Page 4 - Lesson note 4 Equal Chords-equidistant- circle Ch.- 10 Circle

2
2
2
r = (x+3) + (5/2) …..(2)
From equations (1) and (2), we get
2
2
2
2
(x+3) + (5/2) = x + (11/2)
2
2
2
2
2
x + 6x + 9 + 25/4 = x + 121/4 (using identity, ( a+ b) = a + b + 2ab )

6x = 121/4 – 25/4 − 9
6x = 15
Or x = 15/6 = 5/2
Substitute the value of x in equation (1), and find the length of radius,
2
2
2
r = (5/2) + (11/2)
= 25/4 + 121/4
= 146/4
Or r = √146/4 cm
Example 2.
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.
Solution:
Let R, S and M be the position of Ishita, Isha and Nisha respectively.
Since OA is a perpendicular bisector on RS, so AR = AS = 24/2 = 12 cm
Radii of circle = OR = OS = OM = 20 cm (Given)
In ΔOAR:
By Pythagoras theorem,
2
2
2
(OA) +(AR) =(OR) By Pythagoras theorem,

2 2 2
(OA) +(12) =20
(OA) = 400 – 144 = 256
2
Or, OA = 16 m …(1)
From figure, ORSM is a kite since OR = OM and RS = SM.
We know that, diagonals of a kite are perpendicular and the diagonal common to both
the isosceles triangles is bisected by another diagonal.
0
So in ΔRSM, ∠RCS = 90 and RC = CM …(2)
Now, 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