Difference between revisions of "2011 AMC 10B Problems/Problem 9"
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-Solution by Joeya | -Solution by Joeya | ||
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+ | ==Remark (slightly more vigorous than Solution 3)== | ||
+ | The ratio of the areas is equal to twice the ratio of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is <math>\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\sqrt{3}.</math> By similarity, <math>\frac{ED}{DB}=\frac{\sqrt{3}}{DB}=\frac{3}{4}</math>, so solving for DB, we get <math>\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}</math>. | ||
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+ | ~JH. L | ||
== See Also== | == See Also== |
Latest revision as of 05:38, 25 June 2022
Contents
Problem
The area of is one third of the area of . Segment is perpendicular to segment . What is ?
Solution 1
by AA Similarity. Therefore . Find the areas of the triangles. The area of is one third of the area of .
Solution 2
by AA Similarity. Since the area of is of and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus, In order to scale the sides of ED and DB to make (since the ratios of sides are the same), we take the square root of to scale each side by the same amount.
Thus and the answer is
Solution 3 (Shortcut)
The ratio of the areas of and is , meaning the ratio of the sides is . The only answer choice involving is .
-Solution by Joeya
Remark (slightly more vigorous than Solution 3)
The ratio of the areas is equal to twice the ratio of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is By similarity, , so solving for DB, we get .
~JH. L
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
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