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Difference between revisions of "1980 AHSME Problems/Problem 4"
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<asy>
 
<asy>
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size(225);
 
defaultpen(linewidth(0.7)+fontsize(10));
 
defaultpen(linewidth(0.7)+fontsize(10));
 
pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150);
 
pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150);
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== Solution ==
 
== Solution ==
  
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Since <math>ABCD</math> and <math>DEFG</math> are squares, <math>\angle ADC = \angle GDE = 90^\circ</math>.  Since <math>CDE</math> is equilateral, <math>\angle CDE = 60^\circ</math>.  So <math>\angle GDA=360^\circ-90^\circ-60^\circ-90^\circ= \boxed{(\textbf{C})\ 120^\circ} </math>.
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<math> \angle GDA=360^\circ-90^\circ-60^\circ-90^\circ=120^\circ\Rightarrow\boxed{C} </math>
 
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=3|num-a=5}}
 
{{AHSME box|year=1980|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:05, 23 July 2025

Problem

In the adjoining figure, $CDE$ is an equilateral triangle and $ABCD$ and $DEFG$ are squares. The measure of $\angle GDA$ is

$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)} \ 135^\circ \qquad \text{(E)} \ 150^\circ$

[asy] size(225); defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150); draw(E--D--G--F--E--C--D--A--B--C); pair point=(0,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(-15)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G));[/asy]

Solution

Since $ABCD$ and $DEFG$ are squares, $\angle ADC = \angle GDE = 90^\circ$. Since $CDE$ is equilateral, $\angle CDE = 60^\circ$. So $\angle GDA=360^\circ-90^\circ-60^\circ-90^\circ= \boxed{(\textbf{C})\ 120^\circ}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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