = r cm
We know, perpendicular from centre to chord bisects it.
Since OP⊥CD, we have
CP = PD = 11/2 cm
And OQ⊥AB
AQ = BQ = 5/2 cm
In ΔOCP:
By Pythagoras theorem,
OC2 = OP2 + CP2
r2 = x2 + (11/2) 2 …..(1)
In ΔOQA:
By Pythagoras theorem,
OA2=OQ2+AQ2
r2= (x+3) 2 + (5/2) 2 …..(2)
From equations (1) and (2), we get
(x+3) 2 + (5/2) 2 = x2 + (11/2) 2
Solve above equation and find the value of x.
x2 + 6x + 9 + 25/4 = x2 + 121/4
(using identity, (a+b) 2 = a2 + b2 + 2ab )
6x = 121/4 – 25/4 − 9
6x = 15
or x = 15/6 = 5/2
Substitute the value of x in equation (1), and find the length of radius,
r2 = (5/2)2 + (11/2) 2
= 25/4 + 121/4
= 146/4
or r = √146/4 cm
Question 5: Give a method to find the centre of a given circle.
Solution:
Steps of Construction:
Step 1: Consider three points A, B and C on a circle.
Step 2: Join AB and BC.
Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.
Step 4: This point O is a centre of the circle, because we know that, the Perpendicular bisectors of chord always pass through the centre.
Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
Solution:
From figure, Let C is the mid-point of chord AB.
To prove: D is the mid-point of arc AB.
Now, In ΔOAC and ΔOBC
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [C is the mid-point of chord AB (given)]
So, by SSS condition: ΔOAC ≅ ΔOBC
So, ∠AOC = ∠BOC (BY CPCT)
Therefore, D is the mid-point of arc AB. Hence Proved.
Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.
Solution:
Form figure: PQ is a diameter of circle which bisects the chord AB at C. (Given)
To Prove: PQ bisects ∠AOB
Now,
In ΔBOC and ΔAOC
OA = OB [Radius]
OC = OC [Common side]
AC = BC [Given]
Then, by SSS condition: ΔAOC ≅ ΔBOC
So, ∠AOC = ∠BOC [By c.p.c.t.]
Therefore, PQ bisects ∠AOB. Hence proved.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.2
RD Sharma Solutions Class 9 Maths Chapter 16 Circles Exercise 16.2 is based on the following topics and subtopics:
- Congruence of circles and arcs
-Congruent circles
-Congruent arcs
- Some important results on Congruent arcs and chords
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