Which shapes would I use to figure out the total surface area of this composite shape? I am currently working on a problem where I have to figure out the formula(s) to find the total surface area of this composite figure. I have to find the individual areas of the shapes in the figure but I am having a hard time figuring out the shapes. I posted the composite figure and the answers that I already picked for the shapes. Can someone please tell me what the shapes are?

Here is the composite figure:



Here is the shapes question that I tried answering:
I hate these kinds of quizzes, because they teach you to anticipate others' thoughts, and not to solve problems.  We are humans, solvers of problems, investigators of science, manipulators of our environs; not betazeds, the readers of others minds!

My point is, there are many different roads to the same truths.  Every one of them is as valid as the others.  Thus, this quiz is incorrect; stupid, really.

We have an object which resembles a halved log, and we are asked to find its surface area.

The two ends are simple: they are halves of circular discs.  We know the area of a circular disc,
$$A_{○} = \pi r^2 \tag{1a}\label{AC1a}$$
where $r$ is the radius.  We know the diameter $d$, which is twice the radius,
$$d = 2 r \quad \iff \quad r = \frac{d}{2}$$
which means that the area of a full circular disc is
$$A_{○} = \pi \left( \frac{d}{2} \right )^2 = \frac{\pi}{4} d^2 \tag{1b}\label{AC1b}$$
Two halves is one whole: $2 \frac{1}{2} = 1$, so $A_{○}$ describes the sum surface area of the two ends.

The flat side is also simple, as it is just a rectangle.  We know its length/height $h$ and its width ($w = d$, the same as the diameter of the circular disc).  The area of a rectangle is
$$A_{▭} = w h \tag{2}\label{AC2}$$

That leaves the curved side, and here we get to the crux of the matter.

We can either calculate the area of the curved side as half the lateral surface area of the sides of the original "log" (cylinder) this is a half of, or we can think outside the box and consider it just a bent rectangle.

(Think of a sheet of paper.  It is easy to bend into a half cylinder.)

Mathematically, the two end up being the same thing.  The perimeter of a full circle is
$$p_{○} = \pi d = 2 \pi r$$
so the perimeter of a half circle is half that, $\pi r$.  If we take a sheet of paper and bend an edge of length $\pi r$ to a half cylinder, it will have radius $r$.

In other words, the area of the curved side is
$$A_{◡} = \pi h r = \pi h \frac{d}{2} \tag{3}\label{AC3}$$

Combining all these we get the total surface area of the object, given its cross section is exactly half a circular disc, length $h$, and width (diameter of those circular discs) $d$:
$$A = 2 \frac{1}{2} A_{○} ~ + ~ A_{▭} ~ + ~ A_{◡} \tag{4a}\label{AC4a}$$
which is equivalent to
$$A = \frac{\pi}{4} d^2 ~ + ~ d h ~ + ~ \pi h \frac{d}{2} \tag{4b}\label{AC4b}$$
In this particular case, $d = 2 ft$ and $h = 15 ft$, so we can calculate $A \approx 80 {ft}^2$, or about $80$ square foot.

(I like to build the formula like this first, for several reasons.  The main one being that this way I avoid re-rounding already rounded values, which increases the numerical error.  In real life you often get asked a similar question but with slightly different values, so instead of redoing all the work, you just reuse the formula.  It ends up often being less total work, really.)

What did we use to get to this answer?  The area of a rectangle, the area of a circular disc, and the perimeter of a circle.  But the last one wasn't even an option!  Another way to think of the area of the curved part is as the half of the lateral area of a cylinder; that would make the set the area of a rectangle, the area of a circular disc, and half the lateral area of a cylinder.

See?  The hard part of this question is not the math, but trying to discover what the person who set the question was thinking.  And I hate that.  I might be a bit autistic, but I am functional; penalizing people like me just because we find it difficult to discern what the person who stated the quiz question was thinking is evil.
I hate these kinds of quizzes, because they teach you to anticipate others' thoughts, and not to solve problems.  We are humans, solvers of problems, investigators of science, manipulators of our environs; not betazeds, the readers of others minds!

My point is, there are many different roads to the same truths.  Every one of them is as valid as the others.  Thus, this quiz is incorrect; stupid, really.

We have an object which resembles a halved log, and we are asked to find its surface area.

The two ends are simple: they are halves of circular discs.  We know the area of a circular disc,
$$A_{○} = \pi r^2 \tag{1a}\label{AC1a}$$
where $r$ is the radius.  We know the diameter $d$, which is twice the radius,
$$d = 2 r \quad \iff \quad r = \frac{d}{2}$$
which means that the area of a full circular disc is
$$A_{○} = \pi \left( \frac{d}{2} \right )^2 = \frac{\pi}{4} d^2 \tag{1b}\label{AC1b}$$
Two halves is one whole: $2 \frac{1}{2} = 1$, so $A_{○}$ describes the sum surface area of the two ends.

The flat side is also simple, as it is just a rectangle.  We know its length/height $h$ and its width ($w = d$, the same as the diameter of the circular disc).  The area of a rectangle is
$$A_{▭} = w h \tag{2}\label{AC2}$$

That leaves the curved side, and here we get to the crux of the matter.

We can either calculate the area of the curved side as half the lateral surface area of the sides of the original "log" (cylinder) this is a half of, or we can think outside the box and consider it just a bent rectangle.

(Think of a sheet of paper.  It is easy to bend into a half cylinder.)

Mathematically, the two end up being the same thing.  The perimeter of a full circle is
$$p_{○} = \pi d = 2 \pi r$$
so the perimeter of a half circle is half that, $\pi r$.  If we take a sheet of paper and bend an edge of length $\pi r$ to a half cylinder, it will have radius $r$.

In other words, the area of the curved side is
$$A_{◡} = \pi h r = \pi h \frac{d}{2} \tag{3}\label{AC3}$$

Combining all these we get the total surface area of the object, given its cross section is exactly half a circular disc, length $h$, and width (diameter of those circular discs) $d$:
$$A = 2 \frac{1}{2} A_{○} ~ + ~ A_{▭} ~ + ~ A_{◡} \tag{4a}\label{AC4a}$$
which is equivalent to
$$A = \frac{\pi}{4} d^2 ~ + ~ d h ~ + ~ \pi h \frac{d}{2} \tag{4b}\label{AC4b}$$
In this particular case, $d = 2 ft$ and $h = 15 ft$, so we can calculate $A \approx 80 {ft}^2$, or about $80$ square foot.

(I like to build the formula like this first, for several reasons.  The main one being that this way I avoid re-rounding already rounded values, which increases the numerical error.  In real life you often get asked a similar question but with slightly different values, so instead of redoing all the work, you just reuse the formula.  It ends up often being less total work, really.)

What did we use to get to this answer?  The area of a rectangle, the area of a circular disc, and the perimeter of a circle.  But the last one wasn't even an option!  Another way to think of the area of the curved part is as the half of the lateral area of a cylinder; that would make the set the area of a rectangle, the area of a circular disc, and half the lateral area of a cylinder.

See?  The hard part of this question is not the math, but trying to discover what the person who set the question was thinking.  And I hate that.  I might be a bit autistic, but I am functional; penalizing people like me just because we find it difficult to discern what the person who stated the quiz question was thinking is evil.