Difference between revisions of "1980 AHSME Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | + | 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>. | |
| − | |||
| − | <math> \angle GDA=360^\circ-90^\circ-60^\circ-90^\circ=120^\circ | ||
== 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}} | ||
Revision as of 01:25, 14 July 2025
Problem
In the adjoining figure, 
is an equilateral triangle and 
and 
are squares. The measure of 
is
![[asy] 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]](http://latex.artofproblemsolving.com/5/b/7/5b7c84bc2a28d8893ad588ced814f2d8e94bd60d.png)
Solution
Since and are squares, 
. Since is equilateral, 
. So 
.
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. 
