Answer:
Area of the shaded region = 16(π - 2) in²
Perimeter of the shaded region = 4(π + [tex]2\sqrt{2}[/tex]) in
Step-by-step explanation:
Since, BDC is a quarter of the circle with radius = 8 in.
Area of the quarter of the circle = [tex]\frac{1}{4}(\pi)(r)^2[/tex]
= [tex]\frac{1}{4}(64)\pi[/tex]
= 16π in²
Area of ΔBCD = [tex]\frac{1}{2}(\text{Base})(\text{Height})[/tex]
= [tex]\frac{1}{2}(8)(8)[/tex]
= 32 in²
Since, area of the shaded part = Area of quarter of the circle - Area of triangle BCD
= (16π - 32)
= 16(π - 2) in²
Therefore, area of the shade region = 16(π - 2) in²
Similarly, length of arc BD = [tex]\frac{1}{4}(2\pi r)[/tex]
= [tex]\frac{8\pi }{2}[/tex]
= 4π in.
Length of the diagonal of a square = (Side)√2
= 8√2 in
Perimeter of the shaded region = Length of arc BD + Length of diagonal BD
= 4π + 8√2
= 4(π + 2√2) in.
