Răspuns :
[tex]\it Cu\ T.P.\ \Rightarrow AB=15cm\\ \\ Ducem\ AD\ \ \hat\imath n\breve al\c{\it t}ime \Rightarrow AD=\dfrac{AB\cdot AC}{BC}=\dfrac{15\cdot20}{25}=\dfrac{300}{25}=12\ cm\\ \\ Cu\ T.\ P.\ \hat\imath n\ \Delta ADC \Rightarrow CD=16\ cm\\ \\ MN-linie\ mijlocie\ pentru \ \Delta ADC \Rightarrow MA=AC:2=20:2=10cm;\\ \\ MN=AD:2=12:2=6\ cm;\ \ NC=CD:2=16:2=8\ cm;\\ \\ BN=BC-NC=25-8=17\ cm[/tex]
[tex]\it \mathcal{P}_{ABMN}=AB+BN+NM+MA=15+17+6+10= 48\ cm\\ \\ R\breve aspuns\ corect:\ b)\ 48\ cm[/tex]