Question

urgent !!! va roooggggg
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Answer 1

∡(B'C,(ABC))=∡B'CB

CB⊥AB

CB,AB⊂(ABC)

∡B'CB=30°⇒teorema ∡30°⇒ 2BB'=B'C

BB'=9⇒B'C=18

Aplicam Pitagora

B'C²=BB'²+BC²

324=81+BC²

BC²=243

BC=9√3cm

d((ABB'),(DCC'))=BC=9√3cm

BC⊥(ABB')

BC⊥(DCC')

b.

A'B║D'C⇒ ∡(A'B,(ABC))=∡(D'C,(ABC))=∡D'CB

∡D'CB=60°

d((A'AD),(B'BC))=AB=CD

AB⊥(A'AD)

AB⊥(B'BC)

In ΔD'DC

∡D'CD=60°⇒ ∡CD'D=30°⇒2DC=D'C

Aplicam Pitagora

D'C²=D'D²+DC²

4DC²=81+DC²

3DC²=81

DC²=27

DC=3√3cm