Aria trapezului
Moderators: Bogdan Posa, Laurian Filip
Aria trapezului
In trapezul \( ABCD (BC||AD) \) , diagonalele se intersecteaza in \( O \) . Notand \( S_1=\mathb{A}_{BOC} \), \( S_2=\mathb{A}_{AOD}, \) sa se arate ca aria trapezului \( ABCD \) este : \( S_1+S_2+2\cdot\sqrt{S_1\cdot{S_2}} \).
-
Andi Brojbeanu
- Bernoulli
- Posts: 294
- Joined: Sun Mar 22, 2009 6:31 pm
- Location: Targoviste (Dambovita)
Fie \( OE, OF \) proiectiile lui \( O \) pe \( AD \), respectiv \( BC \).
Atunci avem de demonstrat relatia:
\( \frac{(AD+BC)\cdot (OE+OF)}{2}=\frac{BC\cdot OF}{2}+\frac{AD\cdot OE}{2}+2\cdot \sqrt {\frac{AD\cdot OE}{2}\cdot \frac{BC\cdot OF}{2}} \)
\( AD\cdot OE+AD\cdot OF+BC\cdot OE+BC\cdot OF=BC\cdot OF+AD\cdot OE+2\sqrt {AD\cdot OE\cdot BC\cdot OF} \).
Scazand membrul drept obtinem: \( (\sqrt {AD\cdot OF})^2+(\sqrt {BC\cdot OE}^2)-2\sqrt {AD\cdot OF\cdot BC\cdot OE}=0 \)\( \Rightarrow (\sqrt {AD\cdot OF}-\sqrt {BC\cdot OE})^2=0 \)
\( \Rightarrow \sqrt {AD\cdot OF}=\sqrt {BC\cdot OE}\Rightarrow AD\cdot OF=BC\cdot OE \)\( \frac{AD}{BC}=\frac{OE}{OF} \), adevarat deoarece in triunghiurile \( AOD \)respectiv\( BOC \), raportul de asemanare \( k \) este egal cu raportul inaltimilor.
Atunci avem de demonstrat relatia:
\( \frac{(AD+BC)\cdot (OE+OF)}{2}=\frac{BC\cdot OF}{2}+\frac{AD\cdot OE}{2}+2\cdot \sqrt {\frac{AD\cdot OE}{2}\cdot \frac{BC\cdot OF}{2}} \)
\( AD\cdot OE+AD\cdot OF+BC\cdot OE+BC\cdot OF=BC\cdot OF+AD\cdot OE+2\sqrt {AD\cdot OE\cdot BC\cdot OF} \).
Scazand membrul drept obtinem: \( (\sqrt {AD\cdot OF})^2+(\sqrt {BC\cdot OE}^2)-2\sqrt {AD\cdot OF\cdot BC\cdot OE}=0 \)\( \Rightarrow (\sqrt {AD\cdot OF}-\sqrt {BC\cdot OE})^2=0 \)
\( \Rightarrow \sqrt {AD\cdot OF}=\sqrt {BC\cdot OE}\Rightarrow AD\cdot OF=BC\cdot OE \)\( \frac{AD}{BC}=\frac{OE}{OF} \), adevarat deoarece in triunghiurile \( AOD \)respectiv\( BOC \), raportul de asemanare \( k \) este egal cu raportul inaltimilor.