Prove that in two concentric circles, the chord of larger circle, which touches the smaller circle,is br at the point of contact.
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Step-by-step explanation:
Let there is a circle having center O
Let AB is the tangent to the smaller circle and chord to the larger circle.
Let P is the point of contact.
Now, draw a perpendicular OP to AB
Now, since AB is the tangent to the smaller circle,
So, ∠OPA = 90
Now, AB is the chord of the larger circle and OP is perpendicular to AB.
Since the perpendicular drawn from the center of the circle to the chord bisect it.
So, AP = PB
Hence, in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
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