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Difference between revisions of "2001 AMC 8 Problems/Problem 8"

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==Solution==
 
==Solution==
  
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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>
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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 both height and width. 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 20:40, 2 December 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]

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?

$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39$

Solution

Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in both height and width. The diagonals of the small kite are $6$ and $7$, so the diagonals of the large kite are $3\cdot6=18$ and $3\cdot7=21$, and the amount of bracing Genevieve needs is the sum of these lengths, which is $39, \boxed{\text{E}}$

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

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