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Tasher3tapora
Tasher3tapora
24.08.2022 00:31 •  Геометрия

В треугольниках ABC и DEF равны пары сторон AB и DE, BC и EF, а также углы BAC и EDF. При каком дополнительном условии можно утверждать, что треугольники ABC и DEF равны?

Выберите все правильные варианты ответа.

∠BAC — острый

∠BAC — прямой

∠BAC — тупой

∠BCA — острый

∠BCA — прямой

∠BCA — тупой

AB>BC

AB<BC

👇
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Ответ:
ника2346олл
ника2346олл
24.08.2022

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4,8(46 оценок)
Ответ:
pvpgame2345
pvpgame2345
24.08.2022

Построено сечение с учётом расположения линий в каждой плоскости.

Длины линий сечения.

AE = √(8² + 4²) = √(64 + 16) = √80 = 4√5.

Длину В1К находим из пропорции (В1К/8 = (8/(8+4)),

отсюда В1К = (8*8)/12 = 16/3.

Тогда ЕК = √(4² + (16/3)²) = √(400/9) = 20/3.

KP = √((8 - (16/3))² + 4²) = √(208/9) = (4/3)√13.

Длину СТ находим из пропорции.

Так как СМ = КС1 = 8 / (16/3) = 8/3, то СМ/СТ = (ВМ/АВ.

Подставим данные. (8/3)/СТ = (8 + (8/3)/8. Получаем СТ = 2.

РТ = √(4² + 2²) = √20 = 2√5.

ДТ = 8 - 2 = 6.

АТ = √(8² + 6²) = 10.

ответ: Р = 4√5 + (20/3) + ((4/3)√13) + (2√5) + 10 =

               = 6√5 + (20/3) + ((4/3)√13) + 10.


постройте сечение куба abcda1b1c1d1 плоскостью проходящей через середины ребер а1в1 и сс1 и вершину
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