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.
Since, AB = BC = CA. So, ABC is an equilateral triangle
Radius = OA = 40 m (Given)
We know, medians of equilateral triangle pass through thecircumcentre and intersect each other at the ratio 2 : 1.
Here AD is the median of equilateral triangle ABC, we can write:
OA/OD = 2/1
or 40/OD = 2/1
or OD = 20 m
Therefore, AD = OA + OD = (40 + 20) m = 60 m
Now, In ΔADC:
By Pythagoras theorem,
AC2 = AD2 + DC2
AC2 = 602 + (AC/2) 2
AC2 = 3600 + AC2 / 4
3/4 AC2 = 3600
AC2 = 4800
or AC = 40√3 m
Therefore, length of string of each phone will be 40√3 m.