Question

1. solve for $x$ in a right triangle with one leg $3$ mi, hypotenuse $5$ mi, and the other leg $x$.
2. solve for $x$ in a right triangle with one leg $8$ km, hypotenuse $10$ km, and the other leg $x$.
3. solve for $x$ in a right triangle with one leg $9$ ft, hypotenuse $10$ ft, and the other leg $x$.
4. solve for $x$ in a right triangle with one leg $4$ in, hypotenuse $12$ in, and the other leg $x$.

Explanation:

Step1: Apply Pythagorean theorem (1)

For right triangles, $a^2 + b^2 = c^2$, where $c$ is hypotenuse. For triangle 1: $x^2 + 3^2 = 5^2$
$x^2 = 5^2 - 3^2 = 25 - 9 = 16$
$x = \sqrt{16} = 4$

Step2: Apply Pythagorean theorem (2)

For triangle 2: $8^2 + x^2 = 10^2$
$x^2 = 10^2 - 8^2 = 100 - 64 = 36$
$x = \sqrt{36} = 6$

Step3: Apply Pythagorean theorem (3)

For triangle 3: $x^2 + 9^2 = 10^2$
$x^2 = 10^2 - 9^2 = 100 - 81 = 19$
$x = \sqrt{19} \approx 4.36$

Step4: Apply Pythagorean theorem (4)

For triangle 4: $4^2 + x^2 = 12^2$
$x^2 = 12^2 - 4^2 = 144 - 16 = 128$
$x = \sqrt{128} = 8\sqrt{2} \approx 11.31$

Answer:

1. $x = 4$ mi
2. $x = 6$ km
3. $x = \sqrt{19} \approx 4.36$ ft
4. $x = 8\sqrt{2} \approx 11.31$ in