1. **Problem 1(a):** Calculate $x$ in a right triangle with legs 4 cm and 6 cm, hypotenuse $x$. 2. Use the Pythagorean theorem formula: $$x^2 = a^2 + b^2$$ where $a$ and $b$ are legs. 3. Substitute values: $$x^2 = 4^2 + 6^2 = 16 + 36 = 52$$ 4. Solve for $x$: $$x = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$$ --- 1. **Problem 1(b):** Calculate $x$ in a right triangle with base 7 cm, hypotenuse 12 cm, height $x$. 2. Use Pythagorean theorem: $$12^2 = 7^2 + x^2$$ 3. Substitute values: $$144 = 49 + x^2$$ 4. Isolate $x^2$: $$x^2 = 144 - 49 = 95$$ 5. Solve for $x$: $$x = \sqrt{95}$$ --- 1. **Problem 1(c):** Calculate $x$ in a triangle with sides 7 cm, 20 cm, and unknown side $x$. 2. Since no right angle is specified, assume $x$ is the hypotenuse if 7 and 20 are legs. 3. Use Pythagorean theorem: $$x^2 = 7^2 + 20^2 = 49 + 400 = 449$$ 4. Solve for $x$: $$x = \sqrt{449}$$ --- **Final answers:** 1(a) $x = 2\sqrt{13}$ cm 1(b) $x = \sqrt{95}$ cm 1(c) $x = \sqrt{449}$ cm