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| | [[Category: Introductory Geometry Problems]] | | [[Category: Introductory Geometry Problems]] |
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| − | ==Solution==
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| − |
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| − | <asy>
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| − | size(7cm);
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| − | pair A,B,C,D,CC,DD;
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| − | A = (-2,7);
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| − | B = (14,7);
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| − | C = (10,0);
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| − | D = (0,0);
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| − | CC = (10,7);
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| − | DD = (0,7);
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| − | draw(A--B--C--D--cycle);
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| − | //label("33",(A+B)/2,N);
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| − | label("21",(C+D)/2,S);
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| − | label("10",(A+D)/2,W);
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| − | label("14",(B+C)/2,E);
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| − | label("$A$",A,NW);
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| − | label("$B$",B,NE);
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| − | label("$C$",C,SE);
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| − | label("$D$",D,SW);
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| − | draw(C--CC); draw(D--DD);
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| − | </asy>
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| − |
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| − | In the diagram, <math>\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}</math>.
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| − | Denote <math>\overline{AE} = x</math> and <math>\overline{DE} = h</math>. In right triangle <math>AED</math>, we have from the Pythagorean theorem: <math>x^2+h^2=100</math>. Note that since <math>EF = DC</math>, we have <math>BF = 33-DC-x = 12-x</math>. Using the Pythagorean theorem in right triangle <math>BFC</math>, we have <math>(12-x)^2 + h^2 = 196</math>.
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| − |
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| − |
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| − | We isolate the <math>h^2</math> term in both equations, getting <math>h^2= 100-x^2</math> and
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| − | <math>h^2 = 196-(12-x)^2</math>.
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| − |
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| − | Setting these equal, we have <math>100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2</math>. Now, we can determine that <math>h^2 = 100-4 \implies h = \sqrt{96}</math>.
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| − |
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| − | <asy>
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| − | size(7cm);
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| − | pair A,B,C,D,CC,DD;
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| − | A = (-2,7);
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| − | B = (14,7);
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| − | C = (10,0);
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| − | D = (0,0);
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| − | CC = (10,7);
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| − | DD = (0,7);
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| − | draw(A--B--C--D--cycle);
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| − | //label("33",(A+B)/2,N);
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| − | label("21",(C+D)/2,S);
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| − | label("10",(A+D)/2,W);
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| − | label("14",(B+C)/2,E);
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| − | label("$A$",A,NW);
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| − | label("$B$",B,NE);
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| − | label("$C$",C,SE);
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| − | label("$D$",D,SW);
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| − | draw(C--CC); draw(D--DD);
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| − | label("21",(CC+DD)/2,N);
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| − | label("$2$",(A+DD)/2,N);
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| − | label("$10$",(CC+B)/2,N);
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| − | label("$\sqrt{96}$",(C+CC)/2,W);
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| − | label("$\sqrt{96}$",(D+DD)/2,E);
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| − | pair X = (-2,0);
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| − | //draw(X--C--A--cycle,black+2bp);
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| − | </asy>
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| − |
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| − | The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Obviously, <math>25</math> is the shorter length, and thus the answer is <math>\boxed{\textbf{(B) }25}</math>.
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| | ==See Also== | | ==See Also== |
| | {{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}} | | {{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
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 
