Difference between revisions of "2007 AMC 10B Problems/Problem 21"
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<cmath>\frac{ZB}{l} = \frac{4}{5} \rightarrow ZB = \frac{4}{5}l</cmath> | <cmath>\frac{ZB}{l} = \frac{4}{5} \rightarrow ZB = \frac{4}{5}l</cmath> | ||
− | But then <math>\frac{5}{3}l+\frac{4}{5}l = CZ+ZB = CB = 4 \longrightarrow \frac{37}{15}l=4 \longrightarrow l = \frac{60}{37} \Longrightarrow \mathrm{( | + | But then <math>\frac{5}{3}l+\frac{4}{5}l = CZ+ZB = CB = 4 \longrightarrow \frac{37}{15}l=4 \longrightarrow l = \frac{60}{37} \Longrightarrow \mathrm{(B)}</math> |
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==See Also== | ==See Also== |
Revision as of 13:17, 27 July 2015
Contents
Problem
Right has and Square is inscribed in with and on on and on What is the side length of the square?
Solution 1
There are many similar triangles in the diagram, but we will only use If is the altitude from to and is the sidelength of the square, then is the altitude from to By similar triangles,
Find the length of the altitude of Since it is a right triangle, the area of is
The area can also be expressed as so
Substitute back into
Solution 2
Let be the side length of the inscribed square. Note that .
Then we can setup the following ratios:
But then
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.