Предмет: Геометрия, автор: shooltime5873

1. Дано ABCD-прямоугольник, AK перпендикулярно (ABC)
KD=6 KC=9
Определить вид треугольника KDC
Найти DC
2. Дано ABCDA1B1C1D1 куб со стороной 4 см
Найти и обьяснить расположение в пространстве:

Найти углы между:

Приложения:

Ответы

Автор ответа: viktoriamaksimovskih
2

Ответ:

1.ΔKDC-прямоугольный; DC=3√5

Объяснение:

1.

ΔKDC-прямоугольный т.к. по теореме о 3-х перпендикулярах(AD⊥DС и AK⊥AD => DС⊥KD).

DC²=KC²-KD²=81-36=45

DC=√45=√9·5=3√5

2.

1) CD⊥BC:

2) AB⊥B1C1

3) AD1 и AB скрещивающиеся

4) ABCD⊥DD1; A1B1C1D1⊥DD1

5) AA1D1D⊥ABCD; AA1D1D⊥A1B1C1D1

Вроде должно быть так.

shooltime5873: А можешь ещё пожалуйста ничьи углы между
1. Прямой A1C и плоскостью ABC
2. Плоскостями AA1 и ACD
3. Прямыми A1C и BB1
shooltime5873: Найти*
shooltime5873: Пожалуйста
viktoriamaksimovskih: 1. они параллельны=> точки пересечения нет, значит угла нет(но я не уверена на счет этого, но если действовать логике это так). 2. 90 градусов т.к. перпендикулярны. 3. опять же не уверена, но общей точки нет=> угла нет, т.к. угол между прямой и плоскостью, это угол наклонной и ее проекции.
viktoriamaksimovskih: Но могу ошибаться.
shooltime5873: Спасибо большое!
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5th class


I. Read the text and say if the sentence is true or false.


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