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Chapter 16: Circles Exercise – 16.3

Question: 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 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.

Three girls Ishita, Isha and Nisha are playing a game by standing on a circleAR = AS = 24/2 = 12 cm

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

In ΔOAR,

OA2 + AR2 = OR2

OA2 + 122 = 202

OA2 = 400 − 144 = 256 m2

OA = 16 m

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

So in ΔRSM, ∠RCS = 90° and RC = CM

Area of ΔORS = 1/2 × OA × RS

⇒ 1/2 × RC × OS = 1/2 × 16 × 24

⇒ RC × 20 = 16 × 24

⇒ RC = 19.2

⇒ RM = 2(19.2) = 38.4 m

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

 

Question: 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

A circular park of radius 40 mSo, ABC is an equilateral triangle

OA (radius) = 40 m

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:

OA/OD = 2/1

⇒ 4OM/OD = 2/1

⇒ OD = 20m

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

= 60 m

In ΔADC

By using Pythagoras theorem

AC2 = AD2 + DC2

AC2 = 602 + (AC2)2

AC2 = 3600 + AC2/4

⇒ 3/4 AC2 = 3600

⇒ AC2 = 4800

⇒AC = 40√3 m

So, length of string of each phone will be 40√3 m.


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