2007 AMC 8 Problems/Problem 23
Contents
Problem
What is the area of the shaded pinwheel shown in the grid?
Solution
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is which is
Solution 2 (Pick's Theorem)
We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.
We can remedy this be pretending the figure is twice as big:
Now we can safely use Pick's Theorem:
However since we scaled the figure's dimensions by , we scaled its area by (since the area of similar shapes scales quadratically with the scaling factor). Therefore the area of each kite is and the area of all four kites combined is .
~ proloto
Solution 3 (area of a kite)
The area of any kite (concave OR convex) with diagonals , is . Let be the smaller diagonal and be the longer diagonal. Then by Pythagorean Theorem . Similarly, is less than half of the diagonal of the grid, or . Therefore the area of the four kites is just:
~ proloto
Video Solution
https://youtu.be/KOZBOvI9WTs -Happytwin
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=748
~ pi_is_3.14
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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