Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== In the figure, <math>\angle A</math>, <math>\angle B</math>, and <math>\angle C</math> are right angles. If <math>\angle AEB = 40^\circ </math> and <math>\angle BED...")
 
m (Test Fix)
 
(4 intermediate revisions by 4 users not shown)
Line 19: Line 19:
  
 
<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>
 +
 +
==Solution==
 +
 +
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>
 +
 +
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>
 +
 +
==See Also==
 +
{{AJHSME box|year=1995|num-b=12|num-a=14}}
 +
{{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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems/Problem_13&oldid=80813"