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chumaeva791
chumaeva791
26.06.2021 17:58 •  Геометрия

Чому дорівнює площа трикутника авс якщо ас = 9 см, ав 2 корінь 2 см, кут а =135° ( іть будь ласка, дуже треба )

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Ответ:
JonHenryEdem1
JonHenryEdem1
26.06.2021

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Ответ:
римма59
римма59
26.06.2021
прав.шест-к составлен из 6 равносторон. тр-ков)

   S осн = 6*а^2*корень из 3/4.

   Видим, что надо знать сторону основания :а-? 

3)S бок = 6*0,5* а*h, где h- апофема .Таким образом, от цели нас отделяет только нахождение стороны основания а.  Из тр-ка МОS-прям.: SO = 12, SM =15, тогда ОМ=9 ( либо по теореме Пифагора, либо этот тр-к подобен "египетскому" с коэфф 3).

4)Из тр-ка  АОМ-прям: ОМ =9 ,угол ОАМ =60 град., тогда АМ =9/корень из 3 = 3*корень из 3, тогда а = 2*АМ = 6*корень из 3.

5)Sполн = 6*  (6*корень из 3)^2 *корень из 3/4+ 3* 6*корень из 3*15 =  

  = 6*36*3* *корень из 3/4  + 18*15* корень из 3=  6*9*3* *корень из 3  + 18*15* корень из 3  =   18*24* корень из 3 = 432* корень из 3 (кв.ед).

 


Вправильной шестиугольной пирамиде апофема равна 15,а высота 12.вычислить площадь полной поверхности
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