Oblicz pole trapezu ma podstawy 40 i 19, a 17 i 10. Z obliczeniami proszę .
a=40
b=19
c=17
d=10
a+b=40+19=59
a-b=40-19=21
[a+b]/[a-b]/59/21
a-b+c+d=40-19+17+10=48
a-b-c+d=40-19-17+10=14
a-b+c-d=40-19+17-10=28
-a+b+c+d=-40+19+17+10=6
pole=¼× 59/21×√[48×14×28×6]=59/84√112896=59/84×336=236j.
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a=40
b=19
c=17
d=10
a+b=40+19=59
a-b=40-19=21
[a+b]/[a-b]/59/21
a-b+c+d=40-19+17+10=48
a-b-c+d=40-19-17+10=14
a-b+c-d=40-19+17-10=28
-a+b+c+d=-40+19+17+10=6
pole=¼× 59/21×√[48×14×28×6]=59/84√112896=59/84×336=236j.