Question

find the length of side $x$ to the nearest tenth. \
(triangle with right angle, two 45° angles, one leg is $\sqrt{2}$, hypotenuse is $x$)

Explanation:

Step1: Identify triangle type

This is a right isosceles triangle (two 45° angles, one right angle), so the two legs are equal. Given one leg is $\sqrt{2}$, the other leg is also $\sqrt{2}$.

Step2: Apply Pythagorean theorem

Use $x^2 = a^2 + b^2$, where $a = b = \sqrt{2}$.
$$x^2 = (\sqrt{2})^2 + (\sqrt{2})^2$$
$$x^2 = 2 + 2 = 4$$

Step3: Solve for x

Take square root of both sides.
$$x = \sqrt{4} = 2$$

Answer:

2.0