1. **State the problem:** We have a right circular cone where the slant length $l$ is equal to the diameter of its base. We need to find the volume of this cone. 2. **Known formulas and definitions:** - The slant length $l$ of a cone relates to the radius $r$ and height $h$ by the Pythagorean theorem: $$l = \sqrt{r^2 + h^2}$$ - The diameter $d$ of the base is twice the radius: $$d = 2r$$ - The volume $V$ of a cone is given by: $$V = \frac{1}{3} \pi r^2 h$$ 3. **Given condition:** The slant length equals the diameter, so: $$l = d = 2r$$ 4. **Use the Pythagorean relation:** $$l^2 = r^2 + h^2$$ Substitute $l = 2r$: $$ (2r)^2 = r^2 + h^2 $$ $$ 4r^2 = r^2 + h^2 $$ 5. **Solve for height $h$:** $$ h^2 = 4r^2 - r^2 = 3r^2 $$ $$ h = \sqrt{3r^2} = r\sqrt{3} $$ 6. **Calculate the volume:** $$ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (r\sqrt{3}) = \frac{1}{3} \pi r^3 \sqrt{3} = \frac{\pi r^3 \sqrt{3}}{3} $$ **Final answer:** $$ V = \frac{\pi r^3 \sqrt{3}}{3} $$ This expresses the volume in terms of the radius $r$ of the base.