Difference between revisions of "2000 AMC 8 Problems/Problem 6"
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pair A,B,C,D; | pair A,B,C,D; | ||
A = (5,5); B = (5,0); C = (0,0); D = (0,5); | A = (5,5); B = (5,0); C = (0,0); D = (0,5); | ||
| − | fill((0,0)--(0,4)--(,4)--(1,1)--(4,1)--(4,0)--cycle,gray); | + | fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); |
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw((4,0)--(4,4)--(0,4)); | draw((4,0)--(4,4)--(0,4)); | ||
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Chop the entire 5 by 5 region into <math>25</math> squares like a piece of graph paper. When you draw all the lines, you can count that only <math>7</math> of the small 1 by 1 squares will be shaded, giving <math>\boxed{A}</math> as the answer. | Chop the entire 5 by 5 region into <math>25</math> squares like a piece of graph paper. When you draw all the lines, you can count that only <math>7</math> of the small 1 by 1 squares will be shaded, giving <math>\boxed{A}</math> as the answer. | ||
| − | == | + | ==Solution 4== |
In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of <math>4\cdot4=16</math>. In the top right of that 4 by 4 square is a 3 by 3 square with an area of <math>3\cdot3=9</math>. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is <math>16-9=\boxed{\text{(A)}\ 7}</math> | In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of <math>4\cdot4=16</math>. In the top right of that 4 by 4 square is a 3 by 3 square with an area of <math>3\cdot3=9</math>. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is <math>16-9=\boxed{\text{(A)}\ 7}</math> | ||
Latest revision as of 20:02, 28 July 2023
Problem
Figure 
is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded 
-shaped region is
![[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]](http://latex.artofproblemsolving.com/7/c/6/7c6ea431a00e38c9ba0b8513fa0b9a39252cfcef.png)
Solution 1
The side of the large square is 
, so the area of the large square is 
.
The area of the middle square is 
, and the sum of the areas of the two smaller squares is 
.
Thus, the big square minus the three smaller squares is 
. This is the area of the two congruent L-shaped regions.
So the area of one L-shaped region is 
, and the answer is 
Solution 2
The shaded area can be divided into two regions: one rectangle that is 1 by 3, and one rectangle that is 4 by 1. (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.) Either way, the total area of these two regions is 
, and the answer is .
Solution 3
Chop the entire 5 by 5 region into 
squares like a piece of graph paper. When you draw all the lines, you can count that only 
of the small 1 by 1 squares will be shaded, giving as the answer.
Solution 4
In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of 
. In the top right of that 4 by 4 square is a 3 by 3 square with an area of 
. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is 
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
