Question

ABCD-trapez isoscel AB||CD AB=24 cm CD=18cm masura unghiului A=60 de grade
distanta de la C la dreapta AD=?

Answer 1

ducem DM_I_AB    si CN_I_AB
AM=BN=(24-18)/2=3
m(∡A)=60°⇒ m(∡ADM)=30°
teorema unghiului de 30° ⇒ AD=3×2=6
DM=CN=√(6²-3²)=√27=3√3
AN=24-3=21
AC=√[21²+(3√3)²]=√(441+27)=√468=6√13
ducem CP_I_AD
m(∡PDC)=m(∡DAB)=60° ⇒ m(∡DCP)=30°
teorema unghiului de 30° ⇒ DP=CD/2=18/2=9
AP=6+9=15
Teorema lui Pitagora
CP=√(AC²-AP²)=√[(6√13)²-15²]=√(468-225)=√243=9√3