Question

Three girls Reshma,Salma And Mandeep are playing a game by standing on a circle of radius 5 M drawn in a park Reshma throws a ball to Salma, Salma to Mandeep and Mandeep to Reshma, if the distance between Reshma and Salma and between Salma And Mandeep is 6 m each then what is the distance between Reshma and Mandeep?

Answer 1

Answer:

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

Step-by-step explanation:

Given that, Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandeep, Mandeep to Reshma.

Let assume that A, B and C represents the position of theee girls Reshma, Salma and Mandeep 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,

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