Three girls Rejia, Mary and Nilakshi are playing
a game by standing on a circle of radius 5m
drawn in a park. Rejia throws a ball to Mary,
Mary to Nilakshi, Nilakshi to Rejia. If the
distance between Rejia and Mary and between
Mary and Nilakshi is 6m each, what is the
distance between Rejia and Nilakshi?
Answers & Comments
Verified answer
Answer:
[tex]\boxed{ \sf{ \: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]