Difference between revisions of "2001 AMC 8 Problems/Problem 7"
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==Solution 3== | ==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> | + | 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> $\hspace{0.2cm} |
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==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 18:58, 26 December 2023
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.
What is the number of square inches in the area of the small kite?
interesting problem-sleepypuppy
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
$\hspace{0.2cm}
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 |
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