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電子課本網(wǎng) 第82頁

第82頁

信息發(fā)布者:

解:過點(diǎn)?$B$?作?$BE$?垂直于橫格,過點(diǎn)?$D$?作?$DF $?垂直于橫格
由題意可知,?$BE=24\ \mathrm {mm},$??$DF=48\ \mathrm {mm}$?
∵?$∠BAD=90°,$?
∴?$∠EAB+∠FAD=90°$?
∴?$∠EAB=∠ADF=36°$?
∴?$AB=\frac {BE}{sin 36°}≈40\ \mathrm {mm},$??$AD=\frac {DF}{cos 36°}≈60\ \mathrm {mm}$?
?$C_{矩形ABCD}=2(AB+ AD)= 200\ \mathrm {mm}$?
答:矩形卡片?$△BCD$?的周長為?$200\ \mathrm {mm}。$?
解:截取大門部分如圖所示

由題意可知,?$AB=0.5m$?
?$∠ABC=\frac {1}{2}×50°=25°$?
∴?$AC=sin 25°×AB≈0.211m$?
∴大門的寬?$d=9×2AC+ AC≈4.01m$?
答:大門的寬為?$4.01m。$?
解:∵?$∠CED=45°$?
∴?$DC = sin 45°×DE = 40\sqrt{2}\ \mathrm {cm}$?
設(shè)圓?$O$?的半徑為?$r$?
∵?$∠BAC=30°,$
?∴?$OC=sin 30°×OA$?
∴?$2(r+40\sqrt{2})=r+150,$?
?$r= 150 - 80\sqrt{2}$?
?$CF= CD+ 2r= (300- 120\sqrt{2})\ \mathrm {cm}$?
答:熱水器?$CF $?的高度為?$(300 - 120\sqrt{2})\ \mathrm {cm}。$?

解:?$ (1)$?過點(diǎn)?$F $?作?$FH⊥CD$?交于點(diǎn)?$H$?
∵?$∠CDF=30°$?
∴?$FH=DF×sin 30°=15\ \mathrm {cm},$?
?$DH=cos 30°×DF=15\sqrt{3}\ \mathrm {cm}$?
∵?$∠DCF=45°$?
∴?$CH=FH= 15\ \mathrm {cm}$?
∴?$CD= CH+ DH= (15 + 15\sqrt{3})\ \mathrm {cm}$?
∵?$CE :$??$ CD= 1:$??$3$?
∴?$CE= (5+5\sqrt{3})\ \mathrm {cm}$?
∴?$DE= CE+CD= (20+20\sqrt{3})\ \mathrm {cm}$?
∵?$DE= BC= AB$?
∴?$AC= 2DE= (40 + 40\sqrt{3})\ \mathrm {cm}$?
?$(2)$?過點(diǎn)?$A$?作?$AG⊥DE$?于點(diǎn)?$G$?
∵?$∠ACD=45°$?
∴?$AG= sin 45°×AC= 20\sqrt{2}+20\sqrt{6}≈77\ \mathrm {cm}$?
答:拉桿端點(diǎn)?$A$?到水平滑桿?$DE$?的距離是?$77\ \mathrm {cm}。$?