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Cutting a rectangle into two congruent triangles


Alignments to Content Standards: 8.G.A.2

Task

Below is a picture of rectangle $ABCD$ with diagonal $AC$.

Twotriangles1_9d61e180b678dbeda408e4aa43cc45f0

  1. Draw the image of triangle $ACD$ when it is rotated $180^\circ$ about vertex $D$. Call $A'$ the image of point $A$ under the rotation and $C'$ the image of point $C$.
  2. Explain why $\overleftrightarrow{DA'} \cong \overleftrightarrow{DA}$ and why $\overleftrightarrow{DC'}$ is parallel to $\overleftrightarrow{AB}$.
  3. Show that $\triangle A'C'D$ can be translated to $\triangle CAB$. Conclude that $\triangle ACD$ is congruent to $\triangle CAB$.
  4. Show that $\triangle ACD$ is congruent to $\triangle CAB$ with a sequence of translations, rotations, and/or reflections different from those chosen in parts (a) and (c).

IM Commentary

This task shows the congruence of two triangles in a particular geometric context arising by cutting a rectangle in half along the diagonal. The teacher may wish to have students think about part (d) of the problem first before going through parts (a) through (c). The grid has been added to the background as a geometric aid and as a means of determining lengths of sides. For part (d), the grid may need to be extended depending on which translations, rotations, and reflections the students choose to use.

A good way of reasoning about this problem abstractly would be to observe that the triangles are both right triangles and the lengths of the sides that come together to form the right angles are $4$ and $8$ units respectively. So to show that they are congruent we just need to "align" the right angles and the sides with corresponding lengths. Some questions which teachers may wish to have students explore further are:

  • Is it possible to show the congruence using only translations?
  • Is it possible to show the congruence with a reflection and then a translation?
  • Is it possible to show the congruence using only reflections?

 

The task is primarily intended for instructional purposes. Part (b) has been added explicitly to help students visualize how certain $180^\circ$ rotations influence a coordinate grid. Also part (d) has many possible answers, providing an ideal opportunity to have students share their (different) constructions with one another. If the teacher wishes to give students a chance to be creative and explore many different ways to transform $\triangle ACD$ to $\triangle CAB$ then part (d) alone can be used.

Solution

  1. Rotation by $180^\circ$ about $D$ takes any line through $D$ to itself but switches the orientation of the line. So, for example, line $\overleftrightarrow{DA}$ is taken to $\overleftrightarrow{DA}$ but the notions of up and down on this vertical line are reversed. So point $A$, which is four units below $D$ will be mapped to the point $A'$ which is four units above $D$. The points $D$ and $C$, on the other hand, lie on a horizontal line through $D$. So rotation by $180^\circ$ about $D$ has the effect of switching left and right on this line. Since $C$ lies $8$ units to the right of $D$, its image, $C'$, after rotating $180^\circ$ about $D$ will be the $8$ units to the left of $D$. This is pictured below:

    Twotrangles2_36b25399489937155ab4de1b04fcfef2

  2. As explained in part (a), rotation by $180^\circ$ about $D$ maps any line through $D$ to itself, exchanging the two rays through $D$ which make up the line. This means that $\overleftrightarrow{DA'} = \overleftrightarrow{DA}$ since $\overleftrightarrow{DA'}$ is the rotation of line $\overleftrightarrow{DA}$ by $180^\circ$ about $D$. Similarly, $\overleftrightarrow{DC'} = \overleftrightarrow{DC}$. So $\overleftrightarrow{DC'}$ is a horizontal line of the grid, as is $\overleftrightarrow{AB}$, and so these two lines are parallel, remaining four units apart from one another.

  3. To send $D$ to $B$ we need to translate by $8$ units to the right and $4$ units down. Translating by $8$ units to the right and $4$ units down also takes $C'$ to $A$ and $A'$ to $C$ as is seen in the picture below:

    Twotriangles3_6dbe6a8b63750339ec9b96aa2d0a06d1

    Since the vertices of triangle $A'C'D$ map to the vertices of triangle $CAB$ and translations take line segments to line segments, this means that translation $8$ units to the right and $4$ units down takes triangle $A'C'D$ to triangle $CAB$. The combination of a $180^\circ$ rotation about $D$ followed by a translation $8$ units to the right and four units down establishes the congruence of triangles $A'C'D$ and $CAB$.

  4. One of the most interesting ways to show the congruence of $\triangle ACD$ and $\triangle CAB$ is to rotate by $180^\circ$ about the intersection point of the diagonals $\overline{AC}$ and $\overline{BD}$ of rectangle $ABCD$. Using the coordinate grid, this is the point two boxes up and four boxes to the right of $A$, labelled $O$ in the diagram below:

    Rotation180_dc15ffdbdd2b1eb0f8222746734d7580

    To see why rotation by $180^\circ$ about $O$ maps $\triangle ACD$ to $\triangle CAB$, note that we know from parts (a) and (b) that rotation by $180^\circ$ will map line $\overleftrightarrow{AC}$ to itself (and similarly for line $\overleftrightarrow{DB}$ which also passes through $O$). Segments $\overline{OC}$ and $\overline{OA}$ are congruent: each is the hypotenuse of a right triangle with legs of length $2$ and $4$ and so each has length $\sqrt{20}$ by the Pythagorean theorem. Since rotation by $180^\circ$ fixes $O$ and interchanges the two rays $\overrightarrow{OA}$ and $\overrightarrow{OC}$ this means that this rotations maps $A$ to $C$ and $C$ to $A$. Similar reasoning applied to the rays $\overrightarrow{OD}$ and $\overrightarrow{OB}$ shows that rotation by $180^\circ$ maps $B$ to $D$ and $B$ to $D$. Hence rotation by $180^\circ$ about $O$ maps triangle $ACD$ to triangle $CAB$ and shows that they are congruent.

     

    There are many other ways to show the congruence of $\triangle ACD$ and $\triangle CAB$. For example, we could use two reflections to move $\triangle ACD$ to the position of $\triangle PQD$ from part (a): first a reflection about line $\overleftrightarrow{DC}$ and then a reflection about $\overleftrightarrow{AD}$ (or first a reflection about $\overleftrightarrow{AD}$ and then a reflection about $\overleftrightarrow{DC}$). This particular method gives an example of two reflections which commute, that is the order in which they are applied does not influence the resulting transformation of the plane.

Cam says:

over 3 years

Indeed it is. I incorrectly made the change to congruences after not reading the argument carefully enough. Thanks for bringing this up!

Michael Nakamaye says:

over 3 years

The original intent was to show that line DA is the same line as DA' and I think that is what the task originally asked. I will let the others weigh in though rather than changing it back.

redbaron says:

over 3 years

As I understand, "=" signs are sometimes used to show that geometric objects coincide with each other. I think that is what students were asked to do in part (b)--to show that line DA' coincides with line DA. As it reads now, they are asked to show that the lines are congruent to each other. If this were the intent, then the solution for (b) would merely restate that line DA' is a rotation of line DA (if remaining consistent with the Common Core's definition of congruence). But the solution for (b) emphasizes that the line was mapped onto itself, suggesting that the notation was meant to indicate coincidence.

So I think the intent actually WAS to show that line DA is the same as line DA', which I don't see as a problem. Though the labels DA and DA' are different, they refer to the same object. If the bottom line is that such use of the "=" sign is not favored at this level (or at any level), I can understand that. But if it IS appropriate, it should be used here instead of the "is congruent to" symbol.

Laura says:

almost 4 years

I think in part B, you should not have the line icon above DA and DA', because you are talking about the length, not the line. The line DA is not the same as the line DA'. The length of the segment DA is equal to the length of the segment DA'.

Cam says:

almost 4 years

Agreed, thanks. I've changed the equalities to congruences.

Edit: Oops, edited prematurely! See discussion in other comment.