Question

In the given figure AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately:

(a) 15º 

(b) 20º 

(c) 30º 

(d) 25º

Answer 1

(d) 25º.

Let ∠EAD = a. Then

∴ For ΔABC, ext CBD = 2α

In ΔCBD, CB = CD ⇒ ∠CDB = 2α 

For ΔAFG, ext ∠FGC = 2α 

∵ In ΔGFE, GF = EF ⇒ ∠FEG = ∠FGE = 2α 

For ΔEAF, ext. ∠EFD = 3α

∵ EF = ED ∴ ∠EDF = ∠EFD = 3α ⇒ ∠EDP = α

For ΔCAD, ext. ∠DCE = 3α 

In ΔECD, DC = ED ⇒ ∠DEC = ∠DCE = 3α 

⇒ ∠FED = ∠DEC – ∠FEC = 3α – 2α = α.

∴ In ΔEFD, α + 3α + 3α = 180º 

⇒ 7α = 180º ⇒ α = \(\frac{180^0}{7}\) = 26º  or approximately 25º.