2002 AMC 10P Problems/Problem 14
Contents
[hide]Problem 14
The vertex of a square is at the center of square The length of a side of is and the length of a side of is Side intersects at and intersects at If angle the area of quadrilateral is
Solution 1
Draw a diagram. Split quadrilateral into and Let the perpendicular from point intersect at , and let the perpendicular from point intersect at We know because since is a square, as given, and so Since is at the center of square , By Additionally, we know so and we know so From here, we can sum the areas of and to get the area of quadrilateral Therefore,
Thus, our answer is
Solution 2
If we draw a diagram as explained by the prompt, as and are both because they are angles of the squares, and given by the question. Similar to solution 1, if we draw and perpendicular to and respectively, square with a side length of will be formed. Looking at , we will notice . Therefore and are congruent by as they both have a and a angle and . Thus, the area of quadrilateral is the same as the area of square , which equals
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.