В ромбе ABCD AB=10см, меньшая диагональ AC=12см.Найдите площадь ромба. Мне нужно подробное решение!

В ромбе ABCD AB=10см, меньшая диагональ AC=12см.Найдите площадь ромба. Мне нужно подробное решение!
- Предмет:
  Геометрия
- Автор:
  neal
- 6 лет назад

Ответы 1

Пользуясь тем, что в ромбе диагонали взаимно перпендикулярны,а также делятся в точке пересечения пополам, имеем точку пересечения О; ВО=ВД/2=12/2=6; В результате имеем прямоугольный треугольник АВО с прямым углом О; Находим по теореме Пифагора половину диагонали АС, то есть АО=АВ^2-ВО^2=АО^2=100-36=64; Значит АО=8; Площадь S=ВД*АО=12*8=96; Ответ S=96
- Автор:
  zoe25
- 6 лет назад
- 0
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В ромбе ABCD AB=10см, меньшая диагональ AC=12см.Найдите площадь ромба. Мне нужно подробное решение!

Ответы 1

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