Difference between revisions of "2001 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | Each diagonal of the large kite is <math> 3 </math> times the length of the corresponding diagonal of the short kite since it was made with a grid <math> 3 </math> times as long in each direction. The diagonals of the small kite are <math> 6 </math> and <math> 7 </math>, so the diagonals of the large kite are <math> 18 </math> and <math> 21 </math>, and the amount of bracing Genevieve needs is the sum of these lengths, which is <math> 39, \boxed{\text{E}} </math> | + | Each diagonal of the large kite is <math> 3 </math> times the length of the corresponding diagonal of the short kite since it was made with a grid <math> 3 </math> times as long in each direction. The diagonals of the small kite are <math> 6 </math> and <math> 7 </math>, so the diagonals of the large kite are <math> 3\cdot6=18 </math> and <math> 3\cdot7=21 </math>, and the amount of bracing Genevieve needs is the sum of these lengths, which is <math> 39, \boxed{\text{E}} </math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=7|num-a=9}} | {{AMC8 box|year=2001|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 05:05, 8 July 2024
Problem
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
![[asy] size(85); for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]](http://latex.artofproblemsolving.com/2/a/8/2a8dc1e865933bce5ace2fa416aaac621779ff0a.png)
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?
Solution
Each diagonal of the large kite is 
times the length of the corresponding diagonal of the short kite since it was made with a grid times as long in each direction. The diagonals of the small kite are 
and 
, so the diagonals of the large kite are 
and 
, and the amount of bracing Genevieve needs is the sum of these lengths, which is 
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
