To solve for the missing sides $x$ in both right triangles, we will use the Pythagorean theorem:

a^2 + b^2 = c^2 \quad \text{(where c is the hypotenuse).}

Problem 4:

• Given:
  • One leg = $12 \, \text{km}$,
  • Hypotenuse = $13 \, \text{km}$,
  • Other leg = $x$.

Using the Pythagorean theorem:

$x^2 + 12^2 = 13^2$ $x^2 + 144 = 169$ $x^2 = 169 - 144$ $x^2 = 25$ $x = \sqrt{25} = 5$

Answer for Problem 4: $x = 5 \, \text{km}$.

Problem 5:

• Given:
  • One leg = $8 \, \text{cm}$,
  • Hypotenuse = $10 \, \text{cm}$,
  • Other leg = $x$.

Using the Pythagorean theorem:

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

Answer for Problem 5: $x = 6 \, \text{cm}$.

Final Answers:

• Problem 4: $x = 5 \, \text{km}$
• Problem 5: $x = 6 \, \text{cm}$

Let me know if you need further clarification! 😊