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

Latest revision as of 23:27, 28 July 2024

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 1

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 $\boxed{\textbf{(B) 6}}$

Solution 2

The pinwheel is composed of $8$ congruent obtuse triangles whose base measures length $1$ and height measures length $1.5$ . Using the area formula for triangles, the pinwheel has an area of $8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.$

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=748

~ pi_is_3.14

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.

@@ Line 13: / Line 13: @@
 <math> \textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12 </math>
-== Solution ==
+== 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 3 (area of a kite)==
+== 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
-The area of any kite (concave OR convex) with diagonals <math>p</math>, <math>q</math> is <math>\frac{1}{2}pq</math>.  Let <math>p</math> be the smaller diagonal and <math>q</math> be the longer diagonal.  Then by Pythagorean Theorem <math>p=\sqrt{2}</math>. Similarly, <math>q</math> is <math>\sqrt{2}</math> less than half of the diagonal of the <math>5 \times 5</math> grid, or <math>q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}</math>.  Therefore the area of the four kites is just:
+<cmath>8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.</cmath>
-<cmath>A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{\textbf{(B) 6}}</cmath>
-~ proloto
 == Video Solution ==
-https://youtu.be/KOZBOvI9WTs -Happytwin
+https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin
 ==Video Solution by OmegaLearn==

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