| 16√2 | ||
Oblicz wysokość ściany bocznej czworościanu foremnego którego objętość wynosi | ||
| 3 |
| 1 | 16√2 | |||
Pp * H = | ||||
| 3 | 3 |
| 1 | a2 √3 | 16 √2 | |||
* | * H = | dobrze ? co dalej ? | |||
| 3 | 4 | 3 |
| 1 | a2 √3 | |||
a * h = | ||||
| 2 | 4 |
| a √3 | ||
h = | ||
| 2 |
| 2 | ||
( | h)2 + H2 = a2 | |
| 3 |
| 4 | a2 √3 | ||
+ H2 = a2 | |||
| 9 | 4 |
| a2 √3 | ||
H2 = a2 − | ||
| 9 |
| a√3 | 3a2 | |||
( | )2 = | |||
| 2 | 4 |

| a√3 | ||
h= | ||
| 2 |
| a2√3 | ||
Pp= | ||
| 2 |
| 1 | a√3 | ||
h⇒ | |||
| 3 | 6 |
| a√3 | ||
( | )2+H2=h2 | |
| 6 |
| a√3 | a√3 | |||
( | )2+H2=( | )2 | ||
| 6 | 2 |
| 3a2 | 3a2 | |||
H2= | − | |||
| 4 | 36 |
| 27a2 | 3a2 | |||
H2= | − | |||
| 36 | 36 |
| 24a2 | ||
H2= | ||
| 36 |
| 2√6a | ||
H= | ||
| 6 |
| 1 | a2√3 | |||
V= | * | *H | ||
| 3 | 4 |
| 16√2 | a2√3 | ||
= | *H | ||
| 3 | 4 |
| 64√2 | ||
H= | ||
| a2√3 |
| 2√6 | 64√2 | ||
= | |||
| 6 | a2√3 |
| 8√3 | ||
podstawiam do h= | ⇒4√3 | |
| 2 |
| a√3 | ||
racja razor będzie a3=64 stąd a=4 i wstawiamy do h= | ⇒2√3 | |
| 2 |
Co za dużo (w tym przypadku obliczeń) to niezdrowo.
| 1 | ||
Objętość czworościanu foremnego V = | a3√2 (nie ma potrzeby wyprowadzać | |
| 12 |
| 1 | 16 | 12 | |||
a3√2 = | √2 /* | ⇒ a3 = 64 ⇒ a = 4 | |||
| 12 | 3 | √2 |
| 1 | ||
h = | *4√3 = 2√3 | |
| 2 |
A jak się nie zna to trzeba liczyć
| 2 | ||
Z obliczeniami można tak: H = √h2 − h2/9 = | h√2 | |
| 3 |
| 1 | 1 | h | ||||
h = | a√3 ⇒ | a = | ||||
| 2 | 2 | √3 |
| 1 | h | 2 | 16 | ||||
* | *h* | h√2 = | √2 ⇒ h3 = 24√3 = (2√3)3 ⇒ h = 2√3 | ||||
| 3 | √3 | 3 | 3 |