Difference between revisions of "1995 AJHSME Problems/Problem 13"

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<math>\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ</math>
 
<math>\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ</math>
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==Solution==
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Because <math>\angle BED=\angle BDE</math>, <math>\angle B=90^\circ</math>, and <math>\triangle BED</math> is a triangle, we get: <cmath>\angle B + \angle BED +\angle BDE=180</cmath> <cmath>90 +\angle BED +\angle BED=180</cmath> <cmath>2\angle BED=90</cmath> <cmath>\angle BED=\angle BDE=45^\circ</cmath>
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So <math>\angle AED=\angle AEB +\angle BED=40 +45=85^\circ</math> Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: <cmath>\angle A +\angle C +\angle CDE +\angle AED=360</cmath> <cmath>90 +90 +\angle CDE +85=360</cmath> <cmath>\angle CDE=95^\circ \text{(E)}</cmath>
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==See Also==
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{{AJHSME box|year=1995|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 20:08, 27 October 2016

Problem

In the figure, $\angle A$, $\angle B$, and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE$, then $\angle CDE =$

[asy] dot((0,0)); label("$E$",(0,0),SW); dot(dir(85)); label("$A$",dir(85),NW); dot((4,0)); label("$D$",(4,0),SE); dot((4.05677,0.648898)); label("$C$",(4.05677,0.648898),NE); draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle);  dot((2,2)); label("$B$",(2,2),N); draw((0,0)--(2,2)--(4,0)); pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898)); dot(x[0]); dot(x[1]); label("$F$",x[0],SE); label("$G$",x[1],SW); [/asy]

$\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ$

Solution

Because $\angle BED=\angle BDE$, $\angle B=90^\circ$, and $\triangle BED$ is a triangle, we get: \[\angle B + \angle BED +\angle BDE=180\] \[90 +\angle BED +\angle BED=180\] \[2\angle BED=90\] \[\angle BED=\angle BDE=45^\circ\]

So $\angle AED=\angle AEB +\angle BED=40 +45=85^\circ$ Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: \[\angle A +\angle C +\angle CDE +\angle AED=360\] \[90 +90 +\angle CDE +85=360\] \[\angle CDE=95^\circ \text{(E)}\]

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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
All AJHSME/AMC 8 Problems and Solutions

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