Difference between revisions of "2006 AMC 12B Problems/Problem 13"
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== Solution == | == Solution == | ||
− | + | the square of the ratio of any linear length on ABCD to the corresponding linear length on BFDE is equal to the ratio of their areas. because angle BAD=60 degrees, triangle ADB and triangle DBC are equilateral and DB is equal to the other sides of rhombus ABCD. therefore, AC=DB/2*sqrt(3)*2=DB*sqrt(3). DB and AC are the longer diagonal of rhombuses BEDF and ABCD, respectively. so the ratio of their areas is (1/sqrt(3))^2=1/3. area ABCD=24, so area BEDF=8 and the answer is C | |
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}} |
Revision as of 18:17, 15 January 2012
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Problem
Rhombus is similar to rhombus . The area of rhombus is 24, and $\angle BAD \equal{} 60^\circ$ (Error compiling LaTeX. Unknown error_msg). What is the area of rhombus ?
Solution
the square of the ratio of any linear length on ABCD to the corresponding linear length on BFDE is equal to the ratio of their areas. because angle BAD=60 degrees, triangle ADB and triangle DBC are equilateral and DB is equal to the other sides of rhombus ABCD. therefore, AC=DB/2*sqrt(3)*2=DB*sqrt(3). DB and AC are the longer diagonal of rhombuses BEDF and ABCD, respectively. so the ratio of their areas is (1/sqrt(3))^2=1/3. area ABCD=24, so area BEDF=8 and the answer is C
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |