MATHS 問題

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