Find the full surface of an equilateral cone if its height is 3 dm.

The full surface of the cone is made up of the area of ​​the base and the area of ​​the lateral surface: S floor = Sb + S side. There is a circle at the base of the cone, its area is found by the formula Sosn = π ∙ R², the lateral surface area is found by the formula Sbok = π ∙ R ∙ L, where R is the radius of the circle lying at the base, L is the generator of the cone. From the condition of the problem it is known that the height of the cone is H = 3 dm and that it is equilateral, that is, its axial section is an equilateral triangle, then L = 2 ∙ R. We get Spol = π ∙ R² + π ∙ R ∙ L = π ∙ R² + π ∙ R ∙ 2 ∙ R = 3 ∙ π ∙ R². From the triangle obtained in the section by the Pythagorean theorem, we find that R² = Н² / 3. Then Spol = 3 ∙ π ∙ Н² / 3 = π ∙ Н². Substituting the numerical value of the height, we get Spol = π ∙ (3) ² = 9 ∙ π ≈ 28, 26 (dm²).
Answer: the total surface area of ​​the cone is ≈ 28, 26 dm².