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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.
По условию задачи а = 7 см, b = 10 см, где а, b – стороны параллелограмма, а угол между ними равен 180°
Пусть α = 120°, тогда β = 180° - 120° = 60°, так как сумма смежных (соседних) углов параллелограмма равна 180°
По теореме косинусов:
с² = а² + b² - 2ab * cos α, где α - угол между сторонами
Найдем большую диагональ:
c² = 7² + 10² - 2 * 7 * 10 * cos 120° = 49 + 100 - 2 * 70 (-1/2) =
= 149 + 70 = √219; c = √219
Найдем меньшую диагональ:
d² = 7² + 10² - 2 * 7 * 10 * cos 60° = 49 + 100 - 2 * 70 * 1/2 = 149 - 70 = 79;
d = √79
ответ: меньшая диагональ параллелограмма равна √79, большая диагональ равна √219