Middle School
Problem of the Week:

Tiling Triangles - posted March 11, 2002

I'm going to tile the area of the large triangle shown above with 25 copies of this small similar triangle, .

Using the given lengths of the sides of the large triangle, what are the lengths of the sides of the small triangle used in the tiling? Explain how you figured out your answers.

Bonus
If the 150 cm side of the large triangle were 250 cm instead, what would be the length of the corresponding side of the small black triangle?

Comments

The Tiling Triangles problem could just as easily have been named the Tessellating Triangles problem. When we first started writing it we debated the difference between "tiling" and "tessellating." In the end we found that the two words can be used interchangeably.

Most students found that either thinking of a picture or actually drawing the picture helped them figure out their solutions.

Michael Kim of Aptos Middle School referred to the fact that he had drawn a diagram and after noticing how the small triangles fit, realized that if you divided each of the side lengths of the large triangle by 5 that would give you the dimensions of the small triangles. Adam Lohr of Palmyra Area Middle School and Ms. Magnifico's Fifth Grade Class of Saklan Valley School came to the same conclusion and also submitted the drawings with their solutions.

Keri D. of Howell Township Middle School North had an interesting way to find the other sides after finding the 40 cm side by dividing by 5. She thought in terms of proportions to find the other side lengths.

Michael Levitin of St. Margaret's Episcopal School explained why he thought in terms of parallelograms and then described the step-by-step process he followed using his diagram.

Many students sent in solutions that simply said that you should divide each side by 25. It might seem like you can just divide each measurement by the number of small triangles but when you check your answer you should see that it doesn't work. If the base of the small black triangle is 8 cm, as many students answered, 25 small triangles will fit along the base of the large triangle. If all of the small triangles that we were given are used up just along the base of the large triangle, what about tiling the rest of it?

You can, however, use the number 25 to think about an answer. If there are 25 small triangles in the big triangle, you know that the area of the big triangle is 25 times the area of each of the small triangles. What does this tell us about the dimensions of the small triangle compared to the big one?

Let's look at this the other way around. In the picture below, the edgelength of the small triangle is one, and the edgelength of the large triangle is 3.

If the area of the small triangle is 1, then the area of the large triangle is 9 (you can count the triangles). Since area is "two dimensional," we can find the ratio of these two areas by comparing the square of the ratio of their edgelengths (which is "one dimensional").

        1  ^2    1
      (---)   = ---
        3        9

So in our problem, we go the other way. We can find the ratio of the edgelengths by taking the square root of the ratios of the areas.

         1       1
   sqrt(----) = ---
         25      5

Now we know that the "scale factor" is 1/5. Read how Garrett Marcotte of Polytechnic School used this idea in his solution below.

-Suzanne

Highlighted solutions:

The length of the triangle's sides are 150 cm=30 cm, 175 cm=35 cm,
200 cm=40 cm.

I figured out the answer by some hints the problem gave me. It said
that you use 25 triangles which make up a bigger triangle with the
sides 150 cm, 175 cm, and 200 cm. When I got this, I drew inside a
triangle 25 smaller triangles, evenly distributing them. When i did
this, I knew that you could not just divided it by 25 or the small
triangle would be too small. Then I looked carefully at the large
triangle and the smaller ones inside it and found that you had to
divide the sides by 5 because each side had 5 smaller triangles
evenly distributed in them. When I did this dividing process for each
1, I got these numbers
150=30
175=35
200=40
I knew that the number I recieved was the answer to the sides so
after adding the cm sign on them, I checked if it was correct and it
was.
The answer to the bonus was 50.
I figured this out by doing the same process as the one before

The length of the side of the small triangle with measure 200 cm on
the large triangle is 40 cm. The side with length measure 150 cm on
the large triangle is length 30 cm and the last side, the side with
length measure 175 cm on the large triangle has length 35 cm on the
small triangle.

To figure this problem out, I saw that one small triangle would fit
in the top corner where the edges of lines 150 cm and 175 cm meet.
Then, I realized that every triangle is part of a rectangle so I drew
another one directly underneath the one in the top corner. This made
two more face up triangles next to it-one on the right and one on the
left. From this I figured out that it was a pattern and did it
repeatedly all the way to the bottom making sure that the triangles
were about the same size. Then, I counted the total number of small
triangles that I made in the larger one and it came out to be 25.

They made a tessellation where one right-side-up triangle fit next to
an upside-down triangle and then a right-side-up one and so on.
I then saw that there were 5 right-side-up triangles lined along the
bottom line that is 200 cm on the large triangle. Dividing 200 by 5
would give me the individual length of that side on one of the
smaller triangles so I divided them and got 40, which gave me my
answer for that side.

Next, I looked at the lengths on the big triangle and saw that the
length on the left side was 150. Then I figured out that 150 was 3/4 of
200. Then I figured out 3/4 of 40 which is 30-this gave me the length
of that side on the small triangles. I knew that this was right because
the small triangles are all congruent and the large and small triangles
are similar. They're sides are equal, it's just that the large triangle
is a magnified version of the smaller ones.

Next, I did the same thing for the last side. 175 is mid-way between
200 and 150 so I figured out what was mid-way between 30 and 40-which
is 35. This is how I figured out what the lengths of the sides were.

The left side of the small triangle is 30 cm., the right side is 35
cm., and the bottom side is 40 cm.

    I started out by placing triangles (connected) in rows increasing
by 2 (starting at one) until I reched 25.
    Then I counted how many small sides formed one big side. (5) I
divided the lengths of the sides of the big triangle by 5 to get the
lengths of the sides of the small triangle.

Bonus: The length would be 50 cm. I used the same method I used for the
regular problem.

The lengths of the small triangle were 30 cm. on the left side, 35
cm. on the right side, and 40 cm. on the bottom.

To figure this out we:

1. Draw out the triangle with the lengths showing so that you could see how large the triangle was. We used MS paint to put the tile 'puzzle' together as a tessellation.
2. We divided each length by 5 because there are 5 smaller triangles in each side of the larger triangle.

Calculations: 200 cm divided by 5 triangles=40 cm is the bottom portion of the smaller triangles. For the left portion, we divided 150 cm. by 5 triangles=30cm per left side, so the left portion of the triangles' length is 30 cm. For the right portion, 175 divided by 5=35 cm. The smaller triangle is 30 cm. on the left side, 35 cm. on the right side, and 40 cm. on the bottom.

The side lengths are 30 cm., 35 cm., and 40 cm.

This is simply a question of ratios of similitude.
The ratio of the areas of the small triangle to the large triangle is
1/25.
In Ratio of similitude, ratio of area= (ratio of lengths)^2
Therefore, the ratio of the lengths when the ratio of the areas is
1/25 is sqrt(1/25), or 1/5.
Divide 150, 175, and 200 by 5 and you get 30,35, and 40.

Bonus:  Using the same process, divide 250 by 5, and the corresponding
side would be 50.

The sides of the tile are 30 cm, 35 cm, and 40 cm.
Bonus tile has sides of 50 cm, 35 cm, and 40 cm.

If we want to cover the triangle with 25 triangle tiles, how do we
know what size the tiles should be? And if we guess (even correctly)
how could we prove that the area of triangle is covered all over with
those tiles? One-way to prove it is to calculate the area of the
original triangle and the area of the tile and make sure that the sum
of all tile areas is the same as the area of triangle. But do how we
know those tiles will fit? Cut them out of paper and fit them in?

An easy way to avoid those difficulties all together is to notice
that if the number of tiles is a square number (like 4, 16, 25, 36
and so on) than we could build a parallelogram from two triangles and
divide that parallelogram into 4, 16, 25, 36 smaller similar
parallelograms by dividing each side into 2, 4, 5, 6 parts and
connecting the dividing dots.

Than if we draw a diagonal connecting the opposite corners of the
parallelogram, not only the parallelogram will be divided into two
(original) triangles, but each parallelogram along the diagonal will
be divided into two triangles. If we continue to divide the rest of
the small parallelograms into two triangles by drawing the lines
parallel to the diagonal then by our construction we will have: a)
double the number of triangles, b) each half of the parallelogram
will have the number of triangles we need, c) area of the triangle is
covered completely with tiles, d) triangular tiles are similar to the
original triangle. That’s how we are going to solve the problem.

Let's draw a diagram like this:

1. Please, take the picture of the triangle in the problem and label
the left angle A, the top angle B, and the right angle C. Now from
the point A make a line parallel to the side BC and another line from
point C parallel to the side AB. Label the intersection of those
lines point D. You have built a parallelogram that is made from two
original triangles.

2. Divide each side of the parallelogram into five equal parts by
putting four dots on each side of the parallelogram. Now, connect
those dots and you will create 25 parallelograms similar to the
original.

3. Please note, that whole area of the parallelogram is covered with
our tiles and the dilation factor of the small parallelogram is 1/5.
In other words the small parallelogram has the same angles as big
one, but the sides are smaller by five times.

4. Now look at the line AC. It crosses parallelogram into two
triangles but also crosses five small tiles into two equal parts –
triangles. Make lines parallel to that line from each point on the
side of the parallelogram. Now the parallelogram is covered with 50
triangle tiles and each triangle is covered with 25 triangle tiles.

5. Each triangle side is 1/5 of the original triangle. Original
triangle is covered completely. And there are 25 triangle tiles.

6. The tile sides are 150/5 = 30 cm, 175/5 = 35 cm, and 200/5 = 40 cm.

7. For bonus triangle the tile sides are 250/5 = 50 cm, 175/5 = 35
cm, and 200/5 = 40 cm.

Michael Levitin.

27 students received credit this week.

Amber Beauchemin, age 11 - Roland Rogers Elementary School, Absecon, NJ
Jenny Berg, age 13 - Central Middle School, Eden Prairie, MN
Mitchell Berkowitz, age 13 - Howell Township Middle School North, Howell, NJ
Bogdan Bintu, age 12 - School nr. 8 Onesti, Onesti, Romania
Vivian Chui, age 12 - Sperling Elementary School, Burnaby, British Columbia, Canada
Cristy Cobb, age 13 - Aptos Middle School, San Francisco, CA
Keri D., age 13 - Howell Township Middle School North, Howell, NJ
Jay Desai, age 13 - Howell Township Middle School North, Howell, NJ
Colleen Dunn, age 13 - Abigail Adams Intermediate School, East Weymouth, MA
Ning-Jiun Jan, age 13 - Howell Township Middle School North, Howell, NJ
Amanda Kees, age 13 - Howell Township Middle School North, Howell, NJ
Brittany Keller, age 13 - Howell Township Middle School North, Howell, NJ
Michael Kim, age 13 - Aptos Middle School, San Francisco, CA
Andrew Ko, age 13 - Palmyra Area Middle School, Palmyra, PA
Andrei Lazanu, age 13 - School No. 205, Bucharest, Romania
Michael Levitin, age 8 - St. Margaret's Episcopal School, San Juan Capistrano, CA
Adam Lohr, age 13 - Palmyra Area Middle School, Palmyra, PA
Garrett Marcotte, age 13 - Polytechnic School, Pasadena, CA
Jessica Martino, age 13 - Howell Township Middle School North, Howell, NJ
Ms. Magnifico's Fifth Grade Class, average age 11 - Saklan Valley School, Moraga, CA
Jana Peirce, age 25+ - Weller Elementary School, Fairbanks, AK
Maia Peirce, age 10 - Weller Elementary School, Fairbanks, AK
Genia Sokolskaya, age 10 - Penn Valley Elementary School, Narberth, PA
Sashe Sokolsky, age 14 - Welsh Valley Middle School, Narberth, PA
Annie Stiver, age 12 - Roland Rogers Elementary School, Absecon, NJ
Willie Sze, age 13 - Howell Township Middle School North, Howell, NJ
Y6 Sparkies, average age 11 - All Saints Grammar, Sydney, Australia

View most of the solutions submitted by the students above

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