Difference between revisions of "1980 AHSME Problems/Problem 4"

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 $\angle GDA=360^\circ-90^\circ-60^\circ-90^\circ= \boxed{(\textbf{C})\ 120^\circ}$ .

1980 AHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1980 AHSME Problems/Problem 4"

Latest revision as of 02:05, 23 July 2025

Problem

Solution

See also

@@ Line 6: / Line 6: @@
 <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);