Pb=2Pp
pole boczne=2 pole podstawy
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Narysowałam połowę podstawy.
Z tgα
| a | ||
h= | *tgα | |
| 2 |
| a | 1 | |||
b= | * | |||
| 2 | cosα |
| 1 | ||
Pp= | a*h | |
| 2 |
| a3sin2α | ||
w odpowiedziach mam | a mi wyszło coś innego | |
| 8cosα(cosα+1) |
| h | |||||||
= tg α
| |||||||
|
| a | ||
h = | tgα
| |
| 2 |
| |||||||
= cos α
| |||||||
| b |
| a | ||
b = | ||
| 2cos α |
| a*h | a2 | |||
Pp = | = | tgα
| ||
| 2 | 4 |
| a2 | ||
(2b+a)*H = | tg α
| |
| 2 |
| a tgα cosα | ||
H = | ||
| 2(1+cos α) |
| a2 | a tgα cosα | a3sin2α | ||||
V = Pp* H = | tg α * | = | ||||
| 4 | 2(1 + cosα) | 8cos α(1 + cosα) |