Q). On the given figure, ∆OSR- ∆OQP, angle QOR=125° and
angle RS0 = 70. Find angle SOR and angel OPQ
Answers
Answer:
★ If two circles intersect in two points prove that the line through their centres is the perpendicular bisector of the common chord.
\bf\underline{\underline{\blue{Given:-}}}
Given:−
★ Two circles C(O,r) and C(O' , s) intersecting at points A and B.
\bf\underline{\underline{\red{To\:Prove:-}}}
ToProve:−
★ OO', is the perpendicular bisector of AB.
\bf\underline{\underline{\green{Construction:-}}}
Construction:−
★ Draw the line segment OA, OB, O'A and O'B. Let OO' and AB intersects at M.
\bf\underline{\underline{\orange{Proof:-}}}
Proof:−
In ∆<OAO' and ∆OBO' , we have
OA = OB [each equal to r]
O'A = O'B [each equal to s]
OO' = OO' [common]
∴ ∆OAO' ≅ ∆OBO' [SSS-congruence]
==> ∠AOO' = ∠BOO'
==> ∠AOM = ∠BOM [ ∠AOO' = ∠AOM and ∠BOO' = ∠BOM]. ...(i)
In ∆AOM and ∆BOM, we have
OA = OB. [ each equal to r]
∠AOM = ∠BOM [ from (i) ]
OM = OM [comon]
∴ ∆AOM ≅ ∆BOM
==> AM = BM and ∠AMO = ∠BMO
==> AM = BM and ∠AMO = ∠BMO = 90°
==> OO' is the perpendicular bisector of AB
Answer:
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Step-by-step explanation:
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