Se da trapezul \( ABCD \) cu \( AB\parallel CD \), \( m(\angle {A}=90\textdegree) \),\( m(\angle {C}=2\cdot m{\angle{B}}) \) si \( AB=3DC \).
a) Sa se determime pe segmentul \( (AD) \) punctele \( M \) si \( N \) astfel incat \( \angle {DMC}\equiv \angle {AMB} \) si \( \angle{DNC}\equiv \angle{ABN} \).
b) Stiind ca \( MN=3 cm \), sa se determine aria trapezului.
Constantin Apostol, Concusul "Laurentiu Duican" 2004
Problema cu un trapez dreptunghic
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a) Notam pe \( CD \) cu \( b \). Atunci \( AB=3b \).
Asemanarea \( \triangle DMC\sim \triangle AMB \) ofera \( \frac{AM}{DM}=\frac{AB}{CD}=3\ \Longleftrightarrow\ DM=\frac{1}{4}AD \).
Din \( \triangle DNC\sim \triangle ABN\ \Longrightarrow\ \frac{DN}{AB}=\frac{DC}{AN}\ \Longrightarrow\ \sqrt{DN\cdot AN}=b\sqrt{3} \)
Deoarece \( m(\angle C)=2m(\angle B),\ m(\angle B)+m(\angle C)=180^{\circ}\ \Longrightarrow\ m(\angle B)=60^{\circ} \).
De aici deducem \( DN+AN=AD=(AB-CD)\tan 60^{\circ}=2b\sqrt{3} \)
Deci \( \sqrt{DN\cdot AN}=b\sqrt{3}=\frac{DN+AN}{2}\ \Longrightarrow\ DN=AN=b\sqrt{3} \), ceea ce arata ca \( N \) este mijlocul lui \( [AD] \).
b)\( MN=DN-DM=\frac{b\sqrt{3}}{2}=3 cm\ \Longrightarrow\ b=2\sqrt{3}cm\ \Longrightarrow\ S_{ABCD}=48\sqrt 3\ \mbox{cm}^2 \)
Asemanarea \( \triangle DMC\sim \triangle AMB \) ofera \( \frac{AM}{DM}=\frac{AB}{CD}=3\ \Longleftrightarrow\ DM=\frac{1}{4}AD \).
Din \( \triangle DNC\sim \triangle ABN\ \Longrightarrow\ \frac{DN}{AB}=\frac{DC}{AN}\ \Longrightarrow\ \sqrt{DN\cdot AN}=b\sqrt{3} \)
Deoarece \( m(\angle C)=2m(\angle B),\ m(\angle B)+m(\angle C)=180^{\circ}\ \Longrightarrow\ m(\angle B)=60^{\circ} \).
De aici deducem \( DN+AN=AD=(AB-CD)\tan 60^{\circ}=2b\sqrt{3} \)
Deci \( \sqrt{DN\cdot AN}=b\sqrt{3}=\frac{DN+AN}{2}\ \Longrightarrow\ DN=AN=b\sqrt{3} \), ceea ce arata ca \( N \) este mijlocul lui \( [AD] \).
b)\( MN=DN-DM=\frac{b\sqrt{3}}{2}=3 cm\ \Longrightarrow\ b=2\sqrt{3}cm\ \Longrightarrow\ S_{ABCD}=48\sqrt 3\ \mbox{cm}^2 \)