The problem asks us to find the value of $y$ in two right triangles. In the first triangle (a), the two legs have lengths 1 and 1, and $y$ is the length of the hypotenuse. In the second triangle (b), the hypotenuse has length 2, one leg has length 1, and $y$ is the length of the other leg. We are given that $y > 0$. We are asked to use the Pythagorean theorem: $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs and $c$ is the hypotenuse.

1. Problem Description

The problem asks us to find the value of $y$ in two right triangles. In the first triangle (a), the two legs have lengths 1 and 1, and $y$ is the length of the hypotenuse. In the second triangle (b), the hypotenuse has length 2, one leg has length 1, and $y$ is the length of the other leg. We are given that $y > 0$. We are asked to use the Pythagorean theorem: $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs and $c$ is the hypotenuse.

2. Solution Steps

a) In the first triangle, we are given the two legs, both with length

1. We have $y$ as the length of the hypotenuse.

Using the Pythagorean theorem, we have:

$y^2 = 1^2 + 1^2$

$y^2 = 1 + 1$

$y^2 = 2$

Since $y > 0$, we take the positive square root:

$y = \sqrt{2}$

b) In the second triangle, we are given the hypotenuse with length 2 and one leg with length

1. We need to find the length of the other leg, $y$.

Using the Pythagorean theorem, we have:

$1^2 + y^2 = 2^2$

$1 + y^2 = 4$

$y^2 = 4 - 1$

$y^2 = 3$

Since $y > 0$, we take the positive square root:

$y = \sqrt{3}$

3. Final Answer

a) $y = \sqrt{2}$

b) $y = \sqrt{3}$