PA and PB are the tangents drawn to circle from external point P .PA = 6cm ,QR is another tangents touch the circle at S and intersect PA at Q and PB at R .The perimeter of triangle PQR is About the author Lyla

Step-by-step explanation: Correct option is A 20 From the graph, we find some conclusion- PA=PB=10cm CE=CA DE=DB Perimeter of ΔPCD is, =PC+PD+CD ⇒PC+PD+CE+ED ⇒PC+PD+CA+DB ⇒(PC+CA)(PD+DB) ⇒PA+PB ⇒10+10 ⇒20 cm. ￼ Reply

Step-by-step explanation:Correct option is

A

20

From the graph, we find some conclusion-

PA=PB=10cm

CE=CA

DE=DB

Perimeter of ΔPCD is,

=PC+PD+CD

⇒PC+PD+CE+ED

⇒PC+PD+CA+DB

⇒(PC+CA)(PD+DB)

⇒PA+PB

⇒10+10

⇒20 cm.

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