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=\boxed{D}</cmath> | ||
+ | |||
+ | -AOPS81619 | ||
==See Also== | ==See Also== |
Latest revision as of 18:17, 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
Solving,
-AOPS81619
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.