Предмет: Геометрия, автор: Nastya163

в треугольнике со сторанами 20, 34, 42 вписан прямоугольник с прямыми углами, периметр которого 40, и одна из его сторон лежит на большей стороне треугольника. Найти стороны треугольника.

Ответы

Автор ответа: vajny

Видимо надо найти стороны ПРЯМОУГОЛЬНИКА! Так как стороны ТРЕУГОЛЬНИКА в условии даны!

Рисунок смотри во вложении.

Пусть х и у - стороны пр-ка. Проведем дополнительно высоту ВЕ тр-ка АВС.

Найдем ее. Площадь по формуле Герона:

S = корень(48*28*14*6) = 336 (полупериметр р = 48)

С другой стороны:

S = (1/2)*42*BE = 336

Отсюда ВЕ = 16

Из подобия тр-ов ВКМ и АВС:

х/42 = ВК/20

Отсюда ВК = 10х/21, АК = 20 -10х/21 = (420-10х)/21

Из подобия тр-ов АКР и АВЕ:

у/16 = АК/20

Или: у/16 = (42-х)/42

8х + 21у = 336

Другое уравнение системы получим из условия, что периметр пр-ка равен 40:

х + у = 20. Домножим это уравнение на (-8) и сложим с предыдущим.

13у = 176

у = 176/13, тогда х = 20 - 176/13 = 84/13

Ответ: 176/13; 84/13.

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