Derive the constant, do not recall it AB & BC

An equilateral triangle of side $s$ splits down the middle into two $30$-$60$-$90$ triangles, so its height is $\frac{\sqrt{3}}{2}s$ and

$$A = \frac{1}{2}\,s\cdot\frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4}s^{2}.$$

Both factors matter, and the classic error keeps the height and forgets the $\frac{1}{2}$, giving $\frac{\sqrt{3}}{2}s^{2}$. That is exactly twice the truth, so it is not detectable by a plausibility check: the answer looks like a volume and is simply doubled.

Two ways to catch it. First, $\frac{\sqrt{3}}{4} \approx 0.433$ is less than $\frac{1}{2}$, and an equilateral triangle on base $s$ must have less area than the square on the same base and less than half of it, since its height $0.866s$ is under $s$ and it is a triangle. Second, the height is $\frac{\sqrt{3}}{2}s$ and the area is $\frac{\sqrt{3}}{4}s^{2}$: the two constants are not the same number, and writing the height down first keeps them apart.

This shape appears in two forms, and the problem's wording decides which.

1. The base is a leg. Then the other leg is also $s$ and $A = \frac{1}{2}s^{2}$.
2. The base is the hypotenuse. Then each leg is $\frac{s}{\sqrt{2}}$, the height to the hypotenuse is $\frac{s}{2}$, and $A = \frac{1}{2}\cdot s\cdot\frac{s}{2} = \frac{s^{2}}{4}$.

The two answers differ by a factor of $2$, and the phrase that separates them is short enough to skip over: "with one leg in the base" against "with the hypotenuse in the base". Reading it is the whole task.

The height in the second case is worth deriving once. In an isosceles right triangle the altitude to the hypotenuse also bisects it and equals half of it, because the two small triangles it creates are themselves isosceles right triangles. That gives $\frac{s}{2}$ without any trigonometry.

A semicircle standing on a base segment has that segment as its diameter. So $r = \frac{s}{2}$, and

$$A = \frac{1}{2}\pi r^{2} = \frac{1}{2}\pi\left(\frac{s}{2}\right)^{2} = \frac{\pi}{8}s^{2}.$$

Three wrong constants circulate, and each corresponds to skipping one step:

1. $\frac{\pi}{2}s^{2}$: halved the circle but used $s$ as the radius. Four times too large.
2. $\frac{\pi}{4}s^{2}$: used $r = \frac{s}{2}$ and forgot to halve the circle. Twice too large.
3. $\pi s^{2}$: both at once. Eight times too large.

The order that avoids all three: write $r = \frac{s}{2}$ first, on its own line, then write $\frac{1}{2}\pi r^{2}$. Substituting last keeps the two operations from competing for attention.

The same discipline handles the variants. A cross section that is a quarter circle on a radius $s$ has $A = \frac{1}{4}\pi s^{2}$, with no halving of $s$, because there the base is a radius rather than a diameter. What the base is to the shape is the question; the shape's own formula follows.

Take the base region bounded by $y = \sqrt{x}$, the $x$-axis and $x = 4$, so $s = \sqrt{x}$ and $s^{2} = x$. Since $\int_0^4 x\,dx = 8$, all three volumes are one multiplication apart:

1. Equilateral triangles: $\frac{\sqrt{3}}{4}\cdot 8 = 2\sqrt{3} \approx 3.464$.
2. Isosceles right triangles on a leg: $\frac{1}{2}\cdot 8 = 4$.
3. Semicircles: $\frac{\pi}{8}\cdot 8 = \pi \approx 3.142$.

Notice the ordering. The right triangle gives the largest solid and the semicircle the smallest, which matches the shapes: on a fixed base, a right isosceles triangle with legs $s$ reaches height $s$, the equilateral reaches $0.866s$, and the semicircle only $\frac{s}{2}$. If a computed volume violates that ordering, a constant is wrong.

All the work that is not the constant is 8.7's work, unchanged: identify both boundaries, subtract to get $s$, and check $s$ at an interior point. When the base runs between two curves rather than down to the axis, $s$ is their difference and $s^{2}$ has a cross term, which is the CA9 error of 8.7 arriving in this topic wearing a triangle.