1. **State the problem:** We have two right-angled triangles sharing a common hypotenuse $z$. The larger triangle has legs 12 cm and 14 cm, and the smaller triangle inside it has one leg 6 cm and the hypotenuse $z$. We need to find the value of $z$ correct to 1 decimal place. 2. **Use Pythagoras' theorem:** For a right triangle with legs $a$ and $b$ and hypotenuse $c$, the theorem states: $$c^2 = a^2 + b^2$$ 3. **Apply Pythagoras to the larger triangle:** $$z^2 = 12^2 + 14^2$$ $$z^2 = 144 + 196$$ $$z^2 = 340$$ 4. **Calculate $z$ from the larger triangle:** $$z = \sqrt{340}$$ 5. **Apply Pythagoras to the smaller triangle:** The smaller triangle has legs 6 cm and the other leg unknown, call it $x$, and hypotenuse $z$. $$z^2 = 6^2 + x^2$$ $$340 = 36 + x^2$$ 6. **Solve for $x^2$:** $$x^2 = 340 - 36$$ $$x^2 = 304$$ 7. **Calculate $x$:** $$x = \sqrt{304}$$ $$x \approx 17.4$$ 8. **Answer:** The value of $z$ is approximately $$z = \sqrt{340} \approx 18.4$$ **Final answer:** $$z = 18.4\text{ cm}$$