Difference between revisions of "1994 AHSME Problems/Problem 8"
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<math> \textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196 </math> | <math> \textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196 </math> | ||
==Solution== | ==Solution== | ||
| + | Since the perimeter is <math>56</math> and all of the sides are congruent, the length of each side is <math>2</math>. We break the figure into squares as shown below. | ||
| + | |||
| + | <asy> | ||
| + | unitsize(0.8cm); | ||
| + | draw(shift(3,4)*((0,-1)--(1,-1))); | ||
| + | draw(shift(3,4)*((-1,-2)--(2,-2))); | ||
| + | draw(shift(3,4)*((-2,-3)--(3,-3))); | ||
| + | draw(shift(3,4)*((-2,-4)--(3,-4))); | ||
| + | draw(shift(3,4)*((-1,-5)--(2,-5))); | ||
| + | draw(shift(3,4)*((0,-6)--(1,-6))); | ||
| + | draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); | ||
| + | draw((1,0)--(1,1)); | ||
| + | draw((2,-1)--(2,2)); | ||
| + | draw((3,-2)--(3,3)); | ||
| + | draw((4,-2)--(4,3)); | ||
| + | draw((5,-1)--(5,2)); | ||
| + | draw((6,0)--(6,1)); | ||
| + | </asy> | ||
| + | |||
| + | We see that there are a total of <math>2(1+3+5)+7=25</math> squares with side length <math>2</math>. Therefore, the total area is <math>4\cdot 25=\boxed{\textbf{(C) }100.}</math> | ||
| + | |||
| + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME box|year=1994|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:27, 9 January 2021
Problem
In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is 
. The area of the region bounded by the polygon is
![[asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy]](http://latex.artofproblemsolving.com/0/c/8/0c884bd4c065e2bbeb061e2e24e3dd8e906e0fde.png)
Solution
Since the perimeter is and all of the sides are congruent, the length of each side is 
. We break the figure into squares as shown below.
![[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4))); draw(shift(3,4)*((-1,-5)--(2,-5))); draw(shift(3,4)*((0,-6)--(1,-6))); draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,-1)--(2,2)); draw((3,-2)--(3,3)); draw((4,-2)--(4,3)); draw((5,-1)--(5,2)); draw((6,0)--(6,1)); [/asy]](http://latex.artofproblemsolving.com/4/6/a/46a2afa74341f58fcb0fb036689a6e56cae41bee.png)
We see that there are a total of 
squares with side length . Therefore, the total area is 
--Solution by TheMaskedMagician
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
