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5. Three girls Reshma, Salma and Mandip areplaying a game by standing on a circle of radius5m drawn in a park. Reshma throws a ball toSalma, Salma to Mandip, Mandip to Reshma. Ifthe distance between Reshma and Salma andbetween Salma and Mandip is 6m each, what isthe distance between Reshma and Mandip?

Answer»

Let O bethe centre of the circle. A, B and C represent the positions of Reshma, Salmaand Mandip.

AB = 6cm and BC = 6cm.

Radius OA = 5cm.(given)

Draw BM ⊥ AC

ABC is an isosceles triangle as AB = BC, M ismid-point of AC.

BM isperpendicular bisector of AC and thus it passes through the centre of thecircle.

Let AM = y and OM = x then BM = (5-x).

In ΔOAM,

OA²=OM²+AM² ( by Pythagorastheorem)

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 = 25y²=25 – 49/25y²=(625 – 49)/25y²=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 Mandipis 9.6 m.

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thanxx but kuch zyada bada method hai



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