Difference between revisions of "1973 AHSME Problems/Problem 4"
Rockmanex3 (talk | contribs) (Solution to Problem 4) |
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| + | ==Problem== | ||
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| + | Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is | ||
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| + | <math> \textbf{(A)}\ 6\sqrt3\qquad\textbf{(B)}\ 8\sqrt3\qquad\textbf{(C)}\ 9\sqrt3\qquad\textbf{(D)}\ 12\sqrt3\qquad\textbf{(E)}\ 24 </math> | ||
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==Solution== | ==Solution== | ||
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==See Also== | ==See Also== | ||
| − | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=3|num-a=5}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 12:57, 20 February 2020
Problem
Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is
Solution
![[asy] fill((0,3.464)--(-6,0)--(6,0)--cycle,yellow); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-6,0)--(6,0)--(3,5.196)--(-6,0)); draw((0,3.464)--(0,0)); label("$6$",(-3,0),S); label("$6$",(3,0),S); [/asy]](http://latex.artofproblemsolving.com/e/5/6/e563c5943ade4cc11a9747016faa8e1a2496604c.png)
Note that the altitude of the shared region bisects the hypotenuse of the original two right triangles (this can be confirmed by using AAS on the two right triangles with that altitude of the shared region). By using 30-60-90 triangles, the altitude’s length is 
. The area of the shared region is 
.
See Also
| 1973 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
