RD Sharma Chapter 16 Class 9 Maths Exercise 16.2 Solutions

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-

1. Congruence of circles and arcs
• Congruent circles
• Congruent arcs
1. 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

1. A diameter/ radius is perpendicular to a chord, and then it divides the chord and its arc.
2. Two chords are congruent, and then their equal arcs are congruent.
3. A diameter/ radius is perpendicular to a chord, and then it divides the chord and its arc.
4. 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

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