As shown in the figure,overleftrightarrow{AB} | vedclass

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Question

(A) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$. Since tangents drawn from an external point to a circle are equal in leng

Vedclass · Accepted Answer

$2 AB$

Vedclass · Answer

(A) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$. Since tangents drawn from an external point to a circle are equal in length,we have: $AB = AC$ (tangents from point $A$) $PB = PR$ (tangents from point $P$) $QC = QR$ (tangents from point $Q$) The perimeter of $\Delta APQ = AP + PQ + AQ$. Substituting $PQ = PR + QR$,we get: Perimeter $= AP + (PR + QR) + AQ$. Using the equalities $PR = PB$ and $QR = QC$: Perimeter $= AP + PB + QC + AQ$. Since $AP + PB = AB$ and $AQ + QC = AC$: Perimeter $= AB + AC$. Since $AB = AC$,the perimeter $= AB + AB = 2 AB$ (or $2 AC$). Thus,the correct option is $2 AB$.

Part $I$	Part $II$
$1.$ In $\Delta ABC, AB=3, BC=4, AC=5$	$a.$ In-radius $= 1$
$2.$ In $\Delta PQR, PQ=5, QR=12, PR=13$	$b.$ In-radius $= 2$
$3.$ In $\Delta XYZ, XY=8, YZ=15, XZ=17$	$c.$ In-radius $= 3$
$4.$ In $\Delta MNP, MN=20, NP=21, MP=29$	$d.$ In-radius $= 6$

As shown in the figure,$\overleftrightarrow{AB}$,$\overleftrightarrow{AC}$ and $\overleftrightarrow{PQ}$ are the tangents to $\odot(O, r)$. Then the perimeter of $\Delta APQ = \ldots$

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