I. ABCD is a square so ∡BCD=90°
|BC|²+|CD|² = |BD|²
8² + 8² = |BD|²
|BD|² = 8²·2
|BD| = √[8²·2]
|BD| = 8√2
II. Circular segment BCD is a quater of circle (because ∡BCD=90°)
R=8
So the lenght of arc BD:
Perimeter of the figure: 
The area of circular segment BCD:
III. The area of triangle BCD:
IV. The area of figure: 