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

m (→‎See also)
(Solution to Problem 15)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{D}</math>
 
  
−
== See also ==
+
<asy>
 +
pair O = (0,0);
 +
pair A = (-8.660,5);
 +
pair B = (-8.660,-5);
 +
pair M = (-8.660,0);
 +
pair D = (-8.660*0.75,5*0.75);
 +
draw(circle(O,10));
 +
dot(O);
 +
label("$O$",O,E);
 +
dot(A);
 +
label("$A$",A,NW);
 +
dot(B);
 +
label("$B$",B,SW);
 +
dot(M);
 +
label("$M$",M,SE);
 +
dot(D);
 +
label("$D$",D,NE);
 +
draw(A--B--O--A);
 +
draw(O--M);
 +
draw(M--D);
 +
</asy>
 +
 
 +
Because <math>AO = OB = AB</math>, <math>\triangle AOB</math> is an [[equilateral triangle]], and <math>\angle OAB = 60^\circ</math>.  Using 30-60-90 triangles, <math>AM = \tfrac{r}{2}</math>, <math>AD = \tfrac{r}{4}</math>, and <math>DM = \tfrac{r\sqrt{3}}{4}</math>.  Thus, the area of <math>\triangle ADM</math> is <math>\tfrac{1}{2} \cdot \tfrac{r}{4} \cdot \tfrac{r\sqrt{3}}{4} = \boxed{\textbf{(D) } \tfrac{r^2 \sqrt{3}}{32}}</math>.
 +
 
 +
== See Also ==
 
{{AHSME 35p 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}}

Latest revision as of 18:29, 21 June 2018

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

[asy] pair O = (0,0); pair A = (-8.660,5); pair B = (-8.660,-5); pair M = (-8.660,0); pair D = (-8.660*0.75,5*0.75); draw(circle(O,10)); dot(O); label("$O$",O,E); dot(A); label("$A$",A,NW); dot(B); label("$B$",B,SW); dot(M); label("$M$",M,SE); dot(D); label("$D$",D,NE); draw(A--B--O--A); draw(O--M); draw(M--D); [/asy]

Because $AO = OB = AB$, $\triangle AOB$ is an equilateral triangle, and $\angle OAB = 60^\circ$. Using 30-60-90 triangles, $AM = \tfrac{r}{2}$, $AD = \tfrac{r}{4}$, and $DM = \tfrac{r\sqrt{3}}{4}$. Thus, the area of $\triangle ADM$ is $\tfrac{1}{2} \cdot \tfrac{r}{4} \cdot \tfrac{r\sqrt{3}}{4} = \boxed{\textbf{(D) } \tfrac{r^2 \sqrt{3}}{32}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
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

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