Draw perpendiculars OA and OB on RS and SM respectively.
AR=AS= \(\frac{6}{2}\) =3 m
OR = OS = OM = 5 m. (Radii of the circle)
In ∆OAR,
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 − 9) m2 = 16 m2
OA = 4m
ORSM will be a kite (OR = OM and RS = SM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
∠RCS will be of 90° and ∠RC = ∠CM
Area of ∆ORS = \(\frac{1}{2}\)× OA×RS
\(\frac{1}{2}\)×RC×OS= \(\frac{1}{2}\)×4×6
RC×5=24
RC=4.8
RM=2RC
=2(4.8)=9.4
Therefore, the distance between Reshma and Mandip is 9.6 m.
In Fig. 9.23, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig).
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.