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b=16
c=5
(b-a)/2=(16-10)/2=6/2=3 --->czesc dluzszej podstawy trapezu
z pitagorasa
3²+h²=5²
9+h²=25
h²=25-9
h=√16=4 --->wysokosc trapezu
P=1/2·(a+b)·h=1/2·(10+16)·4=1/2·26·4=52 [j²]