幾何既maths題! easy到爆....只有2日! 急!!! 20分!!!
2017/07/04 14:54
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http://i86.photobucket.com/albums/k113/chy684959/DSCN2537.jpg 噤呢個網入去睇果幾條數...(8,10,11) 好少咋!! 識幾多題就做幾多題la...thx....有20分架!! 要步驟...唔該...^^
最佳解答:
hello 8 (a) AB=5 cm BC=(30-2*5)/2=10 cm EC^2+4^2=BC^2 EC^2+16=100 EC=9.17 cm (correct to 3 significant figures) (b) The area of ABE =4*3/2 =6 cm^2 The area of BEC =4*9.17/2 =18.34 cm^2 The area of ABCD =2(6+18.34) =48.7 cm^2 10 angle BCD=180-105=75 angle E =180-angle BCD =180-75 =105 angle F =180-angle E =180-105 =75 11 since angle P= angle QRS=118 a =angle P/2 =118/2=59 angle S =180-angle QRS =180-118 =62 angle Q =angle S =62 a+b+118+angle Q=360 59+b+118+62=360 b=121
其他解答:
8. AB=AD=SQRT(3^2+4^2)=5 AB+AD+BC+CD=30 BC+CD+5+5=30, BC+CD=20, BC=CD=10 BC^2=BE^2+EC^2, EC^2=10^2-4^2 ==> EC=SQRT(84) EC=9.165cm area of ABD=3x4=12cm^2 area of BCD=4x9.165=36.661cm^2 area of ABCD=12+36.661=48.661cm^2 10. FGD=ABC=105 DEF=FGD=105 EFG=180-DEF ==> EFG=180-105=75 E=105 F=75 11. QRS=SPQ ==> 2a=118 ==> a=118 / 2 = 59 PQR=RSP ==> PQR + RSP + 2 * 118 = 360 PQR = (360 - 236) / 2 = 124 / 2 = 62 a + b + 118 + 62 = 360 b = 360 - 59 - 118 - 62 = 121 a = 59 b=121|||||8a. AD^2=AE^2+DE^2 <----畢氏定理 AD^2=3^2+4^2 AD^2=9+16 AD^2=25 AD=5 AE=AD <--從圖所知 AE=5 perimeter of ABCD=30 AD+AE+CB+CD=30 CB+CD=20 asCB=CD <--從圖所知 CB=CD=20/2=10 CB^2=EC^2+BE^2 <----畢氏定理 EC^2=100-16 EC^2=84 EC=9.16515139 EC=9.165cm <--3 significant figures b. area of ABD =3*(4+4)/2 =12cm^2 area of BCD =EC*(4+4)/2 =36.661cm^2 area of ABCD =area of ABD+area of BCD =12+36.661 =48.661cm^2 <--3 significant figures -------------------------------------------------------------------- 10.( *=角 ) as ABCD is parallelogam *ADC=*ABC=105 <----對角相等 as CDE is a straight *GDE+*ADC=180 *GDE=75 as ABCD is parallelogam *F=*GDE=75 <----對角相等 as GF//DE *E+*F=180 <---當一直線cut兩條平行線,兩夾角相加=180度 *E=105 ------------------------------------------------------------------ 11. as PQRS is parallelogam *SPQ=*QRS=118 <----對角相等 as PT bisects *SPQ a=*SPQ/2 a=59 as SR//PQ a+b=180 <---當一直線cut兩條平行線,兩夾角相加=180度 b=121 -------------------------------------------------------------------------------------------------- 這樣子寫下來的步驟其實幾難明... if你唔明可以再問我.... 我可以再解釋一下^^|||||8. a) AB = BD = (AE^2+BE^2)^0.5 = 5 cm [Pythagoras Th.] AB + BC + CD + DA = 30 therefore, BC = CD = 10 cm EC = (BC^2 - BE^2)^0.5 [Pythagoras Th.] (ANSWER) b) total area = area of triangle ABD + area of triangle BCD = 0.5*BD*AE + 0.5*BD*EC (ANSWER) 10 Angle(CBA) + Angle(BAD) = Angle(CDA) + Angle(BAD) = 180 therefore Angle(CBA) = Angle(CDA) = 105 Angle(CDA) = Angle(DEF) = 105 (ANSWER) Angle(GFE) + Angle(DEF) = 180 Angle(GFE) = 180 - Angle(DEF) (ANSWER) 11 Angle(R) + Angle(S) = 180 Angle(S) = 180 - 118 = 62 Angle(STP) = Angle(S) = 62 [Base angle of isosceles triangle] b + Angle(STP) = 180 b = 180 - Angle(STP) (ANSWER) a + b = 180 a = 180 - b (ANSWER)
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幾何既maths題! easy到爆....只有2日! 急!!! 20分!!!發問:
http://i86.photobucket.com/albums/k113/chy684959/DSCN2537.jpg 噤呢個網入去睇果幾條數...(8,10,11) 好少咋!! 識幾多題就做幾多題la...thx....有20分架!! 要步驟...唔該...^^
最佳解答:
hello 8 (a) AB=5 cm BC=(30-2*5)/2=10 cm EC^2+4^2=BC^2 EC^2+16=100 EC=9.17 cm (correct to 3 significant figures) (b) The area of ABE =4*3/2 =6 cm^2 The area of BEC =4*9.17/2 =18.34 cm^2 The area of ABCD =2(6+18.34) =48.7 cm^2 10 angle BCD=180-105=75 angle E =180-angle BCD =180-75 =105 angle F =180-angle E =180-105 =75 11 since angle P= angle QRS=118 a =angle P/2 =118/2=59 angle S =180-angle QRS =180-118 =62 angle Q =angle S =62 a+b+118+angle Q=360 59+b+118+62=360 b=121
其他解答:
8. AB=AD=SQRT(3^2+4^2)=5 AB+AD+BC+CD=30 BC+CD+5+5=30, BC+CD=20, BC=CD=10 BC^2=BE^2+EC^2, EC^2=10^2-4^2 ==> EC=SQRT(84) EC=9.165cm area of ABD=3x4=12cm^2 area of BCD=4x9.165=36.661cm^2 area of ABCD=12+36.661=48.661cm^2 10. FGD=ABC=105 DEF=FGD=105 EFG=180-DEF ==> EFG=180-105=75 E=105 F=75 11. QRS=SPQ ==> 2a=118 ==> a=118 / 2 = 59 PQR=RSP ==> PQR + RSP + 2 * 118 = 360 PQR = (360 - 236) / 2 = 124 / 2 = 62 a + b + 118 + 62 = 360 b = 360 - 59 - 118 - 62 = 121 a = 59 b=121|||||8a. AD^2=AE^2+DE^2 <----畢氏定理 AD^2=3^2+4^2 AD^2=9+16 AD^2=25 AD=5 AE=AD <--從圖所知 AE=5 perimeter of ABCD=30 AD+AE+CB+CD=30 CB+CD=20 asCB=CD <--從圖所知 CB=CD=20/2=10 CB^2=EC^2+BE^2 <----畢氏定理 EC^2=100-16 EC^2=84 EC=9.16515139 EC=9.165cm <--3 significant figures b. area of ABD =3*(4+4)/2 =12cm^2 area of BCD =EC*(4+4)/2 =36.661cm^2 area of ABCD =area of ABD+area of BCD =12+36.661 =48.661cm^2 <--3 significant figures -------------------------------------------------------------------- 10.( *=角 ) as ABCD is parallelogam *ADC=*ABC=105 <----對角相等 as CDE is a straight *GDE+*ADC=180 *GDE=75 as ABCD is parallelogam *F=*GDE=75 <----對角相等 as GF//DE *E+*F=180 <---當一直線cut兩條平行線,兩夾角相加=180度 *E=105 ------------------------------------------------------------------ 11. as PQRS is parallelogam *SPQ=*QRS=118 <----對角相等 as PT bisects *SPQ a=*SPQ/2 a=59 as SR//PQ a+b=180 <---當一直線cut兩條平行線,兩夾角相加=180度 b=121 -------------------------------------------------------------------------------------------------- 這樣子寫下來的步驟其實幾難明... if你唔明可以再問我.... 我可以再解釋一下^^|||||8. a) AB = BD = (AE^2+BE^2)^0.5 = 5 cm [Pythagoras Th.] AB + BC + CD + DA = 30 therefore, BC = CD = 10 cm EC = (BC^2 - BE^2)^0.5 [Pythagoras Th.] (ANSWER) b) total area = area of triangle ABD + area of triangle BCD = 0.5*BD*AE + 0.5*BD*EC (ANSWER) 10 Angle(CBA) + Angle(BAD) = Angle(CDA) + Angle(BAD) = 180 therefore Angle(CBA) = Angle(CDA) = 105 Angle(CDA) = Angle(DEF) = 105 (ANSWER) Angle(GFE) + Angle(DEF) = 180 Angle(GFE) = 180 - Angle(DEF) (ANSWER) 11 Angle(R) + Angle(S) = 180 Angle(S) = 180 - 118 = 62 Angle(STP) = Angle(S) = 62 [Base angle of isosceles triangle] b + Angle(STP) = 180 b = 180 - Angle(STP) (ANSWER) a + b = 180 a = 180 - b (ANSWER)
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