Difference between revisions of "1980 AHSME Problems/Problem 4"
J314andrews (talk | contribs) |
J314andrews (talk | contribs) (Shrank diagram because it was huge.) |
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<asy> | <asy> | ||
| + | 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 == | ||
| − | + | 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}} | ||
Latest revision as of 01:05, 23 July 2025
Problem
In the adjoining figure,
is an equilateral triangle and
and
are squares. The measure of
is
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.