Question

(Q.1). Show that the diagonals of a regular pentagons are equal.

(Q.2). Prove that each interior angle of a regular pentagon is three times of each exterior angle of a regular decagon.

Answer 1

ans.1  if we draw two diagonals, they make with their parallel side a parallelogram. But since the sides of the regular pentagon are equal, the parallelogram is a rhombus. (Opposite sides are equal in any parallelogram; if two adjacent sides are equal, then all four sides are equal.

Answer 2

Answer:Given:  Consider a regular pentagon ABCDE

To prove: All the diagonals of the pentagon are equal

i.e. AC = AD = BD = CE = BE

 

Proof:

ABCDE is a regular pentagon

Þ AB = BC = CD = DE = AE and ∠A = ∠B = ∠C = ∠D = ∠E = 108°

 

Consider ∆ABC and ∆ADE

AB = AE                                     (Given)

BC = DE                                     (Given)

∠ABC = ∠AED = 108°                 (Given)

∴ ∆ABC ≅ ∆ADE                         (SAS congruency rule)

∴ AC = AD              (corresponding sides of congruent triangles)    -------- (1)

 

Similarly ∆BCD ≅ ∆CDE

⇒ BD = CE               -------- (2)

 

∆ ABC ≅ ∆BCD              

⇒ AC = BD                  -------- (3)

 

∆ABC ≅ ∆ABE

⇒ BE = AC   -------- (4)

 

From equations (1), (2), (3) and (4), we get

AC = BD = AD = CE = BE

 

Therefore, all the diagonals are equal

Step-by-step explanation: hence proved