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電子課本網(wǎng) 第41頁(yè)

第41頁(yè)

信息發(fā)布者:
解:∵?$DE//BC$?
∴?$△ADE∽△ABC$?
∴?$\frac {DE}{BC}=\frac {AE}{AC}$?
∵?$\frac {AE}{EC}=\frac 12$?
∴?$\frac {DE}{BC}=\frac {AE}{AC}=\frac 13$?
∵?$DE//BC$?
∴?$∠ODE=∠OCB$?
∵?$∠DOE=∠BOC$?
∴?$△DOE∽△BOC,$?且相似比為?$1 :$??$ 3$?
∴?$△DOE$?與?$△BOC$?的周長(zhǎng)比為?$1 :$??$3,$?面積比為?$1 :$??$9$?
解:???$(1)$???∵???$AD//BC$???
∴???$∠DAO=∠OCB$???
∵???$∠AOD=∠BOC$???
∴???$△AOD∽△COB$???
∵???$△AOD$???的面積與???$△BOC$???的面積之比為???$1:$??????$9$???
∴???$AD:$??????$BC=1:$??????$3$???
???$(2)$???∵???$△AOD∽△COB,$??????$AD:$??????$BC=1:$??????$3$???
∴???$OD:$??????$OB=AD:$??????$BC=1:$??????$3$???
∴???$S_{△AOD}:$??????$S_{△AOB}=1:$??????$3$???
∵???$△AOB$???的面積為???$6$???
∴???$S_{△AOD}=2,$??????$S_{△ABD}=8$???
∵???$S_{△ABD}:$??????$S_{△BCD}=AD:$??????$BC=1:$??????$3$???
∴???$S_{△BCD}=24$???
∴???$S_{梯形ABCD}=S_{△ABD}+S△ BCD=32$???
解:作??$AG⊥BC,$??垂足為??$G$??

∵??$AD=AC$??
∴??$∠ACB=∠FDC$??
∵點(diǎn)??$D$??是??$BC$??的中點(diǎn)
∴??$DB=DC$??
∵??$DE⊥BC$??
∴??$∠EDB=∠EDC=90°$??
在??$△BDE$??和??$△CDE$??中
??$\begin{cases}{DE=DE}\\{∠EDB=∠EDC}\\{DB=DC}\end{cases}$??
∴??$△BDE≌△CDE(\mathrm {SAS})$??
∴??$∠FCD=∠ABC$??
∵??$∠FDC=∠ACB$??
∴??$△FCD∽△ABC,$??且相似比為??$CD:$????$BC=1:$????$2$??
∴??$S_{△ABC}=4S_{△FCD}$??
∵??$S_{△FCD}=5$??
∴??$S_{△ABC}=\frac 12×BC×AG=20$??
∵??$BC=10$??
∴??$AG=4$??
∵點(diǎn)??$D$??為??$BC$??的中點(diǎn)
∴??$BD=CD=5$??
∵??$AD=AC,$????$AG⊥BC$??
∴點(diǎn)??$G$??為??$CD$??的中點(diǎn),??$DG=\frac 12CD=\frac 52$??
∴??$BG=BD+DG=\frac {15}{2}$??
∵??$DE⊥BC$??
∴??$DE//AG$??
∴??$△BDE∽△BGA$??
∴??$\frac {BD}{BG}=\frac {DE}{AG}$??
∵??$BD=5,$????$BG=\frac {15}{2},$????$AG=4$??
∴??$\frac 5{\frac {15}{2}}=\frac {DE}4$??
∴??$DE=\frac 83$??
解:由???$△ADE∽△ABC,$???得???$\frac {S_{△ADE}}5=(\frac {AD}{AB})^2=x^2,$???
???$S_{△ADE}=5x^2$???
又???$\frac y{S_{△ADE}}=\frac {BD}{AD}=\frac {AB-AD}{AD}=\frac {AB}{AD}-1=\frac 1{x}-1$???
∴???$y=(\frac 1{x}-1) · S_{△ADE}=(\frac 1{x}-1) · 5x^2=5x-5x^2$???
∴???$y=5x-5x^2(0<x<1)$???