Difference between revisions of "1973 AHSME Problems/Problem 32"
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==See Also== | ==See Also== | ||
| − | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=31|num-a=33}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 14:04, 20 February 2020
Problem
The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length 
is
Solution
![[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label("6",(4.5,2.598,0),SW); label("$\sqrt{15}$",(4.5,0.866,0.866),N); currentprojection=orthographic(1/6,1/2,1/3); [/asy]](http://latex.artofproblemsolving.com/d/9/1/d9124a410a03d7559abebfc6a428d8cfb2d5f34d.png)
Draw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is 
.
![[asy] draw((0,1.732)--(0,0)--(3.464,0),dotted); draw((0,1.732)--(3.464,0)); label("$2\sqrt{3}$",(1.732,0),S); label("$\sqrt{15}$",(1.732,0.866),NE); [/asy]](http://latex.artofproblemsolving.com/9/6/5/965a47a13fb897421cd3a9a653267ffc5ed17e55.png)
Using the Pythagorean Theorem, the length of the altitude is 
, so the volume of the triangular pyramid is 
.
See Also
| 1973 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 33 | |
| 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 | ||
