2000 AMC 8 Problems/Problem 6
From 
to 
, a house could be mail-ordered from the Sears catalog. Shown here is a floor plan for the Shelburne No. 
model which was sold during the 
s. The dimensions of each room are given in feet and inches, and adjacent walls meet at right angles. In square feet, what is the area of the dining room of the Shelburne No. model? Express your answer to the nearest square foot.
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. 
