RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions help the students better understand the congruence of circles and the arcs. The entire chapter 16 comprises different topics on Circles explained thoroughly with a stepwise approach.
Upon exercising the RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions, the student can get stronger with his/ her fundamentals about the circle while boosting deep understanding to address problems around the circle. Learners have access to download the PDF that consists of complete explanations and examples, which helps students to score well in the exam.
Learn about RD Sharma Chapter 16 Class 9
Download RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions PDF
Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.2
Important Definitions RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions
In this RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions, we will discuss the topics and subtopics mentioned in the following points-
- Congruence of circles and arcs
- Congruent circles
- Congruent arcs
- Some essentials result in Congruent arcs and chords
Congruence of Circles and Arcs
Here we will learn about the formal definition of congruence for each shape. Get down to know the complete information.
Congruent Circles
Congruent Circles are circles when two circles have congruent radii.
Congruent Arcs
Congruent Arcs are the arcs when two arcs are equal in measure and segments of congruent circles.
Some Essentials Results on Congruent Arcs and Chords
- A diameter/ radius is perpendicular to a chord, and then it divides the chord and its arc.
- Two chords are congruent, and then their equal arcs are congruent.
- A diameter/ radius is perpendicular to a chord, and then it divides the chord and its arc.
- In the congruent circle, two chords are congruent if they are equidistant to the center.
Examples of RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions
Ques- The radius of a circle is 8 cm, and the length of one of its chords is 12 cm. Get the extent (distance) of the chord from the center.
Solution-
- Radius of circle (OP) = 8 cm (Given)
- Chord (PQ) = 12cm (Given)
Draw a perpendicular OR on PQ.
We know, perpendicular from center to chord bisects the chord
Which implies, PR = QR = 12/2 = 6 cm
In right ΔORP:
Using Pythagoras theorem,
OP2 = PR2 + OR2
64 = 36 + OR2
OR2 = 64 – 36 = 28
or OR = √28 = 5.291 (approx.)
The extent (distance) of the chord from the center is 5.291 cm.
Ques: Find the length of a chord, which is at an extent (distance) of 5cm from the center of the circle of radius 10cm.
Solution-
Distance of the chord from the centre = OZ = 5 cm (Given)
Radius of the circle = OX = 10 cm (Given)
In ΔOZX:
Using Pythagoras theorem,
OX2 = XZ2 + OZ2
100 = XZ2 + 25
XZ2 = 100 – 25 = 75
XZ = √75 = 8.66
As, perpendicular from the center to chord bisects the chord.
Therefore, XZ = YZ = 8.66 cm
= XY = XZ + YZ = 8.66 + 8.66 = 17.32
Answer: XY = 17.32 cm
Ques: Find the length of a chord, which is at an extent (distance) of 4cm from the center of the circle of radius 6 cm.
Solution-
- Distance of the chord from the center = OR = 4 cm (Given)
- Radius of the circle = OP = 6 cm (Given)
In ΔORP:
Using Pythagoras theorem,
OP2 = PR2 + OR2
36 = PR2 + 16
PR2 = 36 – 16 = 20
PR = √20 = 4.47
Or PR = 4.47cm
As, perpendicular to the centre to chord bisects the chord.
Therefore, PR = QR = 4.47 cm
= PQ = PR + QR = 4.47 + 4.47 = 8.94
Answer- PQ = 8.94 cm
Benefits of RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions
This RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions can be useful for the students in exercising numerous questions revolving around the theories of circles, be it about the radius, sector, segment, and circle diameter. A student can certainly expect a good score upon practicing these questions. Specifically, these solutions can be beneficial for students to practice within a minimum time period before the exam, yet can be hopeful about scoring good marks.