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2019年4月1日 星期一

101學年度國中運動績優生甄試--數學科詳解


101 學年度國民中學運動成績優良學生 
升學輔導甄試學科考試 數學科 試題

  :$$a\div \left( -3\frac { 1 }{ 5 } \right) =a\div \left( -\frac { 16 }{ 5 } \right) =a\times \left( -\frac { 5 }{ 16 } \right) =a\times b\Rightarrow b=-\frac { 5 }{ 16 } ,故選\bbox[red,2pt]{(B)}$$


:$$\left( 4\times { 10 }^{ 9 } \right) \div \left( 8\times { 10 }^{ -5 } \right) =\frac { 4\times { 10 }^{ 9 } }{ 8\times { 10 }^{ -5 } } =\frac { 4 }{ 8 } \times \frac { { 10 }^{ 9 } }{ { 10 }^{ -5 } } =\frac { 1 }{ 2 } \times { 10 }^{ 14 }=\frac { 10 }{ 2 } \times { 10 }^{ 13 }=5\times { 10 }^{ 13 },故選\bbox[red,2pt]{(B)}$$



(A) 1不是質數
(B) \(4=2\times 2\)不是質數
(C) \(91=13\times 7\)不是質數
故選\(\bbox[red,2pt]{(D)}\)



:$$0.6x-1相當於六折後再少一元,故選\bbox[red,2pt]{(B)}$$


:$$\begin{cases} a+3=0 \\ 2b-5=0 \end{cases}\Rightarrow \begin{cases} a=-3 \\ b=5/2 \end{cases}\Rightarrow \begin{cases} a+2b=-3+5=2\neq 3 \\ 常數項2a-4b+9=-6-10+9=-7 \end{cases},故選\bbox[red,2pt]{(C)}$$


:$$x^{ 2 }-2x-1=0\Rightarrow x^{ 2 }-2x+1=2\Rightarrow { \left( x-1 \right)  }^{ 2 }=2\Rightarrow x-1=\pm \sqrt { 2 } \Rightarrow x=1\pm \sqrt { 2 } ,故選\bbox[red,2pt]{(B)}$$



:$$\left( 3x^{ 2 }-2x+1 \right) -A=-3x^{ 2 }+2x-1\Rightarrow A=\left( 3x^{ 2 }-2x+1 \right) -\left( -3x^{ 2 }+2x-1 \right) \\=6x^{ 2 }-4x+2,故選\bbox[red,2pt]{(A)}$$



:$$a_{16}-a_1=(a_1+15d)-a_1=15d=60\Rightarrow d=4,故選\bbox[red,2pt]{(A)}$$



:$$\begin{cases} \angle A+\angle B=180° \\ \angle A+\angle D=180° \end{cases}\Rightarrow \begin{cases} 5y-2+4x=180° \\ 5y-2+32=180° \end{cases}\Rightarrow \begin{cases} 5y+4x=182° \\ 5y=150° \end{cases}\Rightarrow \begin{cases} x=8° \\ y=30° \end{cases}\\ \Rightarrow x+y=8°+30°=38°,故選\bbox[red,2pt]{(D)}$$



:$$(x,y)=(1,5), (2,6),\dots\Rightarrow y=x+4,故選\bbox[red,2pt]{(B)}$$


矩形ABCD的長寬比為10:6=5:3,各選項斜線矩形的長寬如下:
(A) (10-2):(6-2)=8:4=2:1
(B) (10-1-1):(6-1-1)=8:4=2:1
(C) 6:(10-6.4)=6:3.6=60:36=5:3
(D) 6:(10-5)=6:5
故選\(\bbox[red,2pt]{(C)}\)



:$$G為\triangle DFE 的重心\Rightarrow \frac{\overline{DG}}{\overline{GK}}=\frac{3a-1}{a+1}=\frac{2}{1}\Rightarrow 2a+2=3a-1\Rightarrow a=3,故選\bbox[red,2pt]{(C)}$$




由上圖可知O至三線段\(\overline{AB}, \overline{BC}及 \overline{CA}\)的距離都相等,因此O為內心故選\(\bbox[red,2pt]{(A)}\)




$$3k+2k+7k+6k=360\Rightarrow 18k=360\Rightarrow k=20\Rightarrow \angle ADB+\angle ACD =(3k+6k)\div 2\\ =(60+120)\div 2=90,故選\bbox[red,2pt]{(B)}$$




:$$最初位置是a\Rightarrow a+5-7+4=10 \Rightarrow a+2=10 \Rightarrow a=8,故選\bbox[red,2pt]{(C)}$$



$$\frac{98}{99}>\frac{97}{98}>\frac{96}{97}\Rightarrow -\frac{98}{99}<-\frac{97}{98}<-\frac{96}{97},故選\bbox[red,2pt]{(B)}$$


:$$五百萬分之一=\frac{1}{5000000}=\frac{1}{5\times 10^6}=\frac{1}{5}\times 10^{-6}= 0.2\times 10^{-6}=2\times 10^{-7},故選\bbox[red,2pt]{(C)}$$


:$$7+(-2)\times[12-(-3)\times 5]\div 9=7+(-2)\times[12-(-15)]\div 9\\= 7+(-2)\times[27]\div 9= 7+(-54)\div 9=7-6=1,故選\bbox[red,2pt]{(B)}$$


:$$ \frac{3a-5}{4}=a-3\Rightarrow 3a-5=4a-12\Rightarrow a=7,故選\bbox[red,2pt]{(A)}$$


:\((-4)^6>0\),只有(D)是正數,其它均為負數,故選\(\bbox[red,2pt]{(D)}\)

:$$2x-1>5>11-3x\Rightarrow 2x-1>5且5>11-3x \Rightarrow x>3且x>2 \Rightarrow x>3,故選\bbox[red,2pt]{(A)}$$




P落在\(\overline{BC}\)邊上\(\Rightarrow \frac{1}{3}+t=1\Rightarrow t=\frac{2}{3}\),故選\(\bbox[red,2pt]{(B)}\)


:$$ x:y=3:1\Rightarrow x=3k,y=k\Rightarrow 5x+2y=15k+2k=17k=34\Rightarrow k=2 \Rightarrow x=6,y=2 \\\Rightarrow (2x-1):(2y+1)=(12-1):(4+1)=11:5,故選\bbox[red,2pt]{(C)}$$


:$$|x|>|y|\Rightarrow \sqrt{(x-y)^2}+\sqrt{(x+y)^2}=|x-y|+|x+y|=y-x-(x+y)=-2x,故選\bbox[red,2pt]{(B)}$$


:$$m是1+2x-x^2=0的根\Rightarrow 1+2m-m^2=0\Rightarrow m^2-2m=1 \Rightarrow m^2-2m+10=11,故選\bbox[red,2pt]{(A)}$$


:$$a^2+2a+1=(a+1)^2\Rightarrow \overline{AR}=\overline{RQ}=\overline{QP}= \overline{AP}=a+1\\令\overline{AD}=b\Rightarrow b^2-(a+1)^2=10a+35 \Rightarrow b^2=(a+1)^2+10a+35=(a+6)^2 \Rightarrow b=a+6\\\Rightarrow \overline{DR}=b-a=a+6-(a+1)=5,故選\bbox[red,2pt]{(A)}$$



$$L//M\Rightarrow 15^\circ+\angle 4+\angle 5+\angle 2=180^\circ \Rightarrow 15^\circ+\angle 4+\angle 5+30^\circ=180^\circ\\ \Rightarrow \angle 4+\angle 5=180-30-15=135^\circ = \angle 1,故選\bbox[red,2pt]{(D)}$$


:$$首項a_1\Rightarrow 末項a_6=a_1-10\Rightarrow 數列和84=(a_1+a_1-10)\times 6\div 2 \Rightarrow 2a_1-10=28\\ \Rightarrow a_1=19,故選\bbox[red,2pt]{(C)}$$




$$\overline{AD}=a \Rightarrow \overline{BD}=8-a \Rightarrow \overline{DC}=\overline{BD}=8-a (\because L是中垂線)\\在直角\triangle ADC中: a^2+6^2=(8-a)^2\Rightarrow 16a=28 \Rightarrow a=\frac{7}{4},故選\bbox[red,2pt]{(A)}$$




I為內心,也是內切圓的圓心,假設內切圓半徑為\(r\),且與三角形各邊的交點為D、E、F,見上圖;
令\(\overline{AD}=\overline{AE}=a, \overline{AD}=\overline{AE}=a, \overline{BD} =\overline{BF}=b,  \overline{CE}=\overline{CF}=c\);由題意知\(\overline{AB}>\overline{BC} >\overline{AC}\),即\(a+b>b+c>c+a \Rightarrow a>c且b>a,即b>a>c\);
又\(r^2=\overline{AI}^2-a^2=\overline{BI}^2-b^2=\overline{CI}^2-c^2 \Rightarrow \overline{BI} > \overline{AI} >  \overline{CI}\),故選\(\bbox[red,2pt]{(D)}\)


:$$\frac{\triangle AED}{\triangle BDE}=\frac{\overline{AE}}{\overline{BE}} \Rightarrow \frac{\triangle AED}{12}=\frac{6}{4}\Rightarrow \triangle AED=18 \Rightarrow \triangle ABD=12+18=30\\ \frac{\triangle ABD}{\triangle ADC}=\frac{\overline{BD}}{\overline{DC}} \Rightarrow \frac{30}{\triangle ADC}=\frac{6}{9}\Rightarrow \triangle ADC=45 \Rightarrow \triangle ABC=30+45=75,故選\bbox[red,2pt]{(D)}$$




\(\overline{BC}\)的中垂線為Y軸,因此可假設\(P(0,a)\);$$\overline{PA}=\overline{PC} \Rightarrow (a-4)^2=(-3)^2+a^2 \Rightarrow 8a=7\Rightarrow a=\frac{7}{8},故選\bbox[red,2pt]{(D)}$$





$$直角\triangle ABC\Rightarrow \overline{BC}^2=\overline{AB}^2+\overline{AC}^2 = 8^2+15^2 =289 \Rightarrow \overline{BC}=17\\\triangle ABC面積=\overline{AC}\times\overline{AB}\div 2 = 15\times 8\div 2=60=\triangle AOC+\triangle AOB+\triangle BOC = \\ (\overline{AC}+\overline{AB} +\overline{BC})\times r \div 2 = (8+15+17)\times r \div 2 =20r \Rightarrow 20r = 60 \Rightarrow r=3\\ 直角\triangle ADC\Rightarrow \overline{CD}^2=\overline{AD}^2+\overline{AC}^2 = 3^2+15^2=234 \Rightarrow \overline{CD} = \sqrt{234} = 3\sqrt{26},故選\bbox[red,2pt]{(B)}$$
:$$\triangle CDF=\overline{DC}\times\overline{EF}\div 2=\frac{1}{2}ab\\
\triangle BCF=\overline{BC}\times\overline{GF}\div 2=\frac{1}{2}ab=\triangle CDF\\
\Rightarrow \triangle CDF=\triangle BCF\Rightarrow \triangle DHC+\triangle CHF=\triangle BFH+\triangle CHF \\\Rightarrow \triangle DHC=\triangle BFH \Rightarrow \triangle DBF=\triangle DHB+\triangle BFH = \triangle DHB+\triangle DHC\\ = \triangle DBC=\frac{1}{2}a^2,故選\bbox[red,2pt]{(D)}$$





邊長為12的正三角形面積為\(12\times 6\sqrt{3}\div 2=36\sqrt{3}\)
正三角形的內心與重心為同一點,因此\(\overline{AI}:\overline{AG}=2:3 \Rightarrow \frac{\triangle ADE}{\triangle ABC}=\frac{\overline{AI}^2}{\overline{AG}^2}=\frac{4}{9} \Rightarrow \triangle ADE=\frac{4}{9}\times 36\sqrt{3} = 16\sqrt{3}\)故選\(\bbox[red,2pt]{(B)}\)



燃燒\(2\frac{3}{4}\)小時代表還有\(4-2\frac{3}{4}=1\frac{1}{4}\)小時的蠟燭剩下,剩下占全部的\( 1\frac{1}{4}\div 4 =\frac{5}{16}\),故選\(\bbox[red,2pt]{(C)}\)



假設衣服原價為\(a\)元,小丸子、小玉及美環分別花了\(0.9a, 0.7a,0.5a\)的錢買衣服;
小丸子比小玉多花了240元,即\(0.9a=0.7a+240 \Rightarrow a=240\div 0.2=1200\),因此美環花了\(1200\times 0.5=600\)元買衣服,故選\(\bbox[red,2pt]{(B)}\)。


:$$(x^2+2x)+(x+2)=x(x+2)+(x+2)=(x+1)(x+2),故選\bbox[red,2pt]{(D)} $$




作\(\overline{DF}\bot\overline{AB}\)及\(\overline{CE}\bot\overline{AB}\),如上圖。
由於ABCD為等腰梯形,因此\(\overline{BC}=\overline{AD}=15\),並令\(\overline{AF}=\overline{EB}=a\),則\(\overline{FE}=25-2a=\overline{CD}\);
\(\triangle ABC\)為直角三角形\(\Rightarrow \overline{AB}^2= \overline{AC}^2+\overline{BC}^2 \Rightarrow 25^2=\overline{AC}^2+15^2 \Rightarrow \overline{AC}=20\);
又三角形ABC面積\(=\overline{AC}\times\overline{BC}\div 2=\overline{AB}\times\overline{CE}\div 2 \Rightarrow 25\overline{CE}=20\times 15 \Rightarrow \overline{CE}=12\);
在直角三角形CEB中, \(\overline{CB}^2= \overline{CE}^2+\overline{EB}^2 \Rightarrow 15^2=12^2+a^2 \Rightarrow a=9\);
因此梯形面積\(=((25-2a)+25)\times 12\div 2 =32\times 6=192\),故選\(\bbox[red,2pt]{(B)}\)



$$\overline{DE}//\overline{FG}\Rightarrow \frac{\overline{DE}}{\overline{FG}} = \frac{\overline{AD}}{\overline{AF}} =\frac{3}{3+2}=\frac{3}{5}\\ \overline{DE}//\overline{BC}\Rightarrow \frac{\overline{DE}}{\overline{BC}} = \frac{\overline{AD}}{\overline{AB}} =\frac{3}{3+2+1}=\frac{3}{6}\\由上述二式可知: \overline{DE}:\overline{FG}:\overline{BC}=3:5:6,故選\bbox[red,2pt]{(D)}$$--END--

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