2000 AMC 8 Problems/Problem 6

From $1908$ to $1940$, a house could be mail-ordered from the Sears catalog. Shown here is a floor plan for the Shelburne No. $1$ model which was sold during the $1920$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. $1$ model? Express your answer to the nearest square foot.

Solution 1

The side of the large square is $1 + 3 + 1 = 5$, so the area of the large square is $5^2 = 25$.

The area of the middle square is $3^2$, and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$.

Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$. This is the area of the two congruent L-shaped regions.

So the area of one L-shaped region is $\frac{14}{2} = 7$, and the answer is $\boxed{A}$

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 $3 + 4 = 7$, and the answer is $\boxed{A}$.

Solution 3

Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper. When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{A}$ 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 $4\cdot4=16$. In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{\text{(A)}\ 7}$

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
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

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