RD Sharma Solutions -Ex-16.3, Circles, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Class 9 : RD Sharma Solutions -Ex-16.3, Circles, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-16.3, Circles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9

Q1) Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha.

Solution:

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

AR = AS =  = 12cm

OR = OS = OM = 20 cm                     [Radii of circle]

InΔOAR,

OA2+AR2 = OR2

OA2+122 = 202

OA2 = 400−144 = 256m2

OA = 16m

We know that, in an isosceles triangle altitude divides the base.

So in ΔRSM,∠RCS = 900 and RC = CM

Area of ΔORS = ×OA×RS

×RC×OS = ×16×24

⇒ RC×20 = 16×24

⇒ RC = 19.2

⇒ RM = 2(19.2) = 38.4m

So, the distance between Ishita and Nisha is 38.4m.

 

Q2) 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) = 40m

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.

We also know that median intersect each other at the ratio 2 : 1.

As AD is the median of equilateral triangle ABC, we can write:

⇒ OD = 20m

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

= 60 m

InΔADC

By using Pythagoras theorem

AC2 = AD2+DC2

AC2 = 602

AC2 = 3600+

 = 3600

⇒ AC2 = 4800

⇒ AC =

So, length of string of each phone will be 

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