Posted by Ariel on Saturday, June 12, 2010 at 2:29am.
Out of 60 review questions I struggled with these ones. Someone please help!!
1.) The length of one side of a regular polygon is half the length of a similar polygon. What is the ratio of their areas?
2.)A patchwork quilt is made up of two different sized squares. The length of the larger square is twice the length of the smaller square. How many of the smaller squares does it take to equal the area of one larger square?
3.)In any prism, the lateral edges are congruent and parallel.
4.)A measuring cup is shaped like a cylinder and has a radius of 5 cm. It is filled with milk to a height of 2 cm. What is the lateral area of the milk in the measuring cup?
5.)If a pyramid with a square base with its side measuring 6 m and its altitude of 3 m were circumscribed by a cone, what would be the ratio of the square base area to the circular base area?
a.)1 to Ï€
b.)1 to 0.5Ï€
c.)1 to 2Ï€
6.)A magician has a conical hat that has a height of 1 m and a radius of 0.3 m. The hat does not fit his head. In order for the hat to fit perfectly, the radius needs to be 0.5 m. The magician can change the size of the hat with a spell, but the spell only changes the height of the hat, not the volume. After he performs the spell, what is the new height of the magician's hat if it has the desired radius of 0.5 m.
- Geometry Help - drwls, Saturday, June 12, 2010 at 3:57am
1) For two similar shapes, the area ratio is the square of the length ratio for corresponding sides.
(1/2)^2 = __
2) Same principle as the last problem
3) I had to review the definition of prism. The ends have to be parallel, and the sides have to be parallelograms, not necessarily rectangles. The paralleogram-face requirement would make the edge lengths equal or "congruent", and parallel
4)2 pi R H = ___
5) The same as the ratio of the area of a square to that of a circumscribed circle. The height does not matter.
2/pi = 1/(0.5 pi)
6) R^2 H must remain constant
H2/H1 = (R1/R2)^2 = (3/5)^2 = 9/25
H2 = (9/25)*1 = 0.36
- Geometry Help - Ariel, Saturday, June 12, 2010 at 11:48am
- Geometry Help - Ashante, Thursday, July 7, 2011 at 9:15am
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