Question

what is area of the yellow square, a5?
pythagorean theorem
a = a1 + a2 + a3 + a4 + a5
(a+b)² = a1 + a2 + a3 + a4 + a5
4c
c⁴
c² - 2ab
c²

Explanation:

Step1: Recall the Pythagorean theorem and area relations

We know that the large square has side length \(a + b\), so its area is \((a + b)^2\). Also, the four blue triangles (A1, A2, A3, A4) are right - angled triangles with legs \(a\) and \(b\), and the yellow square (A5) has side length \(c\). The area of each right - angled triangle is \(\frac{1}{2}ab\), so the total area of the four triangles is \(4\times\frac{1}{2}ab=2ab\).

Step2: Express the area of the large square in terms of known areas

The area of the large square \((a + b)^2\) can also be expressed as the sum of the areas of the four triangles and the yellow square. So, \((a + b)^2=2ab+A5\).
We also know from the Pythagorean theorem that \(c^{2}=a^{2}+b^{2}\) (since the yellow square has side \(c\) and the right - angled triangles have hypotenuse \(c\)). Let's expand \((a + b)^2\): \((a + b)^2=a^{2}+2ab + b^{2}\). Substituting \(a^{2}+b^{2}=c^{2}\) into the expansion, we get \((a + b)^2=c^{2}+2ab\).
But we also have \((a + b)^2=2ab+A5\). So, by equating the two expressions for \((a + b)^2\), we can solve for \(A5\):
\(c^{2}+2ab=2ab + A5\)
Subtract \(2ab\) from both sides of the equation: \(A5 = c^{2}\)
Alternatively, since the yellow square has side length \(c\), the area of a square is side length squared, so the area of the yellow square \(A5=c\times c = c^{2}\)

Answer:

\(c^{2}\) (corresponding to the option with \(c^{2}\))