1. **State the problem:** We have three right triangles and need to find the unknown side $x$ in each using the Pythagorean theorem. 2. **Recall the Pythagorean theorem:** For a right triangle with legs $a$, $b$ and hypotenuse $c$, the relation is $$a^2 + b^2 = c^2$$ 3. **Triangle 1 (top-right):** Legs are 64 and 16, hypotenuse is $x$. Apply the theorem: $$64^2 + 16^2 = x^2$$ Calculate squares: $$4096 + 256 = x^2$$ $$4352 = x^2$$ Take square root: $$x = \sqrt{4352}$$ Simplify: $$4352 = 64 \times 68 = 64 \times (4 \times 17)$$ $$x = \sqrt{64 \times 4 \times 17} = 8 \times 2 \times \sqrt{17} = 16\sqrt{17}$$ 4. **Triangle 2 (center):** Hypotenuse is 65, one leg is $4\sqrt{65}$, other leg is $x$. Apply the theorem: $$x^2 + (4\sqrt{65})^2 = 65^2$$ Calculate squares: $$(4\sqrt{65})^2 = 16 \times 65 = 1040$$ $$65^2 = 4225$$ Substitute: $$x^2 + 1040 = 4225$$ Subtract 1040: $$x^2 = 4225 - 1040 = 3185$$ Take square root: $$x = \sqrt{3185}$$ Factor 3185: $$3185 = 5 \times 637 = 5 \times 7 \times 91 = 5 \times 7 \times 7 \times 13$$ $$x = \sqrt{5 \times 7^2 \times 13} = 7 \sqrt{65}$$ 5. **Triangle 3 (bottom-left):** Hypotenuse is $18\sqrt{3}$, one leg is 27, other leg is $x$. Apply the theorem: $$27^2 + x^2 = (18\sqrt{3})^2$$ Calculate squares: $$27^2 = 729$$ $$(18\sqrt{3})^2 = 18^2 \times 3 = 324 \times 3 = 972$$ Substitute: $$729 + x^2 = 972$$ Subtract 729: $$x^2 = 972 - 729 = 243$$ Take square root: $$x = \sqrt{243}$$ Simplify: $$243 = 81 \times 3 = 9^2 \times 3$$ $$x = 9\sqrt{3}$$ **Final answers:** - Triangle 1: $x = 16\sqrt{17}$ - Triangle 2: $x = 7\sqrt{65}$ - Triangle 3: $x = 9\sqrt{3}$