A)
In triunghiul ACB, mC=90, m B=60⇒m CAB=30⇒conf th 30-60-90 CB=AB/2
CB=12
In ΔACB, aplicam Pitagora⇒AB²=AC²+CB²
24²=AC²+12²
AC²=576-144=432
AC=12√3
B)
Construiesti inaltimea trapezului pe care o notezi CM
In ΔAMC, mM=90,mA=30, AC=12√3⇒th 30-60-90 CM=6√3
In ΔCMB, mM=90⇒th lui Pitagora MB²=CB²-CM²
MB²=144-108=36
MB=6
MB=AN=6
MN//DC⇒MN≡DC
MN=24-(6+6)=24-12=12
DC=12
P= 12+12+12+24=60
