2000 AMC 8 Problems/Problem 6
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
pair A,B,C,D; 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); 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); (Error compiling LaTeX. fill((0,0)--(0,4)--(,4)--(1,1)--(4,1)--(4,0)--cycle,gray); ^ 8c6f36a8dc6fc12c492654db8beca32d5979b932.asy: 7.21: syntax error error: could not load module '8c6f36a8dc6fc12c492654db8beca32d5979b932.asy')
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.
Solutioi 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 | |
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