# RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions

RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions focuses on the key outcomes based on equal chords concerns with Circle. These are important from the examination point of view to address varieties of problems or questions appearing in the exams based on circles. We have provided some examples in the articles, and students may take the help of those questions while practicing for the exam.

The PDF attached to this article has given learners a variety of questions for practice. RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions are prepared by professionals for the students to score well in the exam. PDF is available for free for the learners.

Learn about RD Sharma Chapter 16 Class 9

## Download RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions PDF

Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.3

## Important Questions RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions

Properties of Equal Chords-

1. Equal Chords of the circle are equivalent to the center.
2. In the circle, equal chords subtend corresponding angles at the center.

### Examples Related to RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions

Ques- Three boys Ram, Shyam, and Mohan are playing a game by standing on a circle of radius of 20 m drawn in a park. Ram throws a ball to Shyam, Shyam to Mohan, and Mohan to Ram. If the distance between Ram and Shyam and between Shyam and Mohan is 24 m each, what is the distance between Ram and Mohan?

Solution-

Let R, S, and M be the position of Ram, Shyam, and Mohan respectively.

Since KA is a perpendicular bisector on RS, so AR = AS = 24/2 = 12 cm

Radii of circle = OR = OS = OM = 20cm (Given)

In ΔOAR:

By Pythagoras theorem,

OA2+ AR2= OR2

OA2+ 122= 202

OA2 = 400 – 144 = 256

Or OA = 16m …(1)

From the figure, OABC is a kite since OA = OC and AB = BC.

ΔRSM, ∠RCS = 900 and RC = CM …(2)

So, Area of ΔORS = Area of ΔORS

=>1/2× OA× RS = 1/2 x RC x OS

=> OA × RS = RC x OS

=> 16 x 24 = RC x 20

=> RC = 19.2

Since RC = CM (from (2), we have

RM = 2(19.2) = 38.4

So, the distance between Ram and Mohan is 38.4 m.

Ques- A circular park of a radius of 40m is located in a colony. Three girls Anita, Babita, and Kavita are sitting at the same distance on its boundary all having a toy telephone in their hands to talk to each other. Find the length of the cord of each phone.

Solution-

Since, AB = BC = CA. So, ABC is an equilateral triangle

Radius = OA = 40m (Given)

We know, medians of the equilateral triangle pass by the circumcentre and meet each other at the ratio 2:1.

Here, AD is the median of equilateral triangle ABC, so-

OA/OD = 2/1

or 40/OD = 2/1

or OD = 20 m

Therefore, AD = OA + OD = (40 + 20) m = 60 m

By Pythagoras theorem,

AC2 = 602 + (AC/2) 2

AC2 = 3600 + AC2 / 4

3/4 AC2 = 3600

AC2 = 4800

or AC = 40√3 m

Therefore, length of cord of each telephone will be 40√3 m.

## Frequently Asked Questions (FAQs) of RD Sharma Chapter 16 Class 9 Maths Exercise 16.3 Solutions

Ques- Is this section explicitly meant for addressing questions in circles only?

Ans- Yes, this section is also primarily about questions on circles. However, the students can expect questions on equal chords. Specifically, one can expect very similar questions as provided above.

Ques- What is the equal chords of a circle?

Ans- Equal chords of a circle are equivalent to its center.

Ques- Can a diameter be a chord?

Ans- A chord that crosses through the center of a circle is known as diameter and is the longest chord.

Ques- How many equal chords are there in a circle?

Ans- In a circle, if two chords are equivalent from the center of the circle, then two chords are the same in length.