Question

6) Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of
radius5m 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 6m each, what is
the distance between Reshma and Mandip?
​

Answer 1

Answer:

$\boxed{ \sf{ \:Distance\:between\:Reshma\:and\:Mandip \: is \: 9. 6 \: m \: }}$

Step-by-step explanation:

Given that, 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.

Let assume that A, B and C represents the position of theee girls Reshma, Salma and Mandip on a circle of radius 5 cm. Let assume that O be the center of circle.

Construction : Join AB, BC, AC, OB and OA.

So, it is given that AB = BC = 6 m and OA = OB = 5m.

Further, As AB = BC, so draw angle bisector of angle ABC.

So, angle bisector passes through center of circle O and is perpendicular bisector of AC. Let it intersects AC at D.

$\sf\implies \sf \: AD = DC = x \: (say)$

Let assume that OD = y

Now, In right-angle triangle OAD

Using Pythagoras Theorem, we have

$\sf \: {OA}^{2} = {OD}^{2} + {AD}^{2}$

$\sf \: {5}^{2} = {y}^{2} + {x}^{2}$

$\sf\implies \sf \: {x}^{2} + {y}^{2} = 25 - - - (1)$

Now, In right-angle triangle ABD

By using Pythagoras Theorem, we have

$\sf \: {AB}^{2} = {BD}^{2} + {AD}^{2}$

$\sf \: {6}^{2} = {(5 - y)}^{2} + {x}^{2}$

$\sf \: 36 = 25 + {y}^{2} - 10y + {x}^{2}$

$\sf \: 36 = 25 - 10y + ({x}^{2} + {y}^{2})$

$\sf \: 36 = 25 - 10y + 25 \: \: \: \: \boxed{ \sf{ \: \because \: {x}^{2} + {y}^{2} = 25 \: }}$

$\sf \: 36 = 50 - 10y$

$\sf \: 10y = 50 - 36$

$\sf \: 10y = 14$

$\sf\implies \: y = \dfrac{14}{10} = \dfrac{7}{5} \: m$

On substituting the value of y in equation (1), we get

$\sf \: {x}^{2} + {\bigg(\dfrac{7}{5} \bigg) }^{2} = 25$

$\sf \: {x}^{2} + \dfrac{49}{25} = 25$

$\sf \: {x}^{2} = 25 - \dfrac{49}{25}$

$\sf \: {x}^{2} = \dfrac{625 - 49}{25}$

$\sf \: {x}^{2} = \dfrac{576}{25}$

$\sf \: {x}^{2} = {\bigg(\dfrac{24}{5} \bigg) }^{2}$

$\sf\implies \: x = \dfrac{24}{5} \: m$

Now,

$\sf \: AC=2AD$

$\sf \: AC=2x$

$\sf \: AC=2 \times \dfrac{24}{5}$

$\sf \: AC= \dfrac{48}{5}$

$\sf\implies \sf \: AC= 9. 6 \: m$

Hence,

$\sf\implies \: Distance\:between\:Reshma\:and\:Mandip \: is \: 9. 6 \: m$