Difference between revisions of "2006 AMC 12B Problems/Problem 13"
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Draw the line <math> \overline{DB}</math> to form an equilateral triangle, since <math>\angle BAD=60</math>, and line segments | Draw the line <math> \overline{DB}</math> to form an equilateral triangle, since <math>\angle BAD=60</math>, and line segments | ||
<math> \overline{AB}</math> and <math> \overline{AD}</math> are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point <math>E</math> at a right angle along line <math> \overline{DB}</math>. Call the connected point <math>P</math>, with triangles <math>DPE</math> and <math>EPB</math> being 30-60-90 triangles, meaning we can find the length of <math> \overline{ED}</math> or <math> \overline{EB}</math>. The base of <math>ABCD</math> must be <math>\sqrt{24}</math>, and half of that length must be <math>\sqrt{6}</math>(triangles <math>DPE</math> and <math>EPB</math> are congruent by <math>SSS</math>). Solving for the third length yields <math>\sqrt{8}</math>, which we square to get the answer <math>\boxed{\text{C}}</math>. | <math> \overline{AB}</math> and <math> \overline{AD}</math> are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point <math>E</math> at a right angle along line <math> \overline{DB}</math>. Call the connected point <math>P</math>, with triangles <math>DPE</math> and <math>EPB</math> being 30-60-90 triangles, meaning we can find the length of <math> \overline{ED}</math> or <math> \overline{EB}</math>. The base of <math>ABCD</math> must be <math>\sqrt{24}</math>, and half of that length must be <math>\sqrt{6}</math>(triangles <math>DPE</math> and <math>EPB</math> are congruent by <math>SSS</math>). Solving for the third length yields <math>\sqrt{8}</math>, which we square to get the answer <math>\boxed{\text{C}}</math>. | ||
+ | ===Solution 3=== | ||
+ | Draw line DB, cutting rhombus BFDE into two triangles which fit nicely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8. | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:35, 12 December 2018
Problem
Rhombus is similar to rhombus . The area of rhombus is 24, and . What is the area of rhombus ?
Solution 1
The ratio of any length on to a corresponding length on is equal to the ratio of their areas. Since , and are equilateral. , which is equal to , is the diagonal of rhombus . Therefore, . and are the longer diagonal of rhombuses and , respectively. So the ratio of their areas is or . One-third the area of is equal to . So the answer is .
Solution 2
Draw the line to form an equilateral triangle, since , and line segments and are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point at a right angle along line . Call the connected point , with triangles and being 30-60-90 triangles, meaning we can find the length of or . The base of must be , and half of that length must be (triangles and are congruent by ). Solving for the third length yields , which we square to get the answer .
Solution 3
Draw line DB, cutting rhombus BFDE into two triangles which fit nicely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8.
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 |
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