Difference between revisions of "2001 AMC 8 Problems/Problem 7"
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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 <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>. | 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 <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>. | ||
+ | ==Solution 3== | ||
+ | Pick's Theorem states: <cmath>\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1</cmath> as the area of a figure on a grid. Counting, we see there are <math>4</math> boundary points and <math>20</math> interior points. Therefore, we have <cmath>\frac{4}{2}+20-1\implies 20+1\implies 21.</cmath> Hence, the answer is <math>\boxed{\text{(A)}}</math> <math>\hspace{0.2cm} \square</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=6|num-a=8}} | {{AMC8 box|year=2001|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:29, 27 March 2019
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.
Contents
[hide]Problem
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: as the area of a figure on a grid. Counting, we see there are boundary points and interior points. Therefore, we have 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.