Предмет: Геометрия, автор: SM2Margosha

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Автор ответа: sfaleev197gmailcom
1
пусть длина короткой стороны 4х
Тогда длина длинной 7х
P = 2*(4x+7x) = 44
2*(4x+7x) = 44
4x+7x = 22
11х = 22
x=22/11
х = 2 см=>
=> короткая сторона 2*4 = 8 см =>
длинная сторона 2*7 = 14 см
DE = FC = 1/2EF
Найдем DE и EF
DE = FC = 14/4 =3.5 см
EF = 14/2 = 7 см
EO = FO = EF/2=7/2=3.5 см
DE = EO.
KOF=FCB =>
Прямоугольник ABCD и треугольник АКВ равновелики друг другу.
S (прямоугольника)= 8*14 = 112 см²
S (треугольника)=112 см²

Andr1806: Можно и так: EF= DE+FC (дано). Тогда EF= DC/2 =7. => EF - средняя линия треугольника АКВ. Значит высота этого треугольника h= 2*AD = 16. Sakb = (1/2)*AB*h = 14*16/2 =112 ед²
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