Question

find a. 2\sqrt{4} 4\sqrt{2} 4 2

Explanation:

Step1: Apply Pythagorean theorem

For a right - angled isosceles triangle (since \(a = b\) in a square divided by a diagonal), the Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\). Substituting \(b = a\) and \(c = 4\sqrt{2}\), we get \(a^{2}+a^{2}=(4\sqrt{2})^{2}\).

Step2: Simplify the equation

\(2a^{2}=32\) (because \((4\sqrt{2})^{2}=4^{2}\times(\sqrt{2})^{2}=16\times2 = 32\)). Then \(a^{2}=\frac{32}{2}=16\).

Step3: Solve for \(a\)

Take the square root of both sides: \(a=\sqrt{16}\). Since \(a>0\) (length), \(a = 4\).

Answer:

\(4\)