Difference between revisions of "2007 AMC 8 Problems/Problem 23"

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==Problem==
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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);
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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>
  
==Solution==
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== Solution ==
 
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>
 
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>
  
==Video Solution==
+
== Video Solution ==
 
 
 
https://youtu.be/KOZBOvI9WTs -Happytwin
 
https://youtu.be/KOZBOvI9WTs -Happytwin
  
==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}}

Latest revision as of 12:35, 19 October 2020

Problem

What is the area of the shaded pinwheel shown in the $5 \times 5$ 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]

$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$

Solution

The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ 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 $25-(15+4)$ which is $\boxed{\textbf{(B) 6}}$

Video Solution

https://youtu.be/KOZBOvI9WTs -Happytwin

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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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