DC=MN=18, AM=NB=(24-18)/2=3
tgA=DM/AM, √3=DM/AM, DM=AM√3=3√3
A=(AB+CD)DM/2=(18+24)3√3/2=63√3 cm²
b. ΔDMB DB²=DM²+MB²=27+441=468, DB=6√3
ΔACN AC²=AN²+CN²=441+27, AC=6√3
c. Distanta de la punctul C la dreapta AD este PC⊥AD
ΔCPD triunghi dreptunghic
∡PDC≡∡DAC (alterne interne)
sin 60=PC/DC, PC=18√3/2=9√3