Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th
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Solution  :- Let R, S and M represent the position of Reshma, Salma and Mandeep respectively. Clearly △RSM is an isosceles triangle as RS = SM = 6m Join OS which intersect RM at A. In △ROS  and △MOS OR = OM (Radii of the same circle) OS = OS (Common) RS = SM (Each 6cm) ∴ △ROS ≅  △MOS (By SSS congruence criterion) ∴ ∠RSO =  ∠MSO (CPCT) In △RAS and △MAS AS = AS (Common) ∴ ∠RSA = ∠MSA (Since, ∠RSO = ∠MSO) RS = MS (Given) ∴ △RAS ≅ △MAS  (By SAS congruence criterion) ∴ ∠RAS = ∠MAS  (CPCT) Since, ∠RAS + ∠MAS = 180°  (Linear pair) ⇒ ∠RAS = ∠MAS = 90° Let OA = x m ⇒ AS = (5 - x)m In right triangle RAS, RS2 = RA2 + AS2 ⇒62 = RA2 + (5 - x)2    ⇒RA2 = 62 - (5 - x)2    ....(i) In right triangle RAO, RO2 = RA2 + OA2 ⇒52 = RA2 + x2 ⇒RA2 = 52 - x2  ...(ii) From equation (i) and (ii), we get 62 - (5-x)2 = 52 - x2 62 - 52 = (5-x)2 - x2 36 - 25 = 25 + x2 - 10x - x2 11 = 25 - 10x ⇒ 10x = 14 ⇒ x = 1.4m From equation (ii), we have RA2 = 52 -(1.4)2 = 25 - 1.96 RA2 = 23.04 ⇒ RA = √23.04 RA = 4.8m As the perpendicular from the centre of a circle bisects the chord. ∴ RM = 2RA RM = 2 x 4.8 = 9.6m Hence, distance between Reshma and Mandeep is 9.6m.
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A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Description : A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Last Answer : Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) ... x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)