In a circle with centre P,AB is a diameter an | vedclass

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(C) Since $AB$ is the diameter of the circle with centre $P$,the angle subtended by the diameter at any point on the circle is a right angle. Therefor

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Since $AB$ is the diameter of the circle with centre $P$,the angle subtended by the diameter at any point on the circle is a right angle. Therefore,$\angle ACB = 90^\circ$.In $	riangle ACB$,by the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.The radius of the circle is $34 \, cm$,so the diameter $AB = 2 	imes 34 = 68 \, cm$.Substituting the values,$32^2 + BC^2 = 68^2$.$1024 + BC^2 = 4624$.$BC^2 = 4624 - 1024 = 3600$.$BC = \sqrt{3600} = 60 \, cm$.

In a circle with centre $P$,$AB$ is a diameter and point $C$ is a point other than $A$ and $B$. If the radius of the circle is $34 \, cm$ and $AC = 32 \, cm$,then $BC = $ .......... $cm$.

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