Difference between revisions of "1964 AHSME Problems/Problem 38"
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==Solution== | ==Solution== | ||
| + | By the [[Median Formula]], <math>PM = \frac12\sqrt{2PQ^2+2PR^2-QR^2}</math> | ||
| + | Plugging in the numbers given in the problem, we get | ||
| + | <cmath>\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 18:14, 9 March 2020
Problem
The sides 
and 
of triangle 
are respectively of lengths 
inches, and 
inches. The median 
is 
inches. Then 
, in inches, is:
Solution
By the Median Formula, 
Plugging in the numbers given in the problem, we get
\[\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2\] (Error compiling LaTeX. Unknown error_msg)
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 37 |
Followed by Problem 39 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
