Difference between revisions of "2014 AMC 10B Problems/Problem 21"
Football017 (talk | contribs) m (Added E and F to diagram.) |
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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);
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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 
.
![[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); 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); label("21",(CC+DD)/2,N); label("$2$",(A+DD)/2,N); label("$10$",(CC+B)/2,N); label("$\sqrt{96}$",(C+CC)/2,W); label("$\sqrt{96}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]](http://latex.artofproblemsolving.com/0/c/6/0c634407e3e4c38a08e2c85ec5dcce4d91bfa582.png)
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. 
