task/24844832
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6.
1) SC =5 ; AC =6 .
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SO -?
OC =AC/2 =6/2 =3
Из ΔSOC по теореме Пифагора :
SO =√(SC² -OC²) = √(5² -3²) =√(25 -9) =√16
=4 .
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2) SO =4 ; SC =5 .
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AC -?
AC =2*OC =2√(SC²- SO²) =2√(5²- 4²) =2√(25- 16) =2
√9 =2*3 =6.
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3) SO =H =4 ; AB =2R =6 .
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L =SA -?
R =AO =AB/2 =6/2 =3
Из ΔSOA по теореме Пифагора :
SA=√(AO²+SO²) =√(R²+H²) =√(3²+4²) =√(9+16) =
√25 =5.
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4) SO =H =4 ;
SA=5
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AB =2R
- ? .
Из ΔSOA :
R =AO =√(SA² - SO²) =√(5² - 4²) =
√(25 - 16) = √9 = 3
AB = 2R =2AO = 2*3
=6.
рисунок см приложения
