2010 AMC 10A Problems/Problem 2
Contents
Problem 2
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?
Solution 1
Let the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is . Thus, the length of the rectangle is times large as the width. The answer is .
Solution 2
We can say the smaller squares area is , so of the area of the larger square is so the large squares are is , so each side is so length is and the width is so
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.