Question

what's the answer?
i​

Answer 1

$\huge\underline\red{10\:cm}$

Let PQ touch the circle at the point R.

We know that tangents drawn from an external point to a circle are equal in length.

AB = AC = 5cm

AP +BP = AQ+ QC = 5cm

AP + PR = AQ + QR = 5cm…………(1)

[ BP = PR & QC = QR]

Now, perimeter of ∆ APQ = AP +PQ+AQ

= AP +RP+QR+AQ

= 5 + 5 [ from eq 1]

Perimeter of ∆ APQ= 10 cm

Answer 2

Step-by-step explanation:

the answer will be option (d)10cm