[tex]ex~1.\\
La~trapezul ~dreptunghic~latura ~perpendiculara~este~inaltimea!\\\\
A_{trapez}= \frac{(B+b)*h}{2} = \frac{(AB+CD)*AD}{2}= \\ \\ = \frac{(24+16)*12}{2}= \frac{40*12}{2}=20*12=240 \,cm^{2} [/tex]
[tex]ex~2.\\Se ~aplica~ teorema~ unghiului~ de ~15~ grade~in~~~ triunghiul\\dreptunghic ~care~se ~refera~la ~faptul~ca ~inaltimea ~opusa~ unghiului\\de~15~grade~este~egala~cu ~sfertul ~ipotenuzei.\\
CB= 2*AM\\
CB=2*24\\
CB=48cm\\\\
Inaltimea~~~AD= \frac{48}{2} \\
AD=12cm\\\\
Deci:\\\\
A_{triunghiului~ABC}= \frac{b*h}{2} \\\\
A_{triunghiului~ABC}= \frac{48*12}{2} \\\\
A_{triunghiului~ABC}=288 cm^2[/tex]
