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BenJost
@BenJost
March 2022
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Помогите решить задачу(фото)
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lkulikova51
S_бок=S_осн/cosφ,где φ угол наклона граней пирамиды к плоскости основания.
S_осн=6^2 sin〖120〗=36∙0,5√3=18√3
S_бок=18√3/
cos30=36.
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Answers & Comments
S_осн=6^2 sin〖120〗=36∙0,5√3=18√3
S_бок=18√3/cos30=36.