Difference between revisions of "1969 AHSME Problems/Problem 15"

(Created page with "== Problem == In a circle with center <math>O</math> and radius <math>r</math>, chord <math>AB</math> is drawn with length equal to <math>r</math> (units). From <math>O</math>, ...")
 
m (See also)
Line 13: Line 13:
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1969|num-b=14|num-a=16}}   
+
{{AHSME 35p box|year=1969|num-b=14|num-a=16}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:01, 30 September 2014

Problem

In a circle with center $O$ and radius $r$, chord $AB$ is drawn with length equal to $r$ (units). From $O$, a perpendicular to $AB$ meets $AB$ at $M$. From $M$ a perpendicular to $OA$ meets $OA$ at $D$. In terms of $r$ the area of triangle $MDA$, in appropriate square units, is:

$\text{(A) } \frac{3r^2}{16}\quad \text{(B) } \frac{\pi r^2}{16}\quad \text{(C) } \frac{\pi r^2\sqrt{2}}{8}\quad \text{(D) } \frac{r^2\sqrt{3}}{32}\quad \text{(E) } \frac{r^2\sqrt{6}}{48}$

Solution

$\fbox{D}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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