Correct Answer: $(24-16\sqrt{2}) $ 𝛑 square cm

Explanation:

Depict the following figure according to question

Area of ABC triangle = $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times4\times4$ = $8cm^{2}$

now, $AC^{2}$= $(AB)^{2}+(BC)^{2}$ = $4^{2}+4^{2}$ =16+16 = 32

AC = $\sqrt{32}$

The radius of a circle inscribed in a triangle =$\frac{2\times \text{area of triangle}}{\text{perimeter of triangle}}$ = $\frac{2\times 8}{(4+4)+\sqrt{32}}=\frac{16}{8+(2\times 4^{2})}$

=$\frac{16}{8+4\sqrt{2}}=\frac{16}{4(2+\sqrt{2})}=\frac{4}{2+\sqrt{2}}=\frac{2\times \sqrt{2\times \sqrt{2}}}{\sqrt{2}(\sqrt{2}+1)}=\frac{2\sqrt{2}}{\sqrt{2}+1}$

= $\frac{2\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2-1})(\sqrt{2}-1)}=\frac{2\sqrt{2}(\sqrt{2}-1)}{2-1}=2\sqrt{2(\sqrt{2}-1)}$

so radius = $2\sqrt{2}(\sqrt{2}-1)$cm

so, area of circle =𝛑 $ R^{2 }=𝛑 { [2\sqrt{2}(\sqrt{2}-1)]}^{2}$

= 𝛑 $ 4\times 2(\sqrt{2}-1)^{2}cm^{2}$

=𝛑 $8(\sqrt{2}-1)^{2}cm^{2}=(24-16\sqrt{2})𝛑 \text{square cm}$