1. **Problem Statement:** Find the ratio between the area of triangle $\triangle ABC$ and $\triangle DEF$ given their side lengths. 2. **Given:** - $\triangle ABC$: $AB=20$, $AC=20$, $BC=10$ - $\triangle DEF$: $DE=60$, $DF=60$, $EF=30$ 3. **Formula:** The area of a triangle with sides $a$, $b$, and $c$ can be found using Heron's formula: $$s = \frac{a+b+c}{2}$$ $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ 4. **Calculate area of $\triangle ABC$:** - Semi-perimeter: $$s_{ABC} = \frac{20 + 20 + 10}{2} = 25$$ - Area: $$A_{ABC} = \sqrt{25(25-20)(25-20)(25-10)} = \sqrt{25 \times 5 \times 5 \times 15}$$ $$= \sqrt{25 \times 25 \times 15} = 25 \sqrt{15}$$ 5. **Calculate area of $\triangle DEF$:** - Semi-perimeter: $$s_{DEF} = \frac{60 + 60 + 30}{2} = 75$$ - Area: $$A_{DEF} = \sqrt{75(75-60)(75-60)(75-30)} = \sqrt{75 \times 15 \times 15 \times 45}$$ $$= \sqrt{75 \times 15 \times 15 \times 45}$$ Simplify stepwise: $$75 = 15 \times 5$$ $$45 = 15 \times 3$$ So, $$A_{DEF} = \sqrt{15 \times 5 \times 15 \times 15 \times 15 \times 3} = \sqrt{15^4 \times 5 \times 3} = 15^2 \sqrt{15} \sqrt{5 \times 3}$$ $$= 225 \sqrt{15} \sqrt{15} = 225 \times 15 = 3375$$ But this is incorrect because we must keep the square root intact: Recalculate carefully: $$A_{DEF} = \sqrt{75 \times 15 \times 15 \times 45}$$ $$= \sqrt{(75 \times 45) \times (15 \times 15)} = \sqrt{3375 \times 225}$$ $$= \sqrt{759375}$$ Factor 759375: $$759375 = 225^2 \times 15$$ Since $225^2 = 50625$, and $50625 \times 15 = 759375$. So, $$A_{DEF} = 225 \sqrt{15}$$ 6. **Ratio of areas:** $$\frac{A_{ABC}}{A_{DEF}} = \frac{25 \sqrt{15}}{225 \sqrt{15}} = \frac{25}{225} = \frac{1}{9}$$ **Final answer:** The ratio of the areas of $\triangle ABC$ to $\triangle DEF$ is $\boxed{\frac{1}{9}}$.