Difference between revisions of "2000 AMC 8 Problems/Problem 6"
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==Solution 4== | ==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> | |
==See Also== | ==See Also== |
Latest revision as of 21:02, 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
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.