In the following figure,if AB = 3,then the pe | vedclass

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(C) According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.From point $A$,$AB$ and $AC$ are tangents to t

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$6$

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(C) According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.From point $A$,$AB$ and $AC$ are tangents to the circle,so $AB = AC = 3$.From point $P$,$PB$ and $PS$ are tangents to the circle,so $PB = PS$.From point $Q$,$QC$ and $QS$ are tangents to the circle,so $QC = QS$.The perimeter of $\Delta APQ = AP + PQ + AQ$.Since $PQ = PS + SQ$,we can substitute $PB$ for $PS$ and $QC$ for $SQ$.Perimeter $= AP + (PB + QC) + AQ$.Perimeter $= (AP + PB) + (AQ + QC) = AB + AC$.Since $AB = 3$ and $AC = 3$,the perimeter $= 3 + 3 = 6$.

In the following figure,if $AB = 3$,then the perimeter of $\Delta APQ$ is...........

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