Please answer this question. ...

In the given figure join OX and OY

∴ OX ⊥ AC, OY ⊥ AB

Since, radius of the incircle is 4cm, OX = OY = 4 cm

Thus, OXAY is a square

∴ AX = AY = 4 cm

Now, CX = AC – AX = 16 cm – 4 cm = 12 cm

Let BY = x cm

AB = BY + AY = (x + 4) cm

It is known that the lengths of tangents drawn from an exterior point to a circle are of equal length.

∴ CZ = CX and BZ = BY

⇒ CZ = 12 cm and BZ = x cm

Now, BC = BZ + CZ = (x + 12) cm

Using Pythagoras theorem in ∆ABC

BC² = AB² + AC²

⇒ (x + 12)² = (x + 4)² + 16²

⇒x² + 24x + 144 = x² + 8x + 16 + 256

⇒16x = 128

⇒ x = 8

∴  AB = (8 + 4) cm = 12 cm

BC = (12 + 8) cm = 20 cm

Thus, the perimeter of ∆ABC = (12 + 16 + 20) cm = 48 cm

 

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