Difference between revisions of "1964 AHSME Problems/Problem 38"
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Plugging in the numbers given in the problem, we get | Plugging in the numbers given in the problem, we get | ||
| − | <cmath>\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2</cmath> | + | <cmath>\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2}</cmath> |
| + | |||
| + | Solving, | ||
| + | <cmath>7=\sqrt{2(16)+2(49)-QR^2}</cmath> | ||
| + | <cmath>49=32+98-QR^2</cmath> | ||
| + | <cmath>QR^2=81</cmath> | ||
| + | <cmath>QR=9</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 18:16, 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
![\[\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2}\]](http://latex.artofproblemsolving.com/5/0/b/50b4be617cd04dd5a1650b10c228240e6446502f.png)
Solving,
![\[7=\sqrt{2(16)+2(49)-QR^2}\]](http://latex.artofproblemsolving.com/0/5/9/05927162753a788c47de2af6c1ef9e12200c87dd.png)
![\[49=32+98-QR^2\]](http://latex.artofproblemsolving.com/a/6/c/a6cfed95c7269cdff404d5ef0044a55c141dd21a.png)
![\[QR^2=81\]](http://latex.artofproblemsolving.com/e/4/e/e4e8f61fa18acb795618aae90aac7d1b4aa47137.png)
![\[QR=9\]](http://latex.artofproblemsolving.com/1/8/4/18424c185f7f0735d341629b6705ab71181ec9e3.png)
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. 
