The length of the bottom side and the side edge of the regular triangular prism abc-a1b1c1 are 2, D and E are the midpoint of BB1 and CC1 respectively. (I) find out the total area of the triangular prism abc-a1b1c1; (II) find out the be ‖ plane ADC1; (III) find out the plane ADC1 ⊥ plane acc1a1

The length of the bottom side and the side edge of the regular triangular prism abc-a1b1c1 are 2, D and E are the midpoint of BB1 and CC1 respectively. (I) find out the total area of the triangular prism abc-a1b1c1; (II) find out the be ‖ plane ADC1; (III) find out the plane ADC1 ⊥ plane acc1a1

(1) The solution is that the triangular prism abc-a1b1c1 is a regular triangular prism, and the length of the prism is 2, so the bottom surface is a regular triangle, and the sides are square, so the total area of the triangular prism abc-a1b1c1 is s = 2 × 34 × 22 + 3 × 22 = 12 + 23. (II) in the regular triangular prism abc-a1b1c1, because D and E are the midpoint of BB1 and CC1 respectively, we can know that BD = 12bb1 = 12cc1 = EC1, and BD ‖ EC1, so the quadrilateral bdc1e is a parallelogram So be ‖ DC1, DC1 ⊂ planar ADC1, be ⊄ planar ADC1. So be ‖ planar ADC1. (III) take AC midpoint h, connect oh and BH ∵ in △ acc1, oh is the median line ‖ oh ‖ & nbsp; CC & nbsp; 1 and oh = 12CC & nbsp; 1, combined with BD ∥ CC1 and BD = 12cc1, the quadrilateral bdoh is parallelogram ∥ BH ⊥ OD ∵ BH ⊥ plane acc1a1 ≁ OD ⊥ plane acc1a1, because od is in plane ADC1 ≁ plane ADC1 ⊥ plane acc1a1

In the triangular prism abc-a1b1c1, ∠ ABC = 90 & ordm;, the three side edges are perpendicular to the ground, m and N are the midpoint of BB1 and a1c1 respectively

Proof: because BB1 ⊥ a1b1c1,
So BB1 ⊥ b1c1,
Because ∠ a1b1c1 = ∠ ABC = 90 & # 186;,
So A1B1 ⊥ b1c1,
So A1B1 ⊥ surface bb1c,
Because ab ‖ A1B1,
So ab ⊥ surface bb1c,
So ab ⊥ CB1;
Take the midpoint e of AC1 and connect en, be,
Because m and N are the midpoint of BB1 and a1c1 respectively,
So en ‖ Aa1 ‖ BB1, en = Aa1 / 2 = BB1 / 2 = BM,
So bmne is a parallelogram,
So Mn ‖ be,
Then Mn ‖ plane ABC1