i) From P we have tangents PA and PB
Hence PA = PB …tangents from same point are equal …(a)
Point C is on PA
From C we have tangents CA and CE
⇒ CA = CE …tangents from same point are equal …(i)
Point D is on PB
From D we have two tangents DE and DB
⇒ DE = DB … tangents from same point are equal …(ii)
Consider ΔPCD
⇒ perimeter of ΔPCD = PC + CD + PD
From figure CD = CE + ED
⇒ perimeter of ΔPCD = PC + CE + ED + PD
Using (i) and (ii)
⇒ perimeter of ΔPCD = PC + CA + DB + PD
From figure we have
PC + CA = PA and DB + PD = PB
⇒ perimeter of ΔPCD = PA + PB
Using (a)
⇒ perimeter of ΔPCD = PA + PA
⇒ perimeter of ΔPCD = 2(PA)
PA is 14 cm given
⇒ perimeter of ΔPCD = 2 × 14
⇒ perimeter of ΔPCD = 28 cm
ii) PA = 11 cm …given
using (a)
PB = 11 cm
From figure
⇒ PB = PD + DB
Using (ii)
⇒ PB = PD + DE
⇒ 11 = 7 + DE …PD is 7 cm given
⇒ DE = 5 cm
Hence DE = 5 cm