+37162 TWO triangle hypotenuses find via pythag theorem
h^2 = 4^2 + 10^2 solve for h rememeber: there are 2
semicircle perimeter = 1/2 pi d = 1/2 pi * 8 add them together
+1262 AREA: 10*8/2 for the area of the triangle and π4^2 so we add them up in a equation to get that the area 80/2+π4^2=40+16π so the area of the shaded region is 40+16π or \(\boxed{40+16π}\)
Circumfrence=8π/2 for the rounded part and 2*(sqrt(10^2+(8/2)^2)) again we make them into an equation 4π+2*(sqrt(10^2+4^2))=4π+2*(sqrt(116)) then we simplify it to get 4π+4*(sqrt(29)) or \(4π+4\sqrt{29}\)
