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Draw any triangle on your paper. Find the three midpoints and connect them in order. Explain why the four triangles created are all similar to the original.

In Problem H1, how do the sides of the new triangles compare to the sides of the original? How do their areas compare?

Draw a square on your paper. Then draw a square with sides 3 times as long. How many times will your original square fit inside the new square?

If two polygons are similar, and one has sides that are r times as long as the sides of the other, how will their areas compare? Explain your answer.

Suppose that a tree’s shadow is 21 feet long and a yardstick’s shadow is 18 inches long. Using the method from Problem B3, find the height of the tree.

**Schifter, Deborah (February, 1999). Learning Geometry: Some Insights Drawn from Teacher Writing. Teaching Children Mathematics, 5, (5), 360-366.
**Reproduced with permission from Teaching Children Mathematics. Copyright © 1999 by the National Council of Teachers of Mathematics. All rights reserved.

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Learning Geometry: Some Insights Drawn from Teacher Writing

Continued…

The midline theorem in Learning Math: Geometry, Session 5 tells us that the segment connecting two midpoints is half as long as the third side of a triangle. So for all four triangles, we can see that the sides are half as long as the sides of the original triangle, so they are similar to it by SSS.

The sides of the new triangles are half the length of the corresponding sides in the original triangle. The areas of the new triangles are one-fourth of the area of the original triangle.

Nine times.

The area of the polygon whose sides are r times as long will have an area that is r^{2} times the area of the original polygon. Proving this assertion involves recalling that any polygon can be divided into triangles. By the way the two polygons relate, the sides of all of the triangles inside one of the polygons will be r times longer than the sides of the triangles inside the other polygon. By Problem H3, each constituent triangle in the polygon whose sides are r times longer will have an area r^{2} times greater. So the entire area will be r^{2} times greater.

We convert all measures into feet and set up a proportion based on the similarity of right triangles whose heights are the yardstick and the tree and whose bases are the respective shadows. So we get that, if x is the height of the tree, then

x/3 = 21/1.5.

It follows that x = 42, so the tree is 42 feet tall.