Part 2: So you have your inverted cup, with the base of your imaginary cone the open top, so r=7/2=3.5. Imagine the cup as a triangle either side, and two rectangles in the middle. The height of the rectangles is the height of the cup (8cm) and the width is �� the width of the base of the cup (5/2=2.5) therefore the triangles must be 1cm at the base and 8cm high. Using good old Pythagoras, the square of the other side of the triangle must be 1��+8��=65, so its length is 8.06cm. What about the length of the slope on the ��extra�� cone? Imagine it��s two triangles, we need to know two sides or a side and one angle. Well, one side is �� the base of the cup (5/2=2.5), and the angle at the top must be the same as the angle as the top of the ��lower�� triangles (180�� in any triangle, and we have 90�� from the rectangle and the angle at the top of the lower triangle). Use trig to get this (7.13�� - the tangent of the angle is 1/8, and you can then find the inverse of this to work out the angle), or cheat and use
http://www.1728.com/pythgorn.htm
. We can use the same methods (to cheat, use
http://www.1728.com/trig.htm
) to determine the sloping side of the top triangle, getting 20.14. So the area of the whole cone is ��rs=3.14*3.5*(20.14+8.06)=309.92cm��. The area of the top cane is 3.14*2.5*20.14=158.10cm��, so the area of the cup excluding its base is 309.92-158.10=151.82cm��. The area of the base is ��r��=3.14*2.5*2.5=19.63, so the total area of the cup is 151.82+19.63=171.45 cm��. Probably too late for your homework, though. P.S. the funny squiggle was supposed to be "pi".