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MATHS 問題
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因為d MATHS... 顯示更多 因為d MATHS 有圖////打吾到出黎.... 所以我scan左入部腦到..... 網址:::::::: 1.) http://i99.photobucket.com/albums/l282/vicki_422/1.jpg 2.) http://i99.photobucket.com/albums/l282/vicki_422/2.jpg 3.) http://i99.photobucket.com/albums/l282/vicki_422/3.jpg 4.) http://i99.photobucket.com/albums/l282/vicki_422/4.jpg 5.) http://i99.photobucket.com/albums/l282/vicki_422/5.jpg 吾該曬..... 更新: 第2題剩係要個answer就得la....
1a). (2πr1)^2-(2πr2)^2 =4π^2[(r1)^2-(r2)^2] b). (2πr1-2πr2)^2 =[2π(r1-r2)]^2 c). (2πr1)^2+(2πr2)^2 =4π^2[(r1)^2+(r2)^2] 3b). By the properties of rectangle, the diagonals meet at the centre. Therefore, EF=25/2=12.5. Since E is the centre and EF perpendicular to BC, then F is the mid-point of BC. Consider the triangle CEF, CF=10, EF=12.5. tan(angle CEF)=10/12.5. EF bisect angle CEB, therefore, angle CEB=2xangle CEF. Angle AED=angle CEB (vertical opposite angle) Angle CED=angle BEA (vertical opposite angle) 4a). Length of AC: Angle CBA= 80+70 (external ange) AC^2=7^2+3^2-2x3x7xcos(angle CBA) (cosine rule) Length of BD: BD/(sin80)=7/(sin70) (sine rule) b). Draw a point E, so that CE perpendicular to BD. Angle CBE=180-80-70=30 sin(30)=CE/7 CE=7/2 Therefore, the area of triangle ACD is ((3+BD)xCE)/2 For question 5, it is a bit hard to explain in words. I hope you can understnad what I mean. Firstly, draw a typical parallelogram, then draw a vertical line from one of the corner. So that, you can get a right angle triangle and a trapezium. The vertical line will then become the height of both the triangle and trapezium. Now, let the height be y and the base of the triangle be x (which means one of the bases of trapezium is 15-x) Area of the triangle=(xy)/2 Area of the trapezium=[(15)+(15-x)]y/2 Area of the parallelogram=area of triangle +area of trapezium 80=[(xy)/2]+[(15)+(15-x)]y/2 80=(xy+30y-xy)/2 16/3=y One of the angle can be found by using: sin(angle A)=y/6 Then, using the interior angles of parallel lines. 180-angle A=angle B
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MATHS 問題
發問:
因為d MATHS... 顯示更多 因為d MATHS 有圖////打吾到出黎.... 所以我scan左入部腦到..... 網址:::::::: 1.) http://i99.photobucket.com/albums/l282/vicki_422/1.jpg 2.) http://i99.photobucket.com/albums/l282/vicki_422/2.jpg 3.) http://i99.photobucket.com/albums/l282/vicki_422/3.jpg 4.) http://i99.photobucket.com/albums/l282/vicki_422/4.jpg 5.) http://i99.photobucket.com/albums/l282/vicki_422/5.jpg 吾該曬..... 更新: 第2題剩係要個answer就得la....
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最佳解答:1a). (2πr1)^2-(2πr2)^2 =4π^2[(r1)^2-(r2)^2] b). (2πr1-2πr2)^2 =[2π(r1-r2)]^2 c). (2πr1)^2+(2πr2)^2 =4π^2[(r1)^2+(r2)^2] 3b). By the properties of rectangle, the diagonals meet at the centre. Therefore, EF=25/2=12.5. Since E is the centre and EF perpendicular to BC, then F is the mid-point of BC. Consider the triangle CEF, CF=10, EF=12.5. tan(angle CEF)=10/12.5. EF bisect angle CEB, therefore, angle CEB=2xangle CEF. Angle AED=angle CEB (vertical opposite angle) Angle CED=angle BEA (vertical opposite angle) 4a). Length of AC: Angle CBA= 80+70 (external ange) AC^2=7^2+3^2-2x3x7xcos(angle CBA) (cosine rule) Length of BD: BD/(sin80)=7/(sin70) (sine rule) b). Draw a point E, so that CE perpendicular to BD. Angle CBE=180-80-70=30 sin(30)=CE/7 CE=7/2 Therefore, the area of triangle ACD is ((3+BD)xCE)/2 For question 5, it is a bit hard to explain in words. I hope you can understnad what I mean. Firstly, draw a typical parallelogram, then draw a vertical line from one of the corner. So that, you can get a right angle triangle and a trapezium. The vertical line will then become the height of both the triangle and trapezium. Now, let the height be y and the base of the triangle be x (which means one of the bases of trapezium is 15-x) Area of the triangle=(xy)/2 Area of the trapezium=[(15)+(15-x)]y/2 Area of the parallelogram=area of triangle +area of trapezium 80=[(xy)/2]+[(15)+(15-x)]y/2 80=(xy+30y-xy)/2 16/3=y One of the angle can be found by using: sin(angle A)=y/6 Then, using the interior angles of parallel lines. 180-angle A=angle B
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