In the given figure if PA=20 cm, what is the perimeter of ΔPQR.

From P we have tangents PA and PB

Hence PA = PB …tangents from same point are equal …(a)

Point Q is on PA

From Q we have tangents QA and QC

⇒ QA = QC …tangents from same point are equal …(i)

Point R is on PB

From R we have two tangents RC and RB

⇒ RC = RB … tangents from same point are equal …(ii)

Consider ΔPQR

⇒ perimeter of ΔPQR = PQ + QR + PR

From figure QR = QC + CR

⇒ perimeter of ΔPQR = PQ + QC + CR + PR

Using (i) and (ii)

⇒ perimeter of ΔPQR = PQ + QA + RB + PR

From figure we have

PQ + QA = PA and RB + PR = PB

⇒ perimeter of ΔPQR = PA + PB

Using (a)

⇒ perimeter of ΔPQR = PA + PA

⇒ perimeter of ΔPQR = 2(PA)

PA is 20 cm given

⇒ perimeter of ΔPQR = 2 × 20

⇒ perimeter of ΔPQR = 40 cm