Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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D = (0,0); | D = (0,0); | ||
E= (0, 7); | E= (0, 7); | ||
− | F= (10, 7) | + | F= (10, 7); |
CC = (10,7); | CC = (10,7); | ||
DD = (0,7); | DD = (0,7); |
Revision as of 10:23, 28 July 2016
Problem
Trapezoid has parallel sides of length and of length . The other two sides are of lengths and . The angles and are acute. What is the length of the shorter diagonal of ?
Solution
size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E= (0, 7); F= (10, 7); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); draw(C--CC); draw(D--DD); (Error compiling LaTeX. E= (0, 7); ^ 13599927fc58e3fdb9115f4fee640e9bd8a3db7c.asy: 9.1: modifying non-public field outside of structure)
In the diagram, . Denote and . In right triangle , we have from the Pythagorean theorem: . Note that since , we have . Using the Pythagorean theorem in right triangle , we have .
We isolate the term in both equations, getting and
.
Setting these equal, we have . Now, we can determine that .
The two diagonals are and . Using the Pythagorean theorem again on and , we can find these lengths to be and . Since , is the shorter length, so the answer is .
See Also
2014 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.