Question

Three girls Reshma , Salma and mandip are playing game by standing on a circle of radius 5cm drawn in a park. Reshma throws a ball to Salma , Salma to mandip and mandip to Reshma. If the distance between Salma and Mandip is 6cm each, What is the distance between Reshma and Mandip
?

Answer 1

Answer:

9.6 cm

Step-by-step explanation:

Let the positions of Reshma, Salma and Mandip be represented as A, B and C respectively.

From the question, we know that AB = BC = 6cm.

So, the radius of the circle i.e. OA = 5cm

Now, draw a perpendicular BM ⊥ AC.

Since AB = BC, ABC can be considered as an isosceles triangle. M is mid-point of AC. BM is the perpendicular bisector of AC and thus it passes through the centre of the circle.

Now,

let AM = y and

OM = x

So, BM will be = (5-x).

By applying Pythagorean theorem in ΔOAM we get,

OA2 = OM2 +AM2

⇒ 52 = x2 +y2 — (i)

Again, by applying Pythagorean theorem in ΔAMB,

AB2 = BM2 +AM2

⇒ 62 = (5-x)2+y2 — (ii)

Subtracting equation (i) from equation (ii), we get

36-25 = (5-x)2 + y2 -x2-y2

Now, solving this equation we get the value of x as

x = 7/5

Substituting the value of x in equation (i), we get

y2 +(49/25) = 25

⇒ y2 = 25 – (49/25)

Solving it we get the value of y as

y = 24/5

Thus,

AC = 2×AM

= 2×y

= 2×(24/5) m

AC = 9.6 m