Question -
Answer -
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,
OA2+AR2=OR2
OA2+122=202
OA2 = 400 – 144 =256
Or OA = 16 m …(1)
From figure, OABC is a kite since OA = OC and AB = BC. We knowthat, diagonals of a kite are perpendicular and the diagonal common to both theisosceles triangles is bisected by another diagonal.
So in ΔRSM, ∠RCS = 900 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
Since RC = CM (from (2), we have
RM = 2(19.2) = 38.4
So, the distance between Ishita and Nisha is 38.4 m.