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 making remote request. Unknown error_msg)

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 bottom 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.