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moicompik
@moicompik
July 2022
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MABC - правильная пирамида. AB = 6, MA = 2 sqrt{7}
Найдите V.
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oganesbagoyan
Verified answer
A=AB=BC =CA =6 ; MA =MB=MC=2√7. MO_высота пирамиды,
H=MO, где O центр основания.
V =(1/3)*S(ABC)*H =(1/3)*a²√3/4*H =(1/3)*6²√3/4*H=3√3*H .
AO =(AB*√3/2)*2/3 =(6*√3/2)*2/3 =2√3.
H= MO =√(MA² - AO²) =√((2√7)² - (2√3)²) =2√(7-3) =4.
Окончательно
:
V =3√3*H =3√3*4 =12
√3.
ответ :12√3.
3 votes
Thanks 11
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Verified answer
A=AB=BC =CA =6 ; MA =MB=MC=2√7. MO_высота пирамиды,H=MO, где O центр основания.
V =(1/3)*S(ABC)*H =(1/3)*a²√3/4*H =(1/3)*6²√3/4*H=3√3*H .
AO =(AB*√3/2)*2/3 =(6*√3/2)*2/3 =2√3.
H= MO =√(MA² - AO²) =√((2√7)² - (2√3)²) =2√(7-3) =4.
Окончательно:
V =3√3*H =3√3*4 =12√3.
ответ :12√3.