Difference between revisions of "2003 AMC 12A Problems/Problem 14"
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Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is <math>2\sqrt{3}</math>. | Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is <math>2\sqrt{3}</math>. | ||
− | The diagonal of the square | + | The diagonal of the square KNML will then be <math>4+4\sqrt{3}</math>. From here there are 2 ways to proceed: |
First: Since the diagonal is <math>4+4\sqrt{3}</math>, the side length is <math>\frac{4+4\sqrt{3}}{\sqrt{2}}</math>, and the area is thus <math>\frac{16+48+32\sqrt{3}}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>. | First: Since the diagonal is <math>4+4\sqrt{3}</math>, the side length is <math>\frac{4+4\sqrt{3}}{\sqrt{2}}</math>, and the area is thus <math>\frac{16+48+32\sqrt{3}}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>. |
Revision as of 18:33, 3 February 2017
Problem
Points and lie in the plane of the square such that , , , and are equilateral triangles. If has an area of 16, find the area of .
Solution
Solution 1
Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is .
The diagonal of the square KNML will then be . From here there are 2 ways to proceed:
First: Since the diagonal is , the side length is , and the area is thus .
Solution 2
Since a square is a rhombus, the area of the square is , where and are the diagonals of the rhombus. Since the diagonal is , the area is .
See Also
2003 AMC 12A (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 12 Problems and Solutions |
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