Difference between revisions of "2011 AMC 10B Problems/Problem 9"
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Thus <math>BD = 4 \times \frac{\sqrt(3)}{3}</math> and the answer is <math>BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}</math> | Thus <math>BD = 4 \times \frac{\sqrt(3)}{3}</math> and the answer is <math>BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}</math> | ||
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+ | ==Solution 3 (Shortcut)== | ||
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+ | The ratio of the areas of <math>\triangle</math><math>EBD</math> and <math>\triangle</math><math>ABC</math> is <math>1 : 3</math>, meaning the ratio of the sides is <math>1 : \sqrt{3}</math>. The only answer choice involving <math>\sqrt{3}</math> is <math>\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}</math> | ||
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+ | -Solution by Joeya | ||
== See Also== | == See Also== |
Revision as of 14:29, 26 January 2021
Problem
The area of is one third of the area of . Segment is perpendicular to segment . What is ?
Solution
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.