Let's go through each of the triangles to find the value of $x$ using the Pythagorean theorem, which is applicable to right triangles.

a) First Triangle

This is a right triangle with both legs having length $x$ and the hypotenuse given as 5.

Using the Pythagorean theorem: $x^2 + x^2 = 5^2$ $2x^2 = 25$ $x^2 = \frac{25}{2}$ $x = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$

So, the exact value of $x$ for triangle (a) is $\frac{5\sqrt{2}}{2}$.

b) Second Triangle

This is an isosceles right triangle where the two legs are equal to $x$, and the hypotenuse is 4.

Again, using the Pythagorean theorem: $x^2 + x^2 = 4^2$ $2x^2 = 16$ $x^2 = 8$ $x = \sqrt{8} = 2\sqrt{2}$

Thus, $x$ for triangle (b) is $2\sqrt{2}$.

c) Third Triangle

This is also an isosceles right triangle, with one leg equal to $3.9$ and the hypotenuse equal to $x$.

Again, applying the Pythagorean theorem: $3.9^2 + 3.9^2 = x^2$ $2(3.9^2) = x^2$ Let's calculate $3.9^2$:

$3.9^2 = 15.21$ $2(15.21) = x^2$ $x^2 = 30.42$ $x = \sqrt{30.42} \approx 5.52$

Thus, $x$ for triangle (c) is approximately $5.52$.

Would you like more details on any of the steps or clarifications? Here are some related questions you might consider:

1. What is the Pythagorean theorem and how is it applied?
2. How can I rationalize square roots, as seen in triangle (a)?
3. What are the properties of isosceles right triangles?
4. How do you approximate square roots like in triangle (c)?
5. What happens if the sides of a right triangle are in different ratios?

Tip: For isosceles right triangles, the ratio of the legs to the hypotenuse is always $1:1:\sqrt{2}$.