Answer:

9.6m

Step-by-step explanation:

let R,M and N be the position . Rejia ,Mary ,Nilakshi

given RM= MN =6

r= 65 m

construction - OR , OM and OL perpendicular RM are drawn

let , RK = X

RM PERPENDICULAR OL bisects it

therfore RL = LM = 3cm

therefore, in rt triangle OLM

OL square + LM square =OM square

= " " + 3 square = square

= ol square = 25 - 9

= root over 16

= 4 cm

now , ar (ORK) = 1/2 × RM ×OL = 1/2 × OM× RK

= 1/2 × 6 ×4=1/2×5× x

= 12 ×2/5 = x

therefore RN = 2 ×24/5 =48/5 = 9.6 cm

hope you understand

plz mark as brainliest

Verified answer

Answer:

[tex]\boxed{ \bf{ \:Distance\:between\:Rejia\:and\:Nilakshi \: is \: 9. 6 \: m \: }}\\ [/tex]

Step - by - step explanation:

Given that, Three girls Rejia, Mary and Nilakshi are playing a game by standing on a circle of radius 5 cm drawn in a park. Rejia throws a ball to Mary, Mary to Nilakshi, Nilakshi to Rejia.

Let assume that A, B and C represents the position of theee girls Rejia, Mary and Nilakshi on a circle of radius 5 m. 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 = 5 m.

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.

[tex]\sf\implies \sf \: AD = DC = x \: (say) \\ [/tex]

Let assume that OD = y

Now, In right-angle triangle OAD

By Using Pythagoras Theorem, we have

[tex]\sf \: {OA}^{2} = {OD}^{2} + {AD}^{2} \\ [/tex]

[tex]\sf \: {5}^{2} = {y}^{2} + {x}^{2} \\ [/tex]

[tex]\sf\implies \sf \: {x}^{2} + {y}^{2} = 25 - - - (1)\\ [/tex]

Now, In right-angle triangle ABD

By using Pythagoras Theorem, we have

[tex]\sf \: {AB}^{2} = {BD}^{2} + {AD}^{2} \\ [/tex]

[tex]\sf \: {6}^{2} = {(5 - y)}^{2} + {x}^{2}\\ [/tex]

[tex]\sf \: 36 = 25 + {y}^{2} - 10y + {x}^{2} \\ [/tex]

[tex]\sf \: 36 = 25 - 10y + ({x}^{2} + {y}^{2})\\ [/tex]

[tex]\sf \: 36 = 25 - 10y + 25 \: \: \: \: \boxed{ \sf{ \: \because \: {x}^{2} + {y}^{2} = 25 \: }}\\ [/tex]

[tex]\sf \: 36 = 50 - 10y\\ [/tex]

[tex]\sf \: 10y = 50 - 36\\ [/tex]

[tex]\sf \: 10y = 14 \\ [/tex]

[tex]\sf\implies \: y = \dfrac{14}{10} = \dfrac{7}{5} \: m\\ [/tex]

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

[tex]\sf \: {x}^{2} + {\bigg(\dfrac{7}{5} \bigg) }^{2} = 25 \\ [/tex]

[tex]\sf \: {x}^{2} + \dfrac{49}{25} = 25\\ [/tex]

[tex]\sf \: {x}^{2} = 25 - \dfrac{49}{25} \\ [/tex]

[tex]\sf \: {x}^{2} = \dfrac{625 - 49}{25} \\ [/tex]

[tex]\sf \: {x}^{2} = \dfrac{576}{25} \\ [/tex]

[tex]\sf \: {x}^{2} = {\bigg(\dfrac{24}{5} \bigg) }^{2}\\ [/tex]

[tex]\sf\implies \: x = \dfrac{24}{5} \: m\\ [/tex]

Now,

[tex]\sf \: AC=2AD\\ [/tex]

[tex]\sf \: AC=2x\\ [/tex]

[tex]\sf \: AC=2 \times \dfrac{24}{5} \\ [/tex]

[tex]\sf \: AC= \dfrac{48}{5} \\ [/tex]

[tex]\sf\implies \sf \: AC= 9. 6 \: m \\ [/tex]

Hence,

[tex]\sf\implies \: Distance\:between\:Rejia\:and\:Nilakshi \: is \: 9. 6 \: m [/tex]