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stas020199
@stas020199
July 2022
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Дано.МАBC-Тетраэдр.М(2;5;7);A(1;-3;2);B(2;3;7);C(3;6;0).Найти KO,где K-середина AM;O-середина BC
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dnepr1
Verified answer
Находим координаты заданных точек:
К = ((1+2)/2=1,5; (-3+5)/2=1; (2+7)/2=4,5) = (1,5;1;4,5).
О = ((2+3)/2=2,5; (3+6)/2=4,5; (7+0)/2=3,5) = (2,5;4,5;3,5).
Длина КО равна √((2,5-1,5)²+(4,5-1)²+(3,5-4,5)²) = √(1+12,25+1) =
= √14,25 = √57/2.
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Verified answer
Находим координаты заданных точек:К = ((1+2)/2=1,5; (-3+5)/2=1; (2+7)/2=4,5) = (1,5;1;4,5).
О = ((2+3)/2=2,5; (3+6)/2=4,5; (7+0)/2=3,5) = (2,5;4,5;3,5).
Длина КО равна √((2,5-1,5)²+(4,5-1)²+(3,5-4,5)²) = √(1+12,25+1) =
= √14,25 = √57/2.