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It is known that the surface area of a tetrahedron ABCD is s, and the centers of its four faces are e, F, G and h. suppose the surface area of tetrahedron efgh is t, then TS is equal to () A. 19B. 49C. 14D. 13
As shown in the figure, the centers of the four faces of the regular tetrahedron ABCD are e, F, G, h, respectively, and the efgh is also a regular tetrahedron. Connect AE and extend the intersection with CD at point m, connect Ag and extend the intersection with BC at point n. ∵ e, G are the centers of the faces, ∵ aeam = Agan = 23. ∵ gemn = 23. Also ∵ Mn = 12bd, ∵ gebd = 13. ∵ the area ratio is the square of the similar ratio, ∵ the area ratio of the two tetrahedrons is; ts = 19 So the answer is: a
It is known that the surface area of a tetrahedron ABCD is s, and the centers of its four faces are e, F, G and h. suppose the surface area of tetrahedron efgh is t, then TS is equal to ()
A. 19B. 49C. 14D. 13
As shown in the figure, the centers of the four faces of the regular tetrahedron ABCD are e, F, G, h, respectively, and the efgh is also a regular tetrahedron. Connect AE and extend the intersection with CD at point m, connect Ag and extend the intersection with BC at point n. ∵ e, G are respectively the centers of the faces, ∵ aeam = Agan = 23. ∵ gemn = 23. Also ∵ Mn = 12bd, ∵ gebd = 13. ∵ the area ratio is the square of the similar ratio, ∵ the area ratio of the two tetrahedrons is; ts = 1 9. So the answer is: a
Translate the diamond ABCD to a1b1c1d1 along the AC direction, a1d1 intersects CD to e, A1B1 intersects BC to F, is the quadrilateral a1fce diamond? Why?
First of all, you have to prove that the quadrilateral a1fce is a parallelogram, then ∵ AC, a1c1 are diagonals, ∵ DAC = ∵ BAC ∵ DCA = ∵ BCA, ad = CD ∵ DAC = ∵ DCA, and △ ADC congruent ∵ a1d1c1 ∵ d1a1c1 = ∵ DCA ∵ a1e = CE ∵ quadrilateral a1fce is a diamond (a group of parallel quadrilateral with equal adjacent sides is a diamond). OK!
In diamond ABCD and diamond a1b1c1d1, the ratio of BD to a1c1 can be calculated by a = angle A1 = 60 degree AB ratio A1B1 = 1 ratio root 3
In diamond ABCD and diamond a1b1c1d1
Because ∠ a = A1 = 60 degree
So ∠ C = ∠ C1 = 60 °, ABC = ∠ a1b1c1, ∠ ADC = ∠ a1d1c1
So in diamond ABCD ∽ diamond a1b1c1d1
So △ abd ∽ a1b1d1, △ BCD ∽ b1c1d1
So BD / b1d1 = AB: A1B1 = 1: radical 3
Connect a1c1 to b1d1 at point o
Because a1c1 ⊥ b1d1
So in RT △ a1b1o
Od1: OC1 = 1: radical 3
So b1d1: a1c1 = 1: root 3
BD: a1c1 = (BD: b1d1) × (b1d1: a1c1) = (1: radical 3): (1: radical 3) = 1:3
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