Cho tứ diện ABCD có \(\left(ABD\right)\perp\left(BCD\right)\) . Tính thể tích khối tứ diện biết AB = BC = a; BD = 2a; \(\widehat{BAD}=\widehat{BCD}=90^o.\)
\(a^3\frac{\sqrt{3}}{2}\)
\(\frac{a^3}{12}\)
\(\frac{a^3}{4}\)
\(a^3\frac{\sqrt{3}}{6}\)
Hướng dẫn giải:
N�t b�t n�tB�t1: DinhKemTapTin[Ch�m???ngTh?ng[(-0.95, -0.18), (-0.89, -0.09), (-0.8, -0.03), (-0.71, 0.03), (-0.62, 0.12), (-0.56, 0.21), (-0.47, 0.12), (-0.41, 0.03), (-0.32, 0), (-0.26, -0.09), (-0.17, -0.18), (-0.08, -0.24), (?, ?), (0.16, 6.27), (0.19, 6.15), (0.25, 6.06), (0.31, 5.97), (0.37, 5.88), (0.43, 5.79), (0.52, 5.73), (0.55, 5.64), (0.64, 5.61), (0.7, 5.7), (0.79, 5.76), (0.82, 5.85), (0.91, 5.94), true], 1, (-4.52, -14.85), (27.4, 8.22), (-13, 756), (1051, -13)]
N�t b�t n�tB�t2: DinhKemTapTin[Ch�m???ngTh?ng[(0.88, 3.09), (1, 3.09), (1.12, 3.09), (1.24, 3.09), (1.36, 3.09), (1.42, 2.97), (1.48, 2.88), (1.51, 2.76), (1.54, 2.67), (1.57, 2.58), (1.6, 2.49), (?, ?), (0.82, 3.12), (0.76, 3.03), (0.73, 2.94), (0.67, 2.85), (0.61, 2.76), (0.52, 2.67), (0.46, 2.58), (0.46, 2.43), (0.49, 2.34), (0.52, 2.22), (0.52, 2.1), (0.55, 2.01), (0.55, 1.89), true], 1, (-4.52, -14.85), (27.4, 8.22), (0, 769), (1064, 0)]
?o?n th?ng f: ?o?n th?ng [B, C]
?o?n th?ng g: ?o?n th?ng [C, D]
?o?n th?ng h: ?o?n th?ng [B, D]
?o?n th?ng j: ?o?n th?ng [A, B]
?o?n th?ng k: ?o?n th?ng [A, C]
?o?n th?ng l: ?o?n th?ng [A, D]
?o?n th?ng i: ?o?n th?ng [A, M]
?o?n th?ng m: ?o?n th?ng [C, M]
B = (-3.19, 2.54)
B = (-3.19, 2.54)
B = (-3.19, 2.54)
C = (-1.01, -0.22)
C = (-1.01, -0.22)
C = (-1.01, -0.22)
D = (3.63, 2.4)
D = (3.63, 2.4)
D = (3.63, 2.4)
A = (-0.01, 7.04)
A = (-0.01, 7.04)
A = (-0.01, 7.04)
?i?m M: ?i?m tr�n h
?i?m M: ?i?m tr�n h
?i?m M: ?i?m tr�n h
Kẻ \(AM\perp BD\Rightarrow AM\perp\left(BCD\right)\) .
Xét tam giác vuông ABD, có \(AD=\sqrt{4a^2-a^2}=a\sqrt{3}\)
Áp dụng hệ thức lượng ta có \(AM=\frac{a.a\sqrt{3}}{2a}=\frac{\sqrt{3}}{2}a.\)
Xét tam giác vuông BCD, có \(CD=\sqrt{4a^2-a^2}=a\sqrt{3}\)
Vậy \(S_{BCD}=\frac{1}{2}.a.a\sqrt{3}=\frac{\sqrt{3}}{2}a^2\)
\(\Rightarrow V_{ABCD}=\frac{1}{3}.\frac{\sqrt{3}}{2}a^2.\frac{\sqrt{3}}{2}a=\frac{a^3}{4}\)