Difference between revisions of "2007 AMC 8 Problems/Problem 23"
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| + | == Problem == | ||
What is the area of the shaded pinwheel shown in the <math>5 \times 5</math> grid? | What is the area of the shaded pinwheel shown in the <math>5 \times 5</math> grid? | ||
| + | |||
<asy> | <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); | 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); | ||
| Line 6: | Line 8: | ||
draw((i,0)--(i,5)); | draw((i,0)--(i,5)); | ||
draw((0,i)--(5,i)); | draw((0,i)--(5,i)); | ||
| − | }</asy> | + | } |
| + | </asy> | ||
<math> \textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12 </math> | <math> \textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12 </math> | ||
| − | ==See Also== | + | == Solution 1 == |
| + | The area of the square around the pinwheel is 25. The area of the pinwheel is equal to <math>\text{the square } - \text{ the white space.}</math> 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}}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | The pinwheel is composed of <math>8</math> congruent obtuse triangles whose base measures length <math>1</math> and height measures length <math>1.5</math>. Using the area formula for triangles, the pinwheel has an area of | ||
| + | <cmath>8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.</cmath> | ||
| + | |||
| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/abSgjn4Qs34?t=748 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/CGpR3VKeVV0 | ||
| + | |||
| + | == 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}} | ||
Latest revision as of 13:17, 29 October 2024
Contents
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/b/b/1/bb1fcead3fb785c17349c43a31ed1e90bfc319e0.png)
Solution 1
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
The pinwheel is composed of 
congruent obtuse triangles whose base measures length 
and height measures length 
. Using the area formula for triangles, the pinwheel has an area of
![\[8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.\]](http://latex.artofproblemsolving.com/d/4/b/d4b79b63d107849d97e03960860f1207ef5228d8.png)
Video Solution
https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=748
~ pi_is_3.14
Video Solution by WhyMath
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. 
