Prvo rješenje.
Zadanom osnom simetrijom dobije se peterokut A A ′ C ′ D C AA'C'DC A A ′ C ′ D C
Trokut B ′ B D B'BD B ′ B D 2 3 a \frac23a 3 2 a
P ( B ′ B D ) = 1 4 ( 2 3 a ) 2 3 = 1 4 ⋅ 4 a 2 3 9 = a 2 3 9 . P(B'BD)=\frac14\left(\frac23a\right)^2\sqrt3 =\frac14\cdot\frac{4a^2\sqrt3}{9} =\frac{a^2\sqrt3}{9}. P ( B ′ B D ) = 4 1 ( 3 2 a ) 2 3 = 4 1 ⋅ 9 4 a 2 3 = 9 a 2 3 . P ( A A ′ C ′ D C ) = P ( A B C ) + P ( A ′ B ′ C ′ ) − P ( B ′ B D ) = 2 ⋅ a 2 3 4 − a 2 3 9 = 7 3 18 a 2 . \begin{aligned} P(AA'C'DC) &=P(ABC)+P(A'B'C')-P(B'BD)\\ &=2\cdot\frac{a^2\sqrt3}{4}-\frac{a^2\sqrt3}{9} =\frac{7\sqrt3}{18}a^2. \end{aligned} P ( A A ′ C ′ D C ) = P ( A B C ) + P ( A ′ B ′ C ′ ) − P ( B ′ B D ) = 2 ⋅ 4 a 2 3 − 9 a 2 3 = 18 7 3 a 2 . Opseg nastalog peterokuta je
3 ∣ A B ∣ + 3 ∣ B A ′ ∣ = 3 a + 3 ⋅ a 3 = 4 a , 3|AB|+3|BA'|=3a+3\cdot\frac a3=4a, 3∣ A B ∣ + 3∣ B A ′ ∣ = 3 a + 3 ⋅ 3 a = 4 a , pa je a = 3 a=3 a = 3
Tražena je površina
P ( A A ′ C ′ D C ) = 7 ⋅ 9 3 18 = 7 3 2 c m 2 . P(AA'C'DC)=\frac{7\cdot9\sqrt3}{18}=\frac{7\sqrt3}{2}\ \mathrm{cm}^2. P ( A A ′ C ′ D C ) = 18 7 ⋅ 9 3 = 2 7 3 cm 2 . Drugo rješenje.
Zadanom osnom simetrijom dobije se peterokut A A ′ C ′ D C AA'C'DC A A ′ C ′ D C
Trokut C D C ′ CDC' C D C ′ a 3 \frac a3 3 a
Površinu peterokuta A A ′ C ′ D C AA'C'DC A A ′ C ′ D C A A ′ C ′ C AA'C'C A A ′ C ′ C C D C ′ CDC' C D C ′
Vrijedi
P ( A A ′ C ′ C ) = ∣ A A ′ ∣ + ∣ C ′ C ∣ 2 ⋅ v = 4 3 a + 1 3 a 2 ⋅ a 3 2 = 5 a 2 3 12 , P(AA'C'C)=\frac{|AA'|+|C'C|}{2}\cdot v =\frac{\frac43a+\frac13a}{2}\cdot\frac{a\sqrt3}{2} =\frac{5a^2\sqrt3}{12}, P ( A A ′ C ′ C ) = 2 ∣ A A ′ ∣ + ∣ C ′ C ∣ ⋅ v = 2 3 4 a + 3 1 a ⋅ 2 a 3 = 12 5 a 2 3 , P ( C D C ′ ) = 1 4 ( a 3 ) 2 3 = a 2 3 36 . P(CDC')=\frac14\left(\frac a3\right)^2\sqrt3 =\frac{a^2\sqrt3}{36}. P ( C D C ′ ) = 4 1 ( 3 a ) 2 3 = 36 a 2 3 . Tada je površina peterokuta
P ( A A ′ C ′ D C ) = 5 a 2 3 12 − a 2 3 36 = 14 a 2 3 36 = 7 a 2 3 18 . P(AA'C'DC)=\frac{5a^2\sqrt3}{12}-\frac{a^2\sqrt3}{36} =\frac{14a^2\sqrt3}{36}=\frac{7a^2\sqrt3}{18}. P ( A A ′ C ′ D C ) = 12 5 a 2 3 − 36 a 2 3 = 36 14 a 2 3 = 18 7 a 2 3 . Opseg nastalog peterokuta je
3 ∣ A B ∣ + 3 ∣ B A ′ ∣ = 3 a + a = 4 a , 3|AB|+3|BA'|=3a+a=4a, 3∣ A B ∣ + 3∣ B A ′ ∣ = 3 a + a = 4 a , pa je a = 3 a=3 a = 3
Tražena je površina jednaka
P ( A A ′ C ′ D C ) = 7 ⋅ 9 3 18 = 7 3 2 c m 2 . P(AA'C'DC)=\frac{7\cdot9\sqrt3}{18}=\frac{7\sqrt3}{2}\ \mathrm{cm}^2. P ( A A ′ C ′ D C ) = 18 7 ⋅ 9 3 = 2 7 3 cm 2 . 
