Arun
Last Activity: 6 Years ago
Let O be
the centre of the circle. A, B and C represent the positions of Reshma, Salma
and Mandip.
AB = 6cm and BC = 6cm.
Radius OA = 5cm.
(given)
Draw BM ⊥ AC
ABC is an isosceles triangle as AB = BC, M is
mid-point of AC.
BM is
perpendicular bisector of AC and thus it passes through the centre of the
circle.
Let AM = y and OM = x then BM = (5-x).
In ΔOAM,
OA²=OM²+AM² ( by Pythagoras
theorem)
5²=x²+ y²—
(i)
In ΔAMB,
AB²=BM² +AM²
(by Pythagoras theorem)
6²= (5-x)²+y² — (ii)
Subtracting (i) from (ii),
36 – 25 = (5-x)² -x²
11 = (25+
x²– 2×5×x) - x²
11= 25+x²-10x - x²
11= 25-10x
10x = 14
x= 7/5
Substituting the value of x in (i), we get
y²+ 49/25 = 25
y² =
25 – 49/25
y² =
(625 – 49)/25
y²=
576/25
y = 24/5
Thus,
AC = 2×AM
AC = 2×y
AC= 2×(24/5) m
= 48/5 m = 9.6 m
Hence, the Distance between Reshma and Mandip
is 9.6 m.
