Janek191:
h = a − 1
czyli
| | √3 | |
a*( |
| − 1) = − 1 / * 2 |
| | 2 | |
a*(
√3 − 2 ) = − 2
| | − 2 | | 2 | | 2 + √3 | | 4 + 2 √3 | |
a = |
| = |
| * |
| = |
| = |
| | √3 − 2 | | 2 − √3 | | 2 + √3 | | 4 − 3 | |
= 4 + 2
√3
−−−−−−−−−
więc
h = 4 + 2
√3 − 1 = 3 + 2
√3
−−−−−−−−−−−−−−−−−−−−−−−−−
Pole trójkąta
P = 0,5*a*h = 0,5*( 4 + 2
√3)*( 3 + 2
√3) = ( 2 +
√3)*( 3 + 2
√3) =
= 6 + 4
√3 + 3
√3 + 2*3 = 12 + 7
√3
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