How are students messing up in this Khan Academy surface area problem? (right triangular prism: 3-4-5 triangular base, height of 11)

Question

The following question appeared in this Khan Academy exercise in January 2017, and the actual data on the ten most frequent incorrect responses is shown below the question. What mistakes did students make to get to each of these wrong answers?

Of student incorrect responses...

• 177 occurred ~8% of the time
• 126 occurred ~7% of the time
• 100 occurred ~7% of the time
• 156 occurred ~6% of the time
• 155 occurred ~4% of the time
• 166 occurred ~3% of the time
• 111 occurred ~2% of the time
• 89 occurred ~2% of the time
• 122 occurred ~2% of the time
• 138 occurred ~2% of the time

Answer 1

One quick way to guess at wrong answers is just to forget a face.

Omitting units, the correct answer is $144$ and the faces have surface areas of: $6$, $6$, $33$, $44$, and $55$.

Missing one face: $144-6 = 138$, $144-33 = 111$, $144-44 = 100$, and $144-55 = 89$.

Indeed, each of $138$, $111$, $100$, and $89$ appears in the list of incorrect responses.

I suppose some might have multiple mistakes; for example, forgetting one of the rectangular faces, and counting both of the other two as $5 \times 11$ for a total of $110$. In this scenario, you could then add on the triangular faces for a total of $122$, which also appears in the list.

Incidentally, even an erroneous approach could produce the correct answer. For example, one might compute the base rectangle as $4 \times 11$ and mistakenly believe all three of the rectangular faces have surface area $44$, for a total of $132$. Adding on the two triangular faces yields $144$, which is the right answer - arrived at in the wrong way!

Meanwhile, computing each rectangular face as $3 \times 11$ yields $99$, which adds with the triangular faces to (again) produce $111$. So, incorrect answers can arise in more than way.

And, as always, there is the simple possibility of altogether not knowing what to do and haphazardly combining numbers. For example, adding two of the edge lengths, $4$ and $5$, for $9$; then adding the other two edge lengths, $3$ and $11$, for $14$. But that results in two numbers, $9$ and $14$, and the final answer is supposed to be just one number. Okay - multiply them together: $9 \times 14 = 126$.

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Note 1. It is a bit tough to guess how an error has occurred, but if one believes the last situation is possible - haphazardly adding two edge lengths, then adding the other two edge lengths, then multiplying their respective sums - then we might expect this to occur for other pairings of edge lengths. Since it does not, I am guessing that the common $126$ error arose in another way:

Compute everything except the back face correctly. This gives a surface area of $$55 + 44 + 6 + 6 = 111$$

Now, mid-solution, forget which label corresponds to which edge, and misread the diagram as labeling the back face as $3 \times 5$, for an additional contribution of $15$. This yields $126$, as (un)desired.

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Note 2. For completeness, observe that many of the incorrect answers are off by a multiple of $11$; i.e., $177$, $166$, $155$, $122$, $111$, $100$, and $89$. Each of these can be explained by mis-computing around the rectangular faces, all of which have a side length of $11$. It is marginally interesting (to me) that $133$ did not appear as a frequent incorrect answer. All wrong responses are now accounted for.

Answer 2

From everyone's comments, here's what we've got so far (see @benjamin's analysis for more details):

$$\begin{align} 177&=\underbrace{3\times55}_{\text{3 times SA big rect}} + \underbrace{6+6}_{\text{area of triangles}} \\\\ 126&=\underbrace{55+44}_{\text{2 rects correct}}+\underbrace{(15)}_{\text{3rd rect wrong (3*5 instead of 3*11)}}+\underbrace{6+6}_{\text{area of triangles}} \\\\ 100&=\underbrace{144-44}_{\text{missing a rect}} \\\\ 156&=\underbrace{121}_{\text{area of 3 rects}}+\underbrace{(3)(4)+(3)(4)}_{\text{didn't divide by 2 for triangles}} \\\\ 155&=\underbrace{144+11}_{\text{messed up on a rect somehow}} \\\\ 166&=\underbrace{144+22}_{\text{messed up on a rect somehow}} \\\\ 111&=\underbrace{144-33}_{\text{missing a rectangle}} \qquad \text{OR} \qquad \underbrace{3\times33 + 6+6}_{\text{similar mistake as for 177}} \\\\ 89&=\underbrace{144-55}_{\text{missing a rectangle}} \\\\ 122&=\underbrace{144-22}_{\text{messed up on a rect somehow}} \\\\ 138&=\underbrace{144-6}_{\text{missing a triangle}} \end{align}$$

Note: As @michael pointed out, these most common wrong answers tend to be in the ballpark of the correct answer (144).