Difference between revisions of "1964 AHSME Problems/Problem 38"
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<math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | 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== | ||
+ | {{AHSME 40p box|year=1964|num-b=37|num-a=39}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 17:17, 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 | |
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All AHSME Problems and Solutions |
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