Найдем сторону ромба АВ=√(АО²+ВО²)=√(225+400)=25, т.к. О- точка пересечения диагоналей. Делит их пополам. Площадь треуг. АОВ равна АВ*ОТ/2, где ОТ - высота треугольника, проведенная к АВ, с другой стороны, т.к. диагонали перпендикулярны, площадь этого же треуг. равна ВО*АО/2⇒ОТ=20*15/25=12, а из треуг. МОТ найдем МО=
√(МТ²-ОТ²)=√(400-144)=√256=16
Здесь расстояние от точки М до АВ - по теореме о трех перпендикулярах, раз проекция МТ на АВО это высота ОТ перпендикулярна АВ, то и МТ ей перпендикулярна.
2. Проведем из точки В высоты на стороны АD И DС соответственно ВО и ВК. Тогда по теореме о трех перпендикулярах МО⊥АD, МК⊥DС, МО=10,ОВ=√(МО²-МВ²)=√(10²-8²)=6, Площадь параллелограмма равна АD*ВО=20*6=120, с др. стороны, площадь равна DС*ВК⇒ВК=2*60/8=15
А расстояние от DС до точки М это МК=√(МВ²+ВК²)=√(64+225)=17
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