Iwill award given: ∆abc≅def find: bc, ba, m∠d

BC = 5

BA11

angle D = 27

Step-by-step explanation:

If the triangles are similar

ABC ≅ DEF

That means AB ≅ DE

BC ≅ EF

and CA ≅ FD

We want to find AB  so it is equal to DE  

BC = EF = 5

BA = AB = DE =11

We also know  the angles are equal

A = D

B = E

C = F

Angle A = angle D = 27

Using CPCTC, we know that congruent parts of the congruent triangles are congruent.

Given: ∆ABC ≅ ∆DEF

BC ≅ FE  ⇒ CPCTC. IF FE = 5 (Given), & FE ≅ BC (CPCTC) ∴ BC = 5 (Transitive Property of Equality.)

BA ≅ DE ⇒ CPCTC. IF DE = 11 (Given) & DE ≅ BA (CPCTC) ∴ BA = 11 (Transitive Property of Equality.)

m∠D ≅ m∠A ⇒ CPCTC. Congruent Parts of Congruent Triangles are Congruent. If ∆ABC ≅ ∆DEF (given), then through CPCTC, m∠D ≅ m∠A.

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12.5 feet

step-by-step explanation:

hi! i think the answer is 1/2 : )