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

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 $ABCD$ ?

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

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.

@@ Line 4: / Line 4: @@
 <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]]
-==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==
 {{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}
 {{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 12:31, 4 January 2019

Problem

See Also