1. S = ½×(4+8)×5 = ½×6×5 = 3×5 = 15 см².
2. S=150, h=S:(½×(a+b)) = 150:(½×(9+11)) = 150:(½×20) = 150:10 = 15 см.
3. Пусть высота будет BH(нужно отметить Н на рисунке). Проведём высоту из точки С, будет она СЕ. Т.к. трапеция равнобедренная, то АН=DE. AH=BH=4 см, ведь угол А=45°, угол Н=90°, соответственно угол В=45° и треугольникк АВН равнобедренный. Из этого, AD=4+5+4 = 13 см.
Найдём площадь: S=½×(5+13)×4 = ½×18×4 = 9×4 = 36 см².
4. Пусть одна часть будет х, тогда BC=3x, AD=4x.
S=½×(3x+4x)×5 = ½×7x×5 = 3,5x×5 = 17,5x -> 17,5x = 35.
x=2 см.
AD=4x = 4×2 = 8 см.
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2(x + x + 8) = 36
2(2x + 8) = 36
2x + 8 = 36 : 2
2x + 8 = 18
2x = 18 - 8
2x = 10
x = 10 : 2
x = 5 (см) - I сторона;
x + 8 = 5 + 8 = 13 (см) - II сторона.
ответ: 5 см, 13 см