Question

two congruent squares are shown in figures 1 and 2 below.
figure 1
figure 2
use the drop - down menus to complete the proof of the pythagorean theorem using the figures.
click the arrows to choose an answer from each menu.
the combined area of the shaded triangles in figure 1 is choose... the combined area of the shaded triangles in figure 2. the area of the unshaded square in figure 1 can be represented by choose.... the combined area of the two unshaded squares in figure 2 can be represented by choose.... the areas of the squares in figure 1 and figure 2 show that choose....

Explanation:

Step1: Analyze the number of shaded triangles

In Figure 1, there are 4 shaded right - angled triangles. In Figure 2, there are also 4 shaded right - angled triangles. Since the two squares are congruent (have the same side length and area) and the triangles are congruent (by the properties of the figures, as the squares are congruent and the triangles are formed in a similar way with legs \(a\) and \(b\)), the combined area of the shaded triangles in Figure 1 is equal to the combined area of the shaded triangles in Figure 2.

Step2: Find the area of the un - shaded square in Figure 1

The side length of the large square (in both figures) is \(a + b\). The area of the large square is \((a + b)^2=a^{2}+2ab + b^{2}\). The area of each of the 4 shaded triangles is \(\frac{1}{2}ab\). The combined area of the 4 shaded triangles is \(4\times\frac{1}{2}ab=2ab\). The area of the un - shaded square in Figure 1 (with side length \(c\)) is \(c^{2}\). Using the formula \(A_{un - shaded\ Figure1}=A_{large\ square}-A_{shaded\ triangles\ Figure1}\), we get \(c^{2}=(a + b)^2-2ab=a^{2}+b^{2}\)

Step3: Find the combined area of the un - shaded squares in Figure 2

The area of the first un - shaded square (with side length \(a\)) is \(a^{2}\) and the area of the second un - shaded square (with side length \(b\)) is \(b^{2}\). The combined area of the two un - shaded squares in Figure 2 is \(a^{2}+b^{2}\)

Step4: Relate the areas

Since the area of the large square is the same for both figures (because the squares are congruent) and the area of the shaded regions is the same (as shown in Step 1), the area of the un - shaded region in Figure 1 (\(c^{2}\)) is equal to the combined area of the un - shaded regions in Figure 2 (\(a^{2}+b^{2}\))

Answer:

The combined area of the shaded triangles in Figure 1 is equal to the combined area of the shaded triangles in Figure 2. The area of the unshaded square in Figure 1 can be represented by \(c^{2}\). The combined area of the two unshaded squares in Figure 2 can be represented by \(a^{2}+b^{2}\). The areas of the squares in Figure 1 and Figure 2 show that \(a^{2}+b^{2}=c^{2}\)