Question

1. solve for x

4.
(there is a right triangle with legs marked as equal (isosceles right triangle) and hypotenuse ( 2sqrt{2} ), one leg is labeled x)

Explanation:

Step1: Identify Triangle Type

The triangle is a right - isosceles triangle (legs are marked equal), so legs \(a = b=x\), hypotenuse \(c = 2\sqrt{2}\).

Step2: Apply Pythagorean Theorem

For a right triangle, \(a^{2}+b^{2}=c^{2}\). Substituting \(a = b=x\) and \(c = 2\sqrt{2}\), we get \(x^{2}+x^{2}=(2\sqrt{2})^{2}\).
Simplify left - hand side: \(2x^{2}\), right - hand side: \((2\sqrt{2})^{2}=2^{2}\times(\sqrt{2})^{2}=4\times2 = 8\).
So the equation becomes \(2x^{2}=8\).

Step3: Solve for \(x^{2}\)

Divide both sides of the equation \(2x^{2}=8\) by 2: \(x^{2}=\frac{8}{2}=4\).

Step4: Solve for \(x\)

Take the square root of both sides. Since \(x\) represents the length of a side of a triangle, \(x>0\). So \(x=\sqrt{4}=2\).

Answer:

\(x = 2\)