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52=30+x
x=22 22-podstawa
po opuszczeniu wysokosci liczymy z pitagorasa:
10²=6²+h²
100=36+h²
h²=64
h=8
pole=(a+b)*1/2*h
P=(22+10)*1/2*8=128
Obw=3a+b
52=3*10+b
52=30+b
b=22
h² =10² -6²
h² =100-36
h² =64
h=8
P=1/2[(a+b)*h]
P=1/2[(10+22)*8]
P=1/2[32*8]
P=1/2[256]
P=128 jednostek²
Odp: Pole tego trapezu wynosi 128 j²