Difference between revisions of "1996 AHSME Problems/Problem 21"
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<math>2\angle C + \angle D + \angle A = 270^\circ</math> | <math>2\angle C + \angle D + \angle A = 270^\circ</math> | ||
− | From isosceles <math> | + | From isosceles <math>\triangle ABD</math>, we get <math>\angle A = \angle D</math>, so: |
<math>2\angle C + 2\angle D = 270^\circ</math> | <math>2\angle C + 2\angle D = 270^\circ</math> | ||
− | + | <math>\angle C + \angle D = 135^\circ</math>, which is answer <math>\boxed{D}</math> | |
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=20|num-a=22}} | {{AHSME box|year=1996|num-b=20|num-a=22}} |
Revision as of 14:02, 20 August 2011
Problem
Triangles and are isosceles with , and intersects at . If is perpendicular to , then is
Solution
Redraw the figure as a concave pentagon :
The angles of the pentagon will still sum to , regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles , , and all make up the angles of the pentagon without overlap.
Since reflex , we have:
.
From isosceles , we get , so:
From isosceles , we get , so:
, which is answer
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |