SQUARE IN THE MIDDLE
This is the easiest part: a square with side length s has area s^2, so the area of the square in the middle is
[tex]A_s = 12^2=144[/tex]
TRIANGLE ON THE LEFT
The base of this triangle is one side of the square, so it is 12 inches long. The height is given, and it is 8 inches long. The area of a triangle is half the product of width and height, and so we have
[tex]A_t = \dfrac{12\cdot 8}{2}=48[/tex]
SEMICIRCLE ON THE RIGHT
The diameter of the semicircle is one side of the square, so it is 12 inches long. The radius is half the diameter, so it's 6 inches long. The area of a circle is [tex]\pi r^2[/tex], so the area of a semicircle will be half this quantity:
[tex]A_c = \dfrac{\pi r^2}{2}\approx \dfrac{3.14\cdot 36}{2}=56.52[/tex]
The total area is the sum of these areas:
[tex]A=A_s+A_t+A_c=144+48+56.52=248.52[/tex]
As for the perimeter, we have:
Since the circumference of a circle is [tex]2\pi r[/tex], the semicircumference will be [tex]\pi r[/tex]
So, the perimeter is
[tex]10+10+12+12+6\pi \approx 44+6\cdot 3.14=62.84[/tex]
