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零五網(wǎng) › 全部參考答案› 經(jīng)綸學(xué)典學(xué)霸 › 2025年學(xué)霸題中題七年級(jí)數(shù)學(xué)上冊(cè)蘇科版 第164頁(yè)解析答案
3. (2025·上海期中) 如圖,$CD // AB$,現(xiàn)將一塊含$30^{\circ}的三角板EFG$按如圖①放置,$\angle G = 90^{\circ}$,$\angle EFG = 30^{\circ}$,使點(diǎn)$E$,$F分別在直線(xiàn)CD$,$AB$上,設(shè)$\angle GFB = \alpha(0^{\circ} ＜ \alpha ＜ 90^{\circ})$。
(1) 求$\angle DEG + \angle GFB$的度數(shù)。
(2) 如果$\angle CEF的平分線(xiàn)EH交直線(xiàn)AB于點(diǎn)H$,如圖②。
①當(dāng)$EH // FG$時(shí),求$\alpha$的度數(shù)；
②在①的條件下,如果點(diǎn)$P是射線(xiàn)EC$上的一點(diǎn),將三角板$EFG繞點(diǎn)E以每秒1^{\circ}$的速度進(jìn)行順時(shí)針旋轉(zhuǎn),同時(shí)射線(xiàn)$PC繞點(diǎn)P以每秒4^{\circ}$的速度進(jìn)行順時(shí)針旋轉(zhuǎn),射線(xiàn)$PC$旋轉(zhuǎn)一周后停止轉(zhuǎn)動(dòng),同時(shí)三角板$EFG$也停止轉(zhuǎn)動(dòng)。當(dāng)旋轉(zhuǎn)多少時(shí)間時(shí),$CP與\triangle EFG$的一邊平行？

答案：
3.(1)如圖①,過(guò)點(diǎn)G作GM//AB,因?yàn)镃D//AB,MG//AB,所以CD//MG,所以∠MGF = ∠BFG,∠DEG = ∠EGM,所以∠DEG + ∠BFG = ∠EGM + ∠MGF = ∠EGF = 90°,所以∠DEG + ∠BFG = 90°.
　　　　　　　　　30
(2)①因?yàn)镋H//FG,所以∠GFB = ∠EHF = α.因?yàn)镋H平分∠CEF,所以∠CEH = ∠FEH.又因?yàn)镃D//AB,所以∠CEH = ∠EHF = α,∠CEF = ∠EFB = 2α,所以2α = 30° + α,解得α = 30°.
②如圖②,當(dāng)PC'//EG時(shí),延長(zhǎng)GE至點(diǎn)Q,因?yàn)镻C'//GQ,所以∠CPC' = ∠CEQ.因?yàn)椤螩EQ = ∠DEG,所以∠CPC' = ∠DEG,由題意知∠CPC' = (4t)°,由①得∠DEG = 60° + t°,所以(4t)° = 60° + t°,解得t = 20.
　　　　 　
如圖③,當(dāng)PC'//EF時(shí),所以∠CPC' = ∠DEF,由題意得∠DEF = ∠DEG + ∠FEG = 120° + t°,所以(4t)° = 120° + t°,解得t = 40;
如圖④,當(dāng)PC'//FG時(shí),延長(zhǎng)GF交CE于點(diǎn)T,過(guò)點(diǎn)G作GV//CD,所以∠VGE = ∠DEG = 60° + t°.因?yàn)椤螰GE = 90°,所以∠VGT = t° - 30°,所以∠ETG = ∠VGT = t° - 30°.因?yàn)镻C'//FG,所以∠DPC' = ∠ETG = t° - 30°.因?yàn)椤螪PC' = (4t)° - 180°,所以(4t)° - 180° = t° - 30°,解得t = 50.
　　　　AB
如圖⑤,當(dāng)PC'//EG(第二次)時(shí),則∠CPC' = ∠CEG,所以360° - (4t)° = 180° - (60° + t°),解得t = 80.
綜上,當(dāng)旋轉(zhuǎn)20秒或40秒或50秒或80秒時(shí),CP與△EFG的一邊平行
4. (2025·成都期末) 已知$\angle AOB = 90^{\circ}$。
(1) 如圖①,若射線(xiàn)$ON$,$OM分別為\angle AOC$,$\angle BOC$的平分線(xiàn),則$\angle MON = $______。
(2) 如圖②,射線(xiàn)$OC從OA出發(fā)繞點(diǎn)O以每秒20^{\circ}$的速度沿逆時(shí)針?lè)较蛐D(zhuǎn),射線(xiàn)$OD從OA出發(fā)繞點(diǎn)O以每秒10^{\circ}$的速度沿順時(shí)針?lè)较蛐D(zhuǎn),設(shè)運(yùn)動(dòng)時(shí)間為$t$秒,且$OM平分\angle BOC$。
①當(dāng)$0 ＜ t ＜ 4.5$時(shí),若$ON分\angle AOC為1:3$兩個(gè)部分,求滿(mǎn)足$\angle DON = \frac{1}{2}\angle MON$時(shí),$t$的值。
②如圖③,若$OP平分\angle AOD$,當(dāng)$0 ＜ t ＜ 6且t \neq 4.5$時(shí),試判斷$2\angle MOP - \angle COD$是否為定值。若是,請(qǐng)求出該定值；若不是,請(qǐng)說(shuō)明理由。

答案：
4.(1)45°　解析:如圖①,∠AOB = 90°,則∠AOC + ∠BOC = 90°.因?yàn)樯渚€(xiàn)ON,OM分別為∠AOC,∠BOC的平分線(xiàn),所以∠CON = $\frac{1}{2}$∠AOC,∠MOC = $\frac{1}{2}$∠BOC,所以∠MON = ∠CON + ∠MOC = $\frac{1}{2}$(∠AOC + ∠BOC) = $\frac{1}{2}$×90° = 45°.
　　　　　　　　　　
(2)①如圖②,因?yàn)樯渚€(xiàn)OC從OA出發(fā)繞點(diǎn)O以每秒20°的速度沿逆時(shí)針?lè)较蛐D(zhuǎn)，射線(xiàn)OD從OA出發(fā)繞點(diǎn)O以每秒10°的速度沿順時(shí)針?lè)较蛐D(zhuǎn)，運(yùn)動(dòng)時(shí)間為t秒,所以∠AOC = (20t)°,∠AOD = (10t)°,所以∠BOC = 90° - (20t)°.因?yàn)镺M平分∠BOC,ON分∠AOC為1:3兩個(gè)部分,所以∠COM = $\frac{1}{2}$∠BOC = 45° - (10t)°,∠AON = $\frac{1}{4}$∠AOC = (5t)°,∠CON = $\frac{3}{4}$∠AOC = (15t)°或∠AON = $\frac{3}{4}$∠AOC = (15t)°,∠CON = $\frac{1}{4}$∠AOC = (5t)°.當(dāng)∠AON = $\frac{1}{4}$∠AOC = (5t)°,∠CON = $\frac{3}{4}$∠AOC = (15t)°時(shí),所以∠DON = ∠AON + ∠AOD = (5t)° + (10t)° = (15t)°,∠MON = ∠COM + ∠CON = 45° - (10t)° + (15t)° = 45° + (5t)°.因?yàn)椤螪ON = $\frac{1}{2}$∠MON,所以(15t)° = $\frac{1}{2}$[45° + (5t)°],解得t = $\frac{9}{5}$;
當(dāng)∠AON = $\frac{3}{4}$∠AOC = (15t)°,∠CON = $\frac{1}{4}$∠AOC = (5t)°時(shí),所以∠DON = ∠AON + ∠AOD = (15t)° + (10t)° = (25t)°,∠MON = ∠COM + ∠CON = 45° - (10t)° + (5t)° = 45° - (5t)°.因?yàn)椤螪ON = $\frac{1}{2}$∠MON,所以(25t)° = $\frac{1}{2}$[45° - (5t)°],解得t = $\frac{9}{11}$
綜上所述，t的值為$\frac{9}{5}$或$\frac{9}{11}$
　　　　 　　
②當(dāng)0 ＜ t ＜ 4.5時(shí),如圖③,∠AOC = (20t)°,∠AOD = (10t)°,∠BOC = 90° - (20t)°,因?yàn)镺M平分∠BOC,OP平分∠AOD,所以∠MOC = $\frac{1}{2}$∠BOC = 45° - (10t)°,∠AOP = $\frac{1}{2}$∠AOD = (5t)°,所以∠MOP = ∠MOC + ∠AOC + ∠AOP = 45° - (10t)° + (20t)° + (5t)° = 45° + (15t)°,∠COD = ∠AOC + ∠AOD = (20t)° + (10t)° = (30t)°,所以2∠MOP - ∠COD = 2[45° + (15t)°] - (30t)° = 90°,為定值;
當(dāng)4.5 ＜ t ＜ 6時(shí),如圖④,∠AOC = (20t)°,∠AOD = (10t)°,∠BOC = (20t)° - 90°,因?yàn)镺M平分∠BOC,OP平分∠AOD,所以∠MOB = $\frac{1}{2}$∠BOC = (10t)° - 45°,∠AOP = $\frac{1}{2}$∠AOD = (5t)°,所以∠MOP = ∠MOB + ∠AOB + ∠AOP = (10t)° - 45° + 90° + (5t)° = 45° + (15t)°,∠COD = ∠AOC + ∠AOD = (20t)° + (10t)° = (30t)°,所以2∠MOP - ∠COD = 2[45° + (15t)°] - (30t)° = 90°,為定值.
綜上所述,2∠MOP - ∠COD = 90°,為定值
　　　　　　　　
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