1.consideram AB II DC
AD=25
AC=26
DC=17
ducem AE_I_DC
AE²=AD²-DE²=25²-DE²=625-DE²
AE²=AC²-EC²=26²-(DC-DE)²=676-(17-DE)²=676-289+34DE-DE²=387+34DE-DE²
625-DE²=387+34DE-DE²
625=387+34DE
34DE=238
DE=238/34=7
AE=√(625-7²)=√(625-49)=√576=24
AB=17-2x7=3
linia mijlocie=(AB+DC)/2=(3+17)/2=10
A=Bxh/2=17x24/2=204