In the figure,stackrel{leftrightarrow}{ AB }, | vedclass

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(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are eq

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are equal,so $AB = AC$.Let the tangent $PQ$ touch the circle at point $M$. Then $PM = PB$ and $QM = QC$ (tangents from points $P$ and $Q$ respectively).The perimeter of $\Delta APQ = AP + PQ + AQ = AP + (PM + MQ) + AQ$.Substituting the equal tangent lengths: Perimeter $= AP + PB + QC + AQ = (AP + PB) + (AQ + QC) = AB + AC$.Since $AB = AC$,the perimeter $= 2 AB$.Given that the perimeter is $16$,we have $2 AB = 16$,which implies $AB = 8$.

In the figure,$\stackrel{\leftrightarrow}{ AB }$,$\stackrel{\leftrightarrow}{ AC }$ and $\stackrel{\leftrightarrow}{ PQ }$ are tangents to $\odot( O , r)$. If the perimeter of $\Delta APQ$ is $16$,then $AB = \ldots$

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