Впараллелограме аbcd угол а равен 60°; вс=10, bd|ab .найти периметр параллелограмма.

Пусть о - точка пересечения диагоналей. угол aob = cod = 60 градусов. тогда угол boc = aod = 120 градусов как смежные. диагонали параллелограмма делятся в точкой пересечения пополам, т.е. ao = oc = 10, bo = ob = 7. рассмотрим треугольник abo. по теореме косинусов ab^2 = ao^2 + bo^2 - 2*ao*bo*cos(60). отсюда находим ab. аналогично из тр-ка boc находим bc (не забываем, что угол уже не 60, а 120 градусов). и, наконец, находим периметр: p = 2*ab + 2*bc.

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Ответ на эту уже  давался отрезок прямой,  проходящей через середины ребер as и bc, обозначим км.медиана основания ам  (она же и высота и  биссектриса основания) равна ав*cos 30° = 12√3 * (√3/2) = 18.точка к на середине ребра sa проецируется на медиану в точку е, находящуюся посредине отрезка ао, равного 2/3 ам.ао = (2/3)*18 = 12, ео = (1/2)*12 = 6.отсюда ем = 6+(1/3)*18 = 6 + 6 = 12.высота пирамиды so =  √(sa²-ao²) =  √(13²-12²) =  √(169-144) =  √25 = 5.отрезок ке равен половине высоты пирамиды: ке = 5/2 = 2,5.угол, образованный плоскостью основания и прямой, проходящей через середины ребер as и bc, - это угол кме =  α.ctg  α = em / ke = 12 / 2.5 = 4.8.α = arc ctg 4.8 =    0.205395 радиан  =  11.76829 градуса