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Given - OD = 6cm and AB= 16cm
now we can say that AD = BD = 1/2 AB ( as perpendicular from centre to the chord bisects the chord into two equal parts).
---> AD = 1/2 × 16 = 8cm.
In ∆AOD, applying pythagoras theorem,
OA^2 = OD^2 + AD^2
= (6)^2 + (8)^2
= 36 + 64 = 100
OA^2 = (10)^2
taking square root
OA = 10 cm..
hope it helps you..
now we can say that AD = BD = 1/2 AB ( as perpendicular from centre to the chord bisects the chord into two equal parts).
---> AD = 1/2 × 16 = 8cm.
In ∆AOD, applying pythagoras theorem,
OA^2 = OD^2 + AD^2
= (6)^2 + (8)^2
= 36 + 64 = 100
OA^2 = (10)^2
taking square root
OA = 10 cm..
hope it helps you..
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