Difference between revisions of "1973 AHSME Problems/Problem 9"
Rockmanex3 (talk | contribs) (Solution to Problem 9) |
Made in 2016 (talk | contribs) (→See Also) |
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==See Also== | ==See Also== | ||
− | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=8|num-a=10}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Latest revision as of 12:58, 20 February 2020
Problem
In with right angle at , altitude and median trisect the right angle. If the area of is , then the area of is
Solution
Draw diagram as shown (note that and can be interchanged, but it doesn’t change the solution).
Note that because is a median, . Also, by ASA Congruency, , so . That means , and since and share an altitude, .
See Also
1973 AHSME (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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |