Difference between revisions of "2001 AMC 8 Problems/Problem 7"
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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. | 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. | ||
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What is the number of square inches in the area of the small kite? | What is the number of square inches in the area of the small kite? | ||
Revision as of 19:50, 11 May 2020
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] 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/f/1/f/f1f9469db4540c2a66686ef837e33ea3746e6e8a.png)
What is the number of square inches in the area of the small kite?
Solution 1
The area of a kite is half the product of its diagonals. The diagonals have lengths of 
and 
, so the area is 
.
Solution 2
Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or .
Solution 3
Pick's Theorem states: ![\[\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1\]](http://latex.artofproblemsolving.com/4/7/1/471200f87ac241b1dba6bf5d692289a7f94ca352.png)
as the area of a figure on a grid. Counting, we see there are 
boundary points and 
interior points. Therefore, we have ![\[\frac{4}{2}+20-1\implies 20+1\implies 21.\]](http://latex.artofproblemsolving.com/f/3/2/f32713ea7de2d561e2042aaaf9adf9e3b8f99714.png)
Hence, the answer is 
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
