Difference between revisions of "2007 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | The area of the square around the pinwheel is 25. The area of the pinwheel is equal to the square - the white space. | + | The area of the square around the pinwheel is 25. The area of the pinwheel is equal to the square - the white space. 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 <math>25-(15+4)</math> which is <math>\boxed{\textbf{(B) 6}\6}</math> |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=22|num-a=24}} | {{AMC8 box|year=2007|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:50, 25 July 2013
Problem
What is the area of the shaded pinwheel shown in the 
grid?
![[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); }[/asy]](http://latex.artofproblemsolving.com/f/c/c/fccd411afd0b28b0f07e081eeb90439e7ff3c2b9.png)
Solution
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to the square - the white space. 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 $\boxed{\textbf{(B) 6}\6}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
