Question

question 2
which statement best explains why the sum of the areas of the two white squares in figure 2 is the same as the area of the white square in figure 1?
in each triangle, the length of side a plus the length of side b equals the length of side c. that means that ( a^{2}+b^{2}=c^{2} ).
the combined area of the four triangles is equal to the area of the large white square.
in each figure the total area is equal and the area of the 4 triangles is equal, so the remaining white area in each figure must also be equal.

Explanation:

Step1: Analyze the total area of each figure

In Figure 1 and Figure 2, assume the side - length of the large square (formed by the outer - most boundaries) is \(a + b\). The area of a square with side - length \(s\) is \(s^{2}\), so the area of the large square in both figures is \((a + b)^{2}=a^{2}+2ab + b^{2}\) (using the formula \((x + y)^{2}=x^{2}+2xy + y^{2}\) where \(x=a\) and \(y = b\)).

Step2: Analyze the area of the four right - angled triangles

The area of a right - angled triangle with legs \(a\) and \(b\) (and hypotenuse \(c\)) is \(A=\frac{1}{2}ab\). The combined area of the four right - angled triangles in both Figure 1 and Figure 2 is \(4\times\frac{1}{2}ab=2ab\).

Step3: Analyze the area of the white region

In Figure 1, the area of the white region (a square with side - length \(c\)) is \(A_{1}=c^{2}\). In Figure 2, the area of the white region is \(A_{2}=a^{2}+b^{2}\). Since the total area of each figure (\((a + b)^{2}\)) is the same and the area of the four triangles (\(2ab\)) is the same in both figures. Using the formula \(A_{white}=A_{total}-A_{triangles}\), we have \(c^{2}=a^{2}+b^{2}\) (because \(A_{total1}-A_{triangles1}=A_{total2}-A_{triangles2}\)).

Answer:

In each figure the total area is equal and the area of the 4 triangles is equal, so the remaining white area in each figure must also be equal.