Geometria triunghiului
Geometria triunghiului
Geometria triunghiului
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<strong>Geometria</strong> <strong>triunghiului</strong> 6<br />
Solutie<br />
✁ ✁✁✁<br />
✁ ✁✁✁✁✁✁<br />
A<br />
❆❆<br />
❆<br />
❆<br />
❆<br />
a<br />
❆ a<br />
❆<br />
B D<br />
b<br />
❆<br />
❆<br />
❆<br />
❆ C<br />
Fie triunghiul isoscel ABC cu AB = AC =<br />
a, BC = b. Fie AD este inaltimea relativa<br />
bazei. Din △ADC dreptunghic gasim AD = a2 <br />
2<br />
b<br />
− = a<br />
2<br />
2 − b2<br />
, atunci<br />
4<br />
sau<br />
sau<br />
Din r = A<br />
p deducem<br />
Din abc<br />
4A<br />
b<br />
<br />
a 2 − b2<br />
4<br />
= R deducem<br />
A△ABC = 1<br />
<br />
b<br />
BC · AD =<br />
2 2<br />
a2 − b2<br />
4 .<br />
<br />
b<br />
2<br />
a2 − b2<br />
4<br />
a + b<br />
2<br />
= 3<br />
,<br />
2<br />
adica<br />
<br />
= 3 a + b<br />
<br />
2<br />
⇔ b 2<br />
<br />
a − b<br />
<br />
a +<br />
2<br />
b<br />
<br />
= 9 a +<br />
2<br />
b<br />
2 2<br />
a + b<br />
2 =<br />
b2 <br />
a − b<br />
<br />
2<br />
. (∗)<br />
9<br />
2b<br />
a2b <br />
a 2 − b2<br />
4<br />
= 25<br />
8<br />
25 a − b<br />
<br />
a +<br />
2<br />
b<br />
<br />
= 4a<br />
2<br />
2 . (∗∗)<br />
Rezolvam sistemul, format din ecuatiile (*) si (**). Substituim a + b<br />
din (*) in (**) si<br />
2<br />
obtinem ecuatia omogena 24a2 − 50ab + 25b2 = 0. Impartim ecuatia la b2 <br />
= 0 si obtinem<br />
a<br />
2 ecuatia 24 − 50<br />
b<br />
a<br />
+ 25 = 0. De aici<br />
b<br />
⎡<br />
⎢<br />
⎣<br />
a 5<br />
=<br />
b 6 ,<br />
a 5<br />
=<br />
b 4 ,<br />
⇔<br />
⎡<br />
⎢ a =<br />
⎢<br />
⎣<br />
5<br />
6 b,<br />
a = 5<br />
4 b.