RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise 16.3

Chapter 16 Circles RD Sharma Solutions Exercise 16.3 Class 9 Maths

Chapter Name	RD Sharma Chapter 16 Circles Exercise 16.3
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 16.1 Exercise 16.2 Exercise 16.4 Exercise 16.5
Related Study	NCERT Solutions for Class 10 Maths

Exercise 16.3 Solutions

1. Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball o Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.

Solution 

Let R, S and M be the position of Ishita, Isha and Nasha respectively

WKT, in an isosceles triangle altitude divides the base, So in ΔRSM , ∠RCS will be 90° and  
RC = CM. 

Area of ΔORS = 1/2 × OA × RS 
⇒ 1/2 × RC × OS = 1/2 × 16 × 24 
⇒ RC × 20 = 16 × 24 ⇒ RC = 192 ⇒ RM = 2(192) = 38.4 m
So, distance between Ishita and Nisha is 384 m. 

2. A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone. 

Solution

Given that AB = BC = CA

So, ABC is an equilateral triangle

OA (radius) =  40 m

Medians of equilaterals triangles pass through the circumcenter (0) of the equilaterals

triangles ABC

We also know that median intersect each other at the 2:1 As AD is the median of equilaterals triangle ABC, we can write:

OA/OD = 2/7 
⇒ 40m/OD = 2/7
⇒OD = 20m. 
∴ AD = OA + OD = (40 + 20)m 
= 60 m

In ΔADC 

By using Pythagoras theorem