109年國中教育會考(補考)數學詳解

109年國中教育會考(補考)數學詳解

解：
$${11 \over 4}-(-1{5 \over 6}) ={11 \over 4}-(-{11 \over 6}) ={11 \over 4}+{11 \over 6} = {33\over 12} +{22\over 12} ={55 \over 12}，故選\bbox[red,2pt]{(D)}$$

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解：

$$圖形為上下對稱，故選\bbox[red, 2pt]{(A)}$$

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解：$$\cases{(A)x=4,y=-7 \Rightarrow x-2y=4+14=18 \ne 10\\ (B)x=4,y=-3 \Rightarrow x-2y=4+6=10 \\(C)x=4,y=5 \Rightarrow x-2y=4-10=-6 \ne 10 \\(D)x=4,y=7 \Rightarrow x-2y=4-14=-10\ne 10}，故選\bbox[red,2pt]{(B)}$$

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解：$$(2x^2+5)-(x+5) =2x^2+5-x-5 =2x^2-x，故選\bbox[red, 2pt]{(C)}$$

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解：$$5^2(25) <29.5 < 5.5^2(30.25) < 6^2(36)，故選\bbox[red,2pt]{(A)}。$$

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解：$$全距=全班最高-最低，因此\\ \cases{蘋果班(最高,最低)=(117,106) \xrightarrow {加入103公分}(117,103) \Rightarrow 全距變大 \\鳳梨班(最高,最低)=(120,99) \xrightarrow{加入119公分)}(120,99) \Rightarrow 全距不變}，故選\bbox[red,2pt]{(B)}。$$

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解：
$$(2^3\times 3^4)^2 \times (2^4\times 3^2) =(2^6\times 3^8) \times (2^4\times 3^2) = 2^{10}\times 3^{10}，故選\bbox[red,2pt]{(B)}。$$

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解：
$$汽水一瓶原價a元\Rightarrow \cases{買2瓶打9折，即售價2a\times 0.9元\\ 買3瓶打8折，即售價3a\times 0.8元} \Rightarrow 2a\times 9+18= 3a\times 8，故選\bbox[red,2pt]{(C)}。$$

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解：$$假設A點坐標為a，則\cases{\overline{AC}=3-a \\ \overline{BC}=3-(-1)=4} \Rightarrow \overline{AC}=3\overline{BC} \Rightarrow 3-a=3\times 4=12 \\\Rightarrow a=3-12=-9，故選\bbox[red,2pt]{(C)}。$$

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解：
$$\cases{315= 3^2\times 5\times 7 \\ 588 = 2^2\times 3\times 7^2} \Rightarrow 最小公倍數=2^2\times 3^2\times 5\times 7^2 =\cases{315\times 2^2\times 7 = 315\times 28 \\588\times 3\times 5=588\times 15}\\，故選\bbox[red,2pt]{(C)}。$$

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解：$$\angle B=50^\circ \Rightarrow \cases{\angle HPG=\angle D= 50^\circ \\ \angle C=180^\circ-50^\circ =130^\circ \\ \angle EPF = \angle B=50^\circ}\Rightarrow \angle FPG = 360^\circ - \angle EPF- \angle HPG- \angle EPH \\ = 360^\circ - 50^\circ - 50^\circ - 110^\circ = 150^\circ ；\\因此在四邊形PGCF中:\angle PFC+\angle PGC = 360^\circ - \angle FPG-\angle C = 360^\circ - 150^\circ -130^\circ = 80^\circ\\，故選\bbox[red, 2pt]{(A)}。$$

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解：$$x^2+5x=0 \Rightarrow x^2+5x + ({5\over 2})^2=({5\over 2})^2 \Rightarrow (x+{5\over 2})^2 ={25 \over 4} \Rightarrow \cases{a=5/2\\ b=25/4} \Rightarrow a+b=35/4\\，故選\bbox[red,2pt]{(B)}。$$

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解：$$與兩水平線有4個交點，必須滿足兩條件之一:\cases{若圖形為凹向上，其頂點 需低於L \\ 若圖形為凹向下，其頂點需高於M};\\\cases{(A)凹向上，頂點(35,-35)在L之上\\ (B)凹向上，頂點(35,-45)在L之上 \\(C)凹向下，頂點(45,-35)在M之上 \\ (D)凹向下，頂點(45,-45)在M之下};故選\bbox[red,2pt]{(C)}。$$

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解：$$\cases{(A)號碼小於3有3+3=6顆球 \\(B)號碼大於4有2+2=4顆球 \\(C)號碼為3有5顆球 \\(D)號碼為4有4顆球} \Rightarrow 號碼小於3的球數最高，故選\bbox[red,2pt]{(A)}$$

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解：

$$\overline{AC}=\overline{AE} \Rightarrow \angle ACE= \angle AEC = (180^\circ -\angle CAD-\angle DAE)\div 2 = (180^\circ -30^\circ -30^\circ)\div 2=60^\circ \\ 同弧的圓周角相等\Rightarrow \angle ABE=\angle ACE=60^\circ;\\在\triangle ABE \Rightarrow \angle AEB =180^\circ-(60^\circ+30^\circ +30^\circ +29^\circ)=31^\circ \\ \Rightarrow \stackrel{\huge \frown}{AB} =2\times \angle AEB=2\times 31^\circ =62^\circ，故選\bbox[red, 2pt]{(D)}$$

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解：$$原館藏\cases{甲:4k \\ 乙:3k} \xrightarrow{兩館皆添購新書書量a}\cases{甲:4k+a\\ 乙:3k+a} \Rightarrow {4k+a \over 3k+a} ={11\over 9} \Rightarrow a={3\over 2}k\\ \Rightarrow \cases{新書總和= 2a= 3k \\ 舊藏書總和=7k} \Rightarrow {新書 \over 舊總和}  ={3k \over 7k} ={3\over 7}，故選\bbox[red,2pt]{(A)}$$

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解：
$$500個饅頭在晚上六點前賣完，收入為500\times 20=10000元；\\某日晚上六點剩下50個饅頭，表示賣出500-50=450個饅頭，收入450\times 20=9000元；\\到了晚上八時剩下x個饅頭，收入(50-x)\times 20\times 0.8=16(50-x)；\\在關店時剩下y個，收入為(x-y)\times 20\times 0.5=10(x-y)；\\兩日收入相差10000-9000-16(50-x)-10(x-y) =200+6x+10y，故選\bbox[red,2pt]{(B)}$$

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解：

$$\cases{\overline{AD}\parallel \overline{BC} \Rightarrow \angle DAC=\angle DCA \\\overline{CA}為\angle BCD的角平分線\Rightarrow \angle DCA=\angle ACB} \\\Rightarrow \angle ACB=\angle DCA=\angle DAC = (180^\circ-\angle ADC)\div 2  =(180^\circ - 110^\circ)\div 2= 35^\circ;\\在\triangle ABC \Rightarrow \angle BAC=180^\circ - \angle ABC-\angle ACB = 180^\circ - 50^\circ -35^\circ = 95^\circ \\ \angle A= \angle BAC+\angle CAD= 95^\circ+35^\circ = 130^\circ \Rightarrow \angle IAD = \angle A \div 2=65^\circ \\\Rightarrow \angle IAC= \angle IAD-\angle CAD=65^\circ - 35^\circ =30^\circ  ，故選\bbox[red,2pt]{(C)}。$$

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解：
$$\cases{甲+乙=30\Rightarrow 甲=30-乙\cdots(1) \\ 10甲+15乙> 375 \cdots(2)},將(1)代入(2)可得10(30-乙)+15乙>375 \\ \Rightarrow 300+5乙> 375 \Rightarrow 乙>15，即乙方案至少執行16天，則游泳至少游了2\times 16=32公里\\，故選\bbox[red,2pt]{(D)}。$$

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解：

$$令\cases{O(-2,1) \\ A(-10,a) \\B(6,b) \\C(8,c)} \Rightarrow \cases{|a-1|=\overline{AE} \\ |b-1|=\overline{BF} \\ |c-1|= \overline{CG}}\\ \cases{\overline{OE}=\overline{OF} =8 \Rightarrow \overline{AE}=\overline{BF} \Rightarrow |a-1|=|b-c| \\ \overline{OG}=6+2=8 > \overline{OE} \Rightarrow \overline{CF} >\overline{AE} \Rightarrow |c-1| >|a-1|}，故選\bbox[red,2pt]{(D)}。$$

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解：

$$\overline{AE}為角平分線\Rightarrow \overline{EF}=\overline{EB}=a；又\overline{BC}為直徑\Rightarrow \angle BDC=90^\circ\\ \overline{EF}\parallel \overline{BD} \Rightarrow {\overline{CE}\over \overline{CB}} ={\overline{EF}\over \overline{BD}} \Rightarrow {9\over 9+a } ={a \over 10} \Rightarrow a^2+9a-90=0 \Rightarrow (a-6)(a+15)=0 \\ \Rightarrow a=6，故選\bbox[red,2pt]{(B)}。$$

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解：

$$\overline{AF} \bot \overline{CD} \Rightarrow \angle CFP = 90^\circ -{x\over 2};又正五邊形每一內角均為108^\circ \Rightarrow \angle B=\angle C=108^\circ \\ 在四邊形BCFP中，108^\circ+108^\circ+90^\circ -{x\over 2}+y=360^\circ \Rightarrow 2y=108^\circ +x，故選\bbox[red, 2pt]{(D)}$$

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解：$$\cases{甲數列:a_1,a_2,\dots,a_n\\ 乙數列:a_1,a_2,\dots,a_n,a_{n+1}} \Rightarrow S_{甲}-S_{乙}=-a_{n+1} =-( a_1+nd)\\= -(45+ n\times (-4))= 4n-45=7 \Rightarrow n=13，故選\bbox[red,2pt]{(D)}。$$

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解：

$$DEBF為菱形 \Rightarrow \overline{EF}\bot \overline{BD} \Rightarrow E、F在對角線\overline{AC}上；又菱形面積為正方形的一半 \Rightarrow \overline{AE} =\overline{EP}=a\\ 在直角\triangle DEP: a^2+({a\over 2})^2=(\sqrt{15})^2 \Rightarrow a=2\sqrt 3 \Rightarrow ABCD面積=2a^2=24，故選\bbox[red,2pt]{(A)}。$$

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解：

$$O為圓心，G為切點\Rightarrow \overline{OG} \bot \overline{AD}  \Rightarrow \overline{OG} \bot \overline{BC} \Rightarrow \cases{\angle GOC=90^\circ \\ \angle EOF=90^\circ} \\\Rightarrow \angle COF+ \angle FOG=90^\circ = \angle FOG+\angle GOE \Rightarrow \angle COF= \angle GOE =a\\ 又\overline{GO}\parallel \overline{AB} \Rightarrow \angle GOE=\angle OEB =a，因此由\cases{\angle OEB=\angle COF=a \\ \angle FCO =\angle EBO =90^\circ} \Rightarrow \triangle FCO\sim \triangle OBE \\ \Rightarrow {\overline{FC} \over \overline{BO}}={\overline{FO} \over \overline{EO}}={\overline{CO} \over \overline{EB}} \Rightarrow {\overline{FC} \over 2}={r \over r}={1 \over \overline{EB}} \Rightarrow \cases{ \overline{FC}= 2 \\ \overline{EB}=1} \Rightarrow r=\sqrt{1^2+2^2} =\sqrt 5 =\overline{CD} \\ \Rightarrow 矩形周長=2 \overline{CD}+ 2\overline{BC} =2\sqrt 5+6，故選\bbox[red, 2pt]{(D)}$$

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解：

$$令\cases{\overline{FC}=a\\ \overline{AF}=b},其中a>b；又\cases{\overline{DF}\parallel \overline{BC} \\ \overline{FE} \parallel \overline{AB}} \Rightarrow BDFE為平行四邊形 \\\Rightarrow \cases{\angle FEC=\angle 1=\angle B=60^\circ \Rightarrow \triangle FEC 為正\triangle \Rightarrow \overline{BD} =\overline{EF}= \overline{EC}=a\\ \angle DEB = \angle 2}；\\ 又\overline{BC} =\overline{AC}=a+b \Rightarrow \overline{BE} =b；\\ 在\triangle DBE中，大角對大邊\Rightarrow
 \angle 2 > \angle 3且\angle 2+\angle 3=120^\circ \Rightarrow \angle 2>60^\circ ，故選\bbox[red,2pt]{(B)}$$

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第二部分：非選擇題 (1 ~ 2 題 )

解：
(1)$$1.9+2.9+38.6+ 4.2= 47.6 \Rightarrow 滿意者占\bbox[red,2pt]{47.6\%}$$(2)$$\cases{BMI肥胖者滿意占比為：{1.9 \over 8.6+1.8+1.9} \approx 0.15\\ BMI過重者滿意占比為：{2.9 \over 6.7+1.5+2.9} ={29\over 111} \approx 0.26\\ BMI正常者滿意占比為：{38.6 \over 23.6+7.4+38.6 } \approx 0.55\\ BMI過輕者滿意占比為：{4.2 \over 1.6+1.2+ 4.2}={3\over 5} = 0.6\\} \Rightarrow 滿意度最高的應該是BMI過輕者
\\ 因此，曉玫的結論\bbox[red, 2pt]{不正確}$$

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解：

(1)

$$\tan \angle MAO = {\overline{MO} \over \overline{MA}} = {5 \over 10\sqrt 3 \div 2}={1\over \sqrt 3} \Rightarrow \angle MAO=\bbox[red, 2pt]{30^\circ}$$ (2)$$\cases{\triangle ANO \cong \triangle AMO \\ \triangle OMA \cong \triangle OMB } \Rightarrow \angle BOM =\angle MOA =\angle AON = 60^\circ \Rightarrow \triangle BON=180^\circ
\\\Rightarrow B點在圓心O的正方上，又B點經旋轉後到地面，也就是旋轉角度為\angle AOB = 120^\circ = {2\over 3}\pi \\因此\cases{ \overline{N'P} = \stackrel{\huge{\frown}}{NP} = r\theta = 5\times {2\over 3}\pi \approx 10.47\\ \overline{AM}= 5\sqrt 3= 5\times 1.732 =8.66 }\Rightarrow \bbox[red, 2pt]{\overline{N'P} > \overline{AM}}$$

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因應新冠肺炎，109年會考多考一次
解題僅供參考