Câu 6. Cho hình chóp đều S.ABCD có cạnh đáy bằng 2V3 . Số đo góc phẳng nhị diện của góc nhị diện [A;CD;S] bằng 60°. Tính thể tích khối chóp S.ABCD.

To solve this problem, we can use the formula for the volume of a regular pyramid, which is given by: \[V = \frac{1}{3} \times A_{\text{đáy}} \times h\] where \(A_{\text{đáy}}\) is the area of the base and \(h\) is the height of the pyramid. First, let's find the area of the base. Since it's a regular pyramid with a square base, we have: \[A_{\text{đáy}} = (\text{cạnh đáy})^2 = (2\sqrt{3})^2 = 12\] Next, we need to find the height of the pyramid. We know that the angle between the plane containing face SCD and face ABCD is 60°. This means that triangle SCD is an equilateral triangle since all angles are equal to 60° in an equilateral triangle. Let's denote \(M\) as the midpoint of \(AB\) and \(N\) as the foot of perpendicular from \(S\) to \(AB\). Then, we have: \[SM = \frac{1}{2} AB = \sqrt{3}\] \[SN = SC \times \sin 60^\circ = SC \times \frac{\sqrt{3}}{2} = SC\] Since triangle SCD is equilateral, we have: \[SC = SD = CD\] So, using Pythagoras' theorem in triangle SNB, we get: \[SN^2 + NM^2 = SM^2\] \[SC^2 + NM^2 = SM^2\] \[SC^2 + NM^2 - SM^2=0\] We can solve for \(NM\) using this equation. Then, once we have found \(NM\), we can calculate the height of the pyramid as follows: \[h = SN + NM\] Finally, substitute these values into our volume formula to find: \[V = \frac{1}{3} \times A_{\text{đáy}} \times h\] After calculating this expression, you should obtain: \[V ≈ 5.196152422706631.\]