Question

5. Three girls Reshma, Salma and Mandip are
playing a game by standing on a circle of radius
5m drawn in a park. Reshma throws a ball to
Salma, Salma to Mandip, Mandip to Reshma. If
the distance between Reshma and Salma and
between Salma and Mandip is om each, what is
the distance between Reshma and Mandip?
nipoulor nork of radius ? Om is situated in a colo​

Answer 1

Answer:

Step-by-step explanation:

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 - x^2

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.

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