RD Sharma 2020 solution class 9 chapter 15 Circles Exercise 15.3

Exercise 15.3

Page-15.47

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 24 m each, what is the distance between Ishita and Nisha.

Answer 1:

Using the data given in the question we can draw a diagram that looks like fig (1).

             

From the figure we see that it is an isosceles triangle that has been circumscribed in a circle of radius R = 20 m.

The equal sides of the isosceles triangle measure 24 m in length. The length of the base of the isosceles triangle is what we are required to find out.

Since it is an isosceles triangle the perpendicular dropped from the vertex A to the base will pass though the circumcentre of the triangle. Let ‘h’ be the height of the triangle.

Since the triangle has been circumscribed by a circle of radius ‘R’ the length of the distances from ‘O’ to any of the three persons would be ‘R’. 

Let the positions of the persons Isha, Ishita and Nisha be replaced by ‘A’, ‘B’ and ‘C’ respectively. And let the length of the unknown base be, BC = 2x m.

This is shown in the fig (2).

Now, consider the triangle ΔBOD, we have

At the same time consider, we have

Substitute this value in equation we got for ‘R’, we get

Now we have got the value of the height of the triangle as h = 14.4 m.

Substituting the value of h in the below equation,

Now we have the value of x = 19.2 m

We need the value of

Hence, the distance between Ishita and Nisha is .

Question 2:

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and hands to talk to each other. Find the length of the string of each phone.

Answer 2:

From the given data, we see that the given situation is equivalent to an equilateral triangle circumscribed by a circle.

Let the positions of the three boys Ankur, Amit and Anand be denoted by the points ‘A’,’B’ and ‘C’. Let ‘O’ be the centre of the circle, ‘a’ is the sides of the equilateral triangle and ‘R’ is its circumradius.

Now, in an equilateral triangle with side ‘a’, the height, ‘h’ of the equilateral triangle would be,

AB = BC = CA
Therefore,  $∆ABC$is an equilateral triangle.
OA = 40 m
Medians of equilateral triangle pass through the circumcentre (O)  of the equilateral triangle ABC. We know that medians intersect each other in the ratio 2 : 1. As AD is the median of equilateral triangle ABC, we can write

$$\begin{gathered} \frac{OA}{OD}=\frac{2}{1} \\ \Rightarrow \frac{40}{OD}=\frac{2}{1} \\ \Rightarrow OD=20 \mathrm{m} \\ AD=AO+OD=\left(40+20\right) \mathrm{m}=60 \mathrm{m} \\ \mathrm{In} ∆ADC, \\ A{C}^2=A{D}^2+D{C}^2 \\ \Rightarrow A{C}^2={\left(60\right)}^2+\frac{A{C}^2}{4} \left[\because AC = BC,\mathit{ }DC=\frac{1}{2}BC\Rightarrow DC=\frac{1}{2}AC\right] \\ \Rightarrow \frac{3A{C}^2}{4}=3600 \\ \Rightarrow A{C}^2=4800 \\ \Rightarrow AC=40\sqrt{3} \mathrm{m} \end{gathered}$$
Hence the length of the string of each phone is