Difference between revisions of "2014 AMC 10B Problems/Problem 21"

(Solution)
(Solution)
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<math> \textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 </math>
 
<math> \textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 </math>
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
==Solution==
 
 
<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);
 
label("$E$",DD,N);
 
label("$F$",CC,N);
 
draw(C--CC); draw(D--DD);
 
</asy>
 
 
In the diagram, <math>\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}</math>.
 
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>.
 
 
 
We isolate the <math>h^2</math> term in both equations, getting <math>h^2= 100-x^2</math> and
 
<math>h^2 = 196-(12-x)^2</math>.
 
 
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>.
 
 
<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);
 
label("$D$",D,SW);
 
label("$E$",DD,SE);
 
label("$F$",CC,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>
 
 
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>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length, so the answer is <math>\boxed{\textbf{(B) }25}</math>.
 
  
 
==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}}

Revision as of 12:31, 4 January 2019

Problem

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?

$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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

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