1. **Problem statement:** Find the length $x$ in each right triangle using the Pythagorean theorem, correct to 1 decimal place. 2. **Formula:** For a right triangle with legs $a$, $b$ and hypotenuse $c$, the Pythagorean theorem states: $$a^2 + b^2 = c^2$$ If $x$ is a leg, then: $$x = \sqrt{c^2 - a^2}$$ If $x$ is the hypotenuse: $$x = \sqrt{a^2 + b^2}$$ 3. **Diagram 1:** Legs 7 cm and 8 cm, hypotenuse $x$. $$x = \sqrt{7^2 + 8^2} = \sqrt{49 + 64} = \sqrt{113} \approx 10.6$$ 4. **Diagram 2:** Legs 4 cm and $x$, hypotenuse 7 cm. $$x = \sqrt{7^2 - 4^2} = \sqrt{49 - 16} = \sqrt{33} \approx 5.7$$ 5. **Diagram 3:** Legs 3.4 cm and 5.7 cm, hypotenuse $x$. $$x = \sqrt{3.4^2 + 5.7^2} = \sqrt{11.56 + 32.49} = \sqrt{44.05} \approx 6.6$$ 6. **Diagram 4:** Legs 5 cm and $x$, hypotenuse 9 cm. $$x = \sqrt{9^2 - 5^2} = \sqrt{81 - 25} = \sqrt{56} \approx 7.5$$ 7. **Diagram 5:** Legs 3.8 cm and $x$, hypotenuse 8.4 cm. $$x = \sqrt{8.4^2 - 3.8^2} = \sqrt{70.56 - 14.44} = \sqrt{56.12} \approx 7.5$$ 8. **Diagram 6:** Legs 7 cm and $x$, hypotenuse 8.8 cm. $$x = \sqrt{8.8^2 - 7^2} = \sqrt{77.44 - 49} = \sqrt{28.44} \approx 5.3$$ **Final answers:** Diagram 1: $x \approx 10.6$ cm Diagram 2: $x \approx 5.7$ cm Diagram 3: $x \approx 6.6$ cm Diagram 4: $x \approx 7.5$ cm Diagram 5: $x \approx 7.5$ cm Diagram 6: $x \approx 5.3$ cm