If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by (A) a = S.R/L (B) a = L.S/R (C) a = L.R/S (D) None of these

If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular
distance of the line of the resultant cohesive force, is given by
(A) a = S.R/L
(B) a = L.S/R
(C) a = L.R/S
(D) None of these
by

1 Answer

Answer :

(A) a = S.R/L
by

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Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

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Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

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AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

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AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

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