1)
We can find the circumference using the formula
[tex]C=2\pi r[/tex]
but remember that the diameter is 2 times the radius
[tex]d=2r[/tex]
So we can use the formula using radius or diameter, the problem gives us the diameter, so let's use it, so the formula will change a little bit
[tex]C=\pi d[/tex]
Where "d" is the diameter.
d = 40 yd, and π = 3.14, so the circumference will be
[tex]\begin{gathered} C=\pi d \\ C=3.14\cdot40=125.6\text{ yd} \end{gathered}[/tex]
And to find out the area we can use this formula
[tex]A=\frac{\pi d^2}{4}[/tex]
Or if you prefer use the radius
[tex]A=\pi r^2[/tex]
Let's use the formula with the diameter again
[tex]\begin{gathered} A=\frac{\pi d^2}{4} \\ \\ A=\frac{3.14\cdot(40)^2}{4} \\ \\ A=1256\text{ yd}^2 \end{gathered}[/tex]
Then the circumference is 125.6 yd and the area is 1256 yd^2
2)
Here we have a compounded figure, we have half of a circle and a triangle, so let's think about how we get the perimeter and the area.
The perimeter will be the sum of the sides of the triangle and half of a circumference, we already know the length of the triangle's side, it's 10.82, we got to find the half of a circle circumference and then sum with the sides.
We know that
[tex]C=\pi d[/tex]
And we can see in the figure that d = 12 mm, then
[tex]C=\pi d=3.14\cdot12=37.68\text{ mm}[/tex]
But that's a full circumference, we just want half of it, so let's divide it by 2.
[tex]\frac{C}{2}=\frac{37.68}{2}=18.84\text{ mm}[/tex]
Now we have half of a circumference we can approximate the perimeter of the figure, it will be
[tex]\begin{gathered} P=10.82+10.82+18.84 \\ \\ P=40.48\text{ mm} \end{gathered}[/tex]
The area will be the area of the triangle sum the area of half of a circle
Then let's find the triangle's area first
[tex]A_{}=\frac{b\cdot h}{2}[/tex]
The base "b" will be the diameter of the circle, and the height "h" will be 9 mm, then
[tex]A_{}=\frac{12\cdot9}{2}=54\text{ mm}^2[/tex]
And the half of a circle's area will be
[tex]A=\frac{1}{2}\cdot\frac{\pi d^2}{4}=\frac{3.14\cdot(12)^2}{8}=$56.52$\text{ mm}^2[/tex]
Then the total area will be
[tex]A_T=56.52+54=110.52\text{ mm}^2[/tex]
Therefore, the perimeter and the area is
[tex]\begin{gathered} P=40.48\text{ mm} \\ \\ A=110.52\text{ mm}^2 \end{gathered}[/tex]
