корень из 6. С решением, будьте любезны...

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Высота правильной треугольной пирамиды равна 2, а двугранный угол при основании 45°. Найдите площадь боковой поверхности пирамиды S. В ответ запишите S/корень из 6. С решением, будьте любезны

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Nangw

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===========================РЕШЕНИЕ=============================

1) Рассмотрим ΔКОМ: ∠О=90°, ∠К=45° ⇒ ∠М=180°-90°-45°=45°, значит треугольник равнобедренный, тогда МО=ОК=2

по т.Пифагора найдём КМ:

$\displaystyle KM=\sqrt{2^2+2^2} =\sqrt{8} =2\sqrt{2}$

КМ=2√2=h - апофема

2) $\displaystyle KC=m=\frac{a\sqrt{3} }{2}$ - медиана

$\displaystyle KO=\frac{1}{3} m$ , так как медианы в точке пересечения делятся в отношении 1:2, значит КО=1 часть, КС=3 части

Подставим значения:

$\displaystyle 2=\frac{1}{3} *\frac{a\sqrt{3} }{2}$ (*6)

$\displaystyle 12=a\sqrt{3}$

$\displaystyle a=\frac{12*\sqrt{3} }{\sqrt{3}*\sqrt{3} }$

$\displaystyle a=\frac{12\sqrt{3} }{3}$

$\displaystyle a=4\sqrt{3}$ - сторона основания ΔАВС

3) Sбок= $\displaystyle \frac{Pocn*h}{2}$

Sбок= $\displaystyle \frac{3*4\sqrt{3}*2\sqrt{2} }{2}$

Sбок= $\displaystyle \frac{24\sqrt{6} }{2}$

Sбок= $\displaystyle 12\sqrt{6} (cm^2)$

Ответ: $\displaystyle \frac{S}{\sqrt{6} } =\frac{12\sqrt{6} }{\sqrt{6} } =12(cm^2)$

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