Difference between revisions of "1958 AHSME Problems/Problem 43"

(Created page with "== Problem == <math> \overline{AB}</math> is the hypotenuse of a right triangle <math> ABC</math>. Median <math> \overline{AD}</math> has length <math> 7</math> and median <math>...")
 
(Solution)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{}</math>
+
<asy>
 +
import geometry;
 +
unitsize(50);
 +
pair A = (0,0), B = (3,0), C = (0, 4);
 +
pair AB = midpoint(A--B), AC = midpoint(A--C);
 +
draw(A--B--C--A);
 +
draw(A--B, StickIntervalMarker(2, 1));
 +
draw(A--C, StickIntervalMarker(2, 2));
 +
draw(C--AB);
 +
draw(B--AC);
 +
dot(AB);
 +
dot(AC);
 +
MP("$A$", A, W);
 +
MP("$B$", B, E);
 +
MP("$C$", C, W);
 +
MP("$M$", AB, S);
 +
MP("$N$", AC, W);
 +
label("$x$", A--AB, S);
 +
label("$x$", AB--B, S);
 +
label("$y$", A--AC, SWW);
 +
label("$y$", AC--C, W);
 +
draw(rightanglemark(C, A, B));
 +
</asy>
 +
<math>\fbox{D}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 11:30, 23 February 2018

Problem

$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$. The length of $\overline{AB}$ is:

$\textbf{(A)}\ 10\qquad  \textbf{(B)}\ 5\sqrt{3}\qquad  \textbf{(C)}\ 5\sqrt{2}\qquad  \textbf{(D)}\ 2\sqrt{13}\qquad  \textbf{(E)}\ 2\sqrt{15}$

Solution

import geometry;
unitsize(50);
pair A = (0,0), B = (3,0), C = (0, 4);
pair AB = midpoint(A--B), AC = midpoint(A--C);
draw(A--B--C--A);
draw(A--B, StickIntervalMarker(2, 1));
draw(A--C, StickIntervalMarker(2, 2));
draw(C--AB);
draw(B--AC);
dot(AB);
dot(AC);
MP("$A$", A, W);
MP("$B$", B, E);
MP("$C$", C, W);
MP("$M$", AB, S);
MP("$N$", AC, W);
label("$x$", A--AB, S);
label("$x$", AB--B, S);
label("$y$", A--AC, SWW);
label("$y$", AC--C, W);
draw(rightanglemark(C, A, B));
 (Error making remote request. Unknown error_msg)

$\fbox{D}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 42
Followed by
Problem 44
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 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png