Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Solution 1: LaTeX plus a short note)
Line 70: Line 70:
 
</asy>
 
</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>.
+
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>.
 +
 
 +
*Or, alternatively, one can notice that the two triangles have the same height but <math>\bigtriangleleup AFC</math> has a shorter base than <math>\bigtriangleup BED</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:59, 27 December 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 $ABCD$?

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

Solution 1

[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, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$. Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x)^2 + h^2 = 196$.


We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$.

Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$. Now, we can determine that $h^2 = 100-4 \implies h = 4\sqrt{6}$.

[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("$4\sqrt{6}$",(C+CC)/2,W); label("$4\sqrt{6}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]

The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$, we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$. Since $\sqrt{96+529}<\sqrt{96+961}$, $25$ is the shorter length*, so the answer is $\boxed{\textbf{(B) }25}$.

  • Or, alternatively, one can notice that the two triangles have the same height but $\bigtriangleleup AFC$ (Error compiling LaTeX. Unknown error_msg) has a shorter base than $\bigtriangleup BED$.

Solution 2

[asy] size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",E,N); [/asy]

The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$. This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$. One may now proceed as in Solution $1$ to obtain an answer of $\boxed{\textbf{(B) }25}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_21&oldid=113576"