Difference between revisions of "1970 AHSME Problems/Problem 28"

(Created page with "== Problem == In triangle <math>ABC</math>, the median from vertex <math>A</math> is perpendicular to the median from vertex <math>B</math>. If the lengths of sides <math>AC</ma...")
 
(Solution)
 
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
 
<math>\fbox{A}</math>
 
<math>\fbox{A}</math>
 +
let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN
  
 
== See also ==
 
== See also ==

Latest revision as of 01:36, 19 December 2020

Problem

In triangle $ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is

$\text{(A) } \sqrt{17}\quad \text{(B) } 4\quad \text{(C) } 4\tfrac{1}{2}\quad \text{(D) } 2\sqrt{5}\quad \text{(E) } 4\tfrac{1}{4}$

Solution

$\fbox{A}$ let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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
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