1. **Problem statement:** Use Pythagoras' theorem to find the unknown length in each right triangle given the two legs. 2. **Formula:** For a right triangle with legs $a$ and $b$, and hypotenuse $c$, Pythagoras' theorem states: $$c = \sqrt{a^2 + b^2}$$ If the unknown side is a leg, and the hypotenuse and one leg are known, use: $$a = \sqrt{c^2 - b^2}$$ 3. **Triangle a:** Legs are 2 cm and 4 cm, find hypotenuse $c$. $$c = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \text{ cm}$$ 4. **Triangle b:** Legs are 6 m and 3 m, find hypotenuse $c$. $$c = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \text{ m}$$ 5. **Triangle c:** Legs are 1 mm and 12 mm, find hypotenuse $c$. $$c = \sqrt{1^2 + 12^2} = \sqrt{1 + 144} = \sqrt{145} \text{ mm}$$ 6. **Triangle d:** Legs are $\sqrt{7}$ m and 2 m, find hypotenuse $c$. $$c = \sqrt{(\sqrt{7})^2 + 2^2} = \sqrt{7 + 4} = \sqrt{11} \text{ m}$$ 7. **Triangle e:** Legs are $\sqrt{20}$ mm and 3 mm, find hypotenuse $c$. $$c = \sqrt{(\sqrt{20})^2 + 3^2} = \sqrt{20 + 9} = \sqrt{29} \text{ mm}$$ 8. **Triangle f:** Legs are 4 cm and 10 cm, find hypotenuse $c$. $$c = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29} \text{ cm}$$ **Final answers:** - a: $2\sqrt{5}$ cm - b: $3\sqrt{5}$ m - c: $\sqrt{145}$ mm - d: $\sqrt{11}$ m - e: $\sqrt{29}$ mm - f: $2\sqrt{29}$ cm