ABCD-прямоугольник , MB-перпендикулярно ABC. MA=10. MC=15, MD=17 MB
За т. Пифагора:
BD2 = AB2 +BC2
MD2 – MB2 = AM2 –HB2 + MC2 – BH2
BH2 = AM2 + HC2 – MD2
BH2 = 102 + 152 - 172 =100+225-289=36
BH = 6
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За т. Пифагора:
BD2 = AB2 +BC2
MD2 – MB2 = AM2 –HB2 + MC2 – BH2
BH2 = AM2 + HC2 – MD2
BH2 = 102 + 152 - 172 =100+225-289=36
BH = 6