Both figures shown here are squares with a side length of . Notice that the first figure is divided into two squares and two rectangles. The second figure is divided into a square and four right triangles with legs of lengths and . Let’s call the hypotenuse of these triangles .
F
First of two squares of the same area. This square is divided into the following: A square with side lengths “a”. Two rectangles with side lengths “a” and “b”. A square with side lengths “b”.
G
Second of two squares of the same area. This square is divided into the following: Four identical triangles on each corner of the square with sides labeled “a” and “b”. A square in the center with unlabeled side lengths.
What is the total area of each figure?
Find the area of each of the 9 smaller regions shown in the figures and label them.
Add up the area of the 4 regions in Figure F and set this expression equal to the sum of the areas of the 5 regions in Figure G. If you rewrite this equation using as few terms as possible, what do you have?
8.3
Activity
Let’s Take It for a Spin
Find the unknown side lengths in these right triangles.
8.4
Activity
A Transformational Proof
Your teacher will give your group a sheet with 4 figures. Cut out the 5 shapes in Figure 1.
Arrange the 5 cut out shapes to fit inside Figure 2.
Now arrange the shapes to fit inside Figure 3.
Check to see that Figure 3 is congruent to the large square in Figure 4.
Check to see that the 5 cut out shapes fit inside the two smaller squares in Figure 4.
If the right triangle in Figure 4 has legs and and hypotenuse , what have you just demonstrated to be true?
Student Lesson Summary
The figures shown can be used to see why the Pythagorean Theorem is true. Both large squares have the same area, but they are broken up in different ways. When the sum of the four areas in Square F is set equal to the sum of the 5 areas in Square G, the result is , where is the hypotenuse of the triangles in Square G and also the side length of the square in the middle.
F
First of two squares of the same area. This square is divided into the following: A square with side lengths “a”. Two rectangles with side lengths “a” and “b”. A square with side lengths “b”.
G
Second of two squares of the same area. This square is divided into the following: Four identical triangles on each corner of the square with sides labeled “a” and “b”. A square in the center with unlabeled side lengths.
This is true for any right triangle. If the legs are and and the hypotenuse is , then .
For example, to find the length of side in this right triangle, we know that . The solution to this equation (and the length of the side) is .
Glossary
None
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