The area is given by the area of the circular sector (⌔): $$\frac{\theta}{2} \times r^2$$ minus the are of triangle (△): $$\frac{b \times h}{2}=\frac{r \times r sin(\theta)}{2}.$$
For both equations, $\theta$ is the absolute value of the angle, given by
$$\theta= \frac{71^{\circ}}{360^{\circ}}\times 2 \pi.$$
Thus, the required area is $$\frac{\theta}{2} \times r^2 - \frac{r \times r sin(\theta)}{2} = \frac{\theta-\sin(\theta)}{2} \times r^2 = \frac{\frac{71^{\circ}}{360^{\circ}}\times 2 \pi - \sin(\frac{71^{\circ}}{360^{\circ}}\times 2 \pi)}{2} \times (10[m])^2 = 14.7 \,m^2.$$
For the perimeter of this region, we just have to add the arc (⌒) lengh (fraction of the the total perimeter of the circumference) and the straight line length (given by the Law of Cosines): $$\theta \times r + \sqrt{r^2 + r^2 - 2 \times r \times r\times \cos(\theta)} = 24.0 \,m.$$
How wide is the shaded region? If the length of the larger rectangle is $a^3$, and the length of the right is $b^3$...
The width of the shaded region is $a^3-b^3$
How tall is the shaded region?
I assume the picture is indicating that the height is $a^2+ab+b^2$
What is the area of a rectangle described as in terms of width and height?
width times height
And what is that in this case?
$(a^3-b^3)(a^2+ab+b^2)$
What is the perimeter of a rectangle described as in terms of width and height?
twice the sum of width and height
And what is that in this case?
$2(a^3-b^3 + a^2+ab+b^2)$
Now... can we express any of these above in a more compact way via factoring? That is debatable, but I do recognize at least one simplification you can do for area:
Knowing that $a^3-b^3=(a-b)(a^2+ab+b^2)$ we have that the area can be described as $(a-b)(a^2+ab+b^2)^2$. Is that really more preferred than $(a^3-b^3)(a^2+ab+b^2)$? I don't see a big difference between the two, so personally I wouldn't care which. They are the same to me.
Using the same for perimeter, we can do the following:
$2(a^3-b^3 + a^2+ab+b^2) = 2((a-b)(a^2+ab+b^2)+1(a^2+ab+b^2)) = 2(a-b+1)(a^2+ab+b^2)$. Again, is this more useful than the first way of writing it? That is debatable...They are not fundamentally different.
Best Answer
Your work on the area is correct except the $37$'s should be $74$'s. There is no need to divide the angle by $2$. (Think through what the answer would be if the angle were $90$ instead of $74$. You'd be looking at a quarter circle with a right isosceles triangle removed, so the area would be $r^2({\pi\over4}-{1\over2})$.)