1. **Problem statement:** We have an isosceles triangle with base length $20\sqrt{3}$ cm. The two equal legs decrease at a rate of 3 cm/h. We want to find the rate at which the triangle's area decreases when the equal legs are equal to the base length. 2. **Known values:** - Base $b = 20\sqrt{3}$ cm (constant) - Length of equal legs $x$ (variable) - Rate of change of legs $\frac{dx}{dt} = -3$ cm/h (negative because length decreases) 3. **Formula for area of isosceles triangle:** $$ A = \frac{1}{2} b h $$ where $h$ is the height. 4. **Find height $h$ in terms of $x$ and $b$:** Using Pythagoras theorem in the triangle formed by height, half base, and leg: $$ h = \sqrt{x^2 - \left(\frac{b}{2}\right)^2} $$ 5. **Express area $A$ as a function of $x$:** $$ A = \frac{1}{2} b \sqrt{x^2 - \left(\frac{b}{2}\right)^2} $$ 6. **Differentiate $A$ with respect to time $t$ using chain rule:** $$ \frac{dA}{dt} = \frac{1}{2} b \cdot \frac{1}{2} \left(x^2 - \left(\frac{b}{2}\right)^2\right)^{-\frac{1}{2}} \cdot 2x \cdot \frac{dx}{dt} $$ Simplify: $$ \frac{dA}{dt} = \frac{b x}{2 \sqrt{x^2 - \left(\frac{b}{2}\right)^2}} \cdot \frac{dx}{dt} $$ 7. **Substitute $b = 20\sqrt{3}$ and $x = b = 20\sqrt{3}$:** Calculate denominator: $$ \sqrt{x^2 - \left(\frac{b}{2}\right)^2} = \sqrt{(20\sqrt{3})^2 - (10\sqrt{3})^2} = \sqrt{1200 - 300} = \sqrt{900} = 30 $$ 8. **Calculate $\frac{dA}{dt}$:** $$ \frac{dA}{dt} = \frac{(20\sqrt{3})(20\sqrt{3})}{2 \times 30} \times (-3) = \frac{400 \times 3}{60} \times (-3) = \frac{1200}{60} \times (-3) = 20 \times (-3) = -60 $$ 9. **Interpretation:** The area decreases at a rate of 60 cm²/h at the moment when the equal legs equal the base length. **Final answer:** $$ \boxed{\frac{dA}{dt} = -60 \text{ cm}^2/\text{h}} $$