1.точка с - середина отрезка ав. найдите координаты точки в, если с(-2; 3) и а(-6; -5).2.а) ав - диаметр окружности с центром о. найдите координаты центра окружности, если а(8; -3) и в(-2; -5) b) запишите уравнение окружности, используя условия пункта а.3. выполните построение, выясните взаимное расположение двух окружностей, заданных уравнений (х+3)²+(у-4)²=9 и (х-2)²+(у-4)²=4 4. точки а(-3; 5) в(3; 5) с(6; -1) d(-3; -1) - вершины прямоугольной трапеции с основаниями ав и сd. найдите длину средней линии и площадь трапеции. 45
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Все рёбра треугольной пирамиды равны. Найти угол наклона:
а) Бокового ребра к плоскости основы.
б) боковой грани к площине основы/
Объяснение:
АВСМ -пирамида, пусть ребро равно х.
a)Угол наклона бокового ребра к плоскости основания это ∠МАО.
Т.к АВ=ВС=АС, то высота проецируется в центр основания О , точку пересечения медиан.Тогда АО=2/3*АН, где АН медиана, ВН=х/2 .
Из ΔАВН-прямоугольного, АН=√(х²-х²/4)=(х√3)/2. Тогда АО=( х√3)/3.
ΔАОМ-прямоугольный, cos∠МАО=АО/АМ , cos∠МАО=( х√3)/3:х=√3/3,
∠МАО=arccos(√3/3) .
ОМ=√(х²-( х√3)/3)² )=(х√6)/3
б)В равностороннем ΔАВС , медиана АН является высотой . Тогда МН⊥ВС по т. о трех перпендикулярах и ∠АНМ-линейный угол между боковой гранью и плоскостью основания.
ОН=1/3*АН , ОН=(х√3)/6.
ΔОНМ-прямоугольный ,tg∠AHM=MO/OH , tg∠AHM=2√2 , ∠AHM=arctg(2√2).