1979 AHSME Problems/Problem 3: Difference between revisions

Revision as of 11:49, 5 January 2017

Problem 3

$[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]$

In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ . What is the measure of $\measuredangle AED$ in degrees?

$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution

Solution by e_power_pi_times_i

Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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1979 AHSME Problems/Problem 3: Difference between revisions

Revision as of 11:49, 5 January 2017

Problem 3

Solution

See also