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Points $M$, $N$ and $P$ are the midpoints of the joined edges of the cube. The length of an edge is $4cm$. What is the area of the triangle $\triangle MNP$?


I think it's $4cm^2$. The base of the triangle is the distance from one midpoint to another midpoint which are aligned horizontally. Since the distance from one midpoint to the common point (where edges join) is half of the length of the edge of the cube, which is $2cm$, the distance from one to another horizontally aligned midpoint must be $2cm+2cm=4cm$. The height of the triangle is the distance from the common point to the highest point, which is $2cm$. The formula for the area of a triangle is $\frac{bh}{2}$. Then the area of the triangle $\triangle MNP$ must be $\frac{4cm \cdot 2cm}{2}=\frac{8cm^2}{2}=4cm^2$.

But the only allowed answers are:

  1. $8\sqrt{2}cm^2$
  2. $\sqrt{2}cm^2$
  3. $8\sqrt{3}cm^2$
  4. $8cm^2$
  5. $6\sqrt{3}cm^2$

Am I wrong or the question is erroneous?

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You're right that there seems to be something wrong with the given options. But your answer is wrong as well.

If I read the question correctly, $M,N,P$ are the midpoints of a set of three edges that meet at a single vertex of the cube. That means that the side of the triangle can be calculated using Pythagoras' Theorem as $\sqrt{2^2 + 2^2} = \sqrt 8$. The triangle is obviously equilateral, and its vertex angles are all $\frac{\pi}{3}$, so its area is $\frac 12{(\sqrt 8)}^2 \sin \frac{\pi}{3} = 2\sqrt 3$. But this is not among the listed options.

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  • $\begingroup$ The triangle isn't equilateral. Two horizontally aligned midpoints are $2cm$ apart from the common point. Therefore, they are apart from each other $2cm+2cm=4cm$. But the length of the side that you calculated using Pythagoras' Theorem is $\sqrt{8}cm$. $4cm \neq \sqrt{8}cm$, therefore the triangle isn't equilateral since all sides aren't of the same length. $\endgroup$ – Hanlon May 20 '18 at 11:12
  • $\begingroup$ @Vuk How do you mean "horizontally aligned"? The line segments joining any two edge midpoints (which are the sides of the triangle) are parallel to the diagonals of the square faces of the cube. $\endgroup$ – Deepak May 20 '18 at 11:38
  • $\begingroup$ When I say "horizontally aligned points" I mean that if you take out the triangle and place it in 2D plane coordinates of those points will have same $x$ value. $\endgroup$ – Hanlon May 20 '18 at 13:08
  • $\begingroup$ In my example, three edges of the cube have a common point $Q$. Let's call these edges $m$, $n$ and $p$. Points $M$, $N$ and $P$ bisect these edges, respectively. If we rotate the cube such that we can see $m$, $n$ and $p$, they would look like the letter T. $Q$ is where they meet. $M$, $N$ and $P$ are the midpoints of these edges. Let's say that $m$ is left, $n$ is right and $p$ is down. Then $\overline{MQ}=\overline{QN}=2cm$. Then $\overline{MN}=4cm$. And $\overline{QP}=2cm$, but $\overline{NP}$ and $\overline{MP}$ aren't equal to $2cm$. $\endgroup$ – Hanlon May 20 '18 at 13:20
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    $\begingroup$ Quoting you,"In my example, three edges of the cube have a common point Q. Let's call these edges m, n and p. Points M, N and P bisect these edges, respectively. If we rotate the cube such that we can see m, n and p, they would look like the letter T." Not a planar T, and that's the key."Q is where they meet. M, N and P are the midpoints of these edges. Let's say that m is left, n is right and p is down. Then MQ=QN=2cm." Agree with you so far."Then MN=4cm." Disagree. You can't expect to flatten out the line segment MN and maintain the correct length. You got MP and NP right but MN wrong. $\endgroup$ – Deepak May 21 '18 at 5:43

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