Question

click the arrows to choose an answer from each menu. the expression choose... represents the area of the figure as the sum of the area of the shaded triangles and the area of the white square. the equivalent expressions choose... use the length of the figure to represent the area. setting two of these area expressions equal to each other and subtracting choose...

Explanation:

Step1: Analyze the figure's components

The figure has a white square with side \( c \) (area \( c^2 \)) and four right - angled triangles? Wait, no, looking at the figure, there are two pairs of right - angled triangles? Wait, actually, the shaded regions: let's see, the big figure is a square with side \( a + b \)? Wait, no, the base is \( a + b \), and the height? Wait, the white square has side \( c \), and the shaded triangles: each triangle has legs \( a \) and \( b \). Wait, how many shaded triangles? Let's count: there are four? No, looking at the figure, maybe two pairs. Wait, actually, the area of the shaded triangles: each triangle has area \( \frac{1}{2}ab \), and there are four? Wait, no, in the figure, the shaded regions: let's see, the white square is in the middle, and the shaded parts are four triangles? Wait, no, maybe two triangles with legs \( a \) and \( b \) each? Wait, no, the first part: the expression for the area as the sum of shaded triangles and white square.

The white square has area \( c^2 \). The shaded triangles: let's see, each triangle has area \( \frac{1}{2}ab \), and there are four? Wait, no, in the figure, the base of the big figure is \( a + b \), and the height is also \( a + b \)? Wait, no, the big figure is a square with side \( a + b \)? Wait, no, the figure is a square with side \( a + b \), and inside there is a white square of side \( c \), and four right - angled triangles? Wait, no, maybe two triangles with legs \( a \) and \( b \) each? Wait, no, let's re - examine.

Wait, the first expression: sum of shaded triangles and white square. The white square area is \( c^2 \). The shaded triangles: each has area \( \frac{1}{2}ab \), and there are four? No, in the figure, I think there are four right - angled triangles? Wait, no, the figure shows two red triangles on the top and bottom, and two on the left and right? Wait, maybe the shaded triangles: each has area \( \frac{1}{2}ab \), and there are four? Wait, no, let's calculate the area of the big square (side \( a + b \)): \( (a + b)^2=a^{2}+2ab + b^{2} \). The white square has area \( c^2 \), and the shaded area is the area of the big square minus the white square: \( (a + b)^2-c^{2}=a^{2}+2ab + b^{2}-c^{2} \). But the other way: the shaded triangles and the white square. Wait, maybe the shaded triangles: each has area \( \frac{1}{2}ab \), and there are four? No, wait, in the figure, the shaded regions: let's see, the white square is in the middle, and the shaded parts are four triangles? Wait, no, maybe two triangles with legs \( a \) and \( b \) each? Wait, no, let's start over.

The first part: the expression that represents the area as sum of shaded triangles and white square. The white square area is \( c^2 \). The shaded triangles: each triangle has area \( \frac{1}{2}ab \), and there are four? Wait, no, in the figure, I think there are four right - angled triangles? Wait, no, the figure has a white square with side \( c \), and four right - angled triangles with legs \( a \) and \( b \)? Wait, no, the base of the big figure is \( a + b \), and the height is \( a + b \), so the area of the big figure is \( (a + b)^2 \). The white square has area \( c^2 \), and the shaded area is \( (a + b)^2 - c^2 \). But the other way: the shaded triangles and the white square. Wait, maybe the shaded triangles: each has area \( \frac{1}{2}ab \), and there are four? No, wait, if we look at the figure, the shaded regions are four triangles? Wait, no, maybe two triangles with legs \( a \) and \( b \) each? Wait, no, let's calculate the area of the shaded triangles. Suppose there are four right - angled triangles, each with area \( \frac{1}{2}ab \), so total area of shaded triangles is \( 4\times\frac{1}{2}ab = 2ab \). Then the area of the figure as sum of shaded triangles and white square is \( c^2+2ab \).

The equivalent expression using the length of the figure (the big square with side \( a + b \)) is \( (a + b)^2=a^{2}+2ab + b^{2} \).

Setting these two expressions equal: \( c^2 + 2ab=a^{2}+2ab + b^{2} \). Then subtracting \( 2ab \) from both sides gives \( c^2=a^{2}+b^{2} \), which is the Pythagorean theorem.

But let's answer the first part: the expression that represents the area as sum of shaded triangles and white square is \( c^{2}+2ab \) (assuming there are two pairs of triangles, each with area \( \frac{1}{2}ab \), so total \( 2ab \)). The equivalent expression using the length (side \( a + b \)) is \( (a + b)^{2}=a^{2}+2ab + b^{2} \). Setting them equal and subtracting \( 2ab \) gives \( c^{2}=a^{2}+b^{2} \).

Step2: First dropdown (expression as sum)

The area of the white square is \( c^{2} \). The area of each shaded triangle: looking at the figure, each triangle has legs \( a \) and \( b \), and there are four? No, wait, in the figure, the shaded regions: let's count the number of triangles. There are four right - angled triangles? Wait, no, the figure shows two red triangles on the top, two on the bottom? Wait, no, maybe two triangles with area \( \frac{1}{2}ab \) each, so total shaded area is \( 2\times\frac{1}{2}ab\times2 = 2ab \)? Wait, no, maybe four triangles, each with area \( \frac{1}{2}ab \), so total \( 4\times\frac{1}{2}ab=2ab \). So the area as sum of shaded triangles and white square is \( c^{2}+2ab \).

Step3: Second dropdown (equivalent expression)

The length of the figure (the big square) is \( a + b \), so its area is \( (a + b)^{2}=a^{2}+2ab + b^{2} \).

Step4: Third dropdown (subtracting)

When we set \( c^{2}+2ab=a^{2}+2ab + b^{2} \), we subtract \( 2ab \) from both sides to get \( c^{2}=a^{2}+b^{2} \).

Answer:

1. For the first "Choose...": \( c^{2}+2ab \)
2. For the second "Choose...": \( (a + b)^{2} \) (or \( a^{2}+2ab + b^{2} \))
3. For the third "Choose...": \( 2ab \) (from both sides)