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

(Solution 3)
 
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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==Solution 1==
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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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label("$E$",DD,N);
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label("$F$",CC,N);
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draw(C--CC); draw(D--DD);
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</asy>
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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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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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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 = 4\sqrt{6}</math>.
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<asy>
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size(7cm);
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pair A,B,C,D,CC,DD;
 +
A = (-2,7);
 +
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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label("$D$",D,SW);
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label("$E$",DD,SE);
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label("$F$",CC,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("$4\sqrt{6}$",(C+CC)/2,W);
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label("$4\sqrt{6}$",(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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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>.
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*Or, alternatively, one can notice that the two triangles have the same height but <math>\bigtriangleup AFC</math> has a shorter base than <math>\bigtriangleup BED</math>.
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==Solution 2==
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<asy>
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size(7cm);
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pair A,B,C,D,E;
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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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E = (4,7);
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draw(A--B--C--D--cycle);
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draw(D--E);
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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",(12,1),E);
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label("14",(2,1),E);
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label("12",(A+E)/2,N);
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label("21",(E+B)/2,N);
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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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label("$D$",D,SW);
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label("$E$",E,N);
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</asy>
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The area of <math>\Delta AED</math> is by Heron's, <math>4\sqrt{9(4)(3)(2)}=24\sqrt{6}</math>. This makes the length of the altitude from <math>D</math> onto <math>\overline{AE}</math> equal to <math>4\sqrt{6}</math>. One may now proceed as in Solution <math>1</math> to obtain an answer of <math>\boxed{\textbf{(B) }25}</math>.
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==Solution 3==
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Using the same way as Solution 1, obtain that AE=2. Let us assume that point G is point A moving in a straight line down until it touches DC if DC was extended both ways by infinity. We know that GC=23. Thus the answer would be the square root of (23 square and (4*square root of 6) square). Thus we get 625. Square root of 625 would be 25.
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-Reality Writes
  
 
==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}}

Latest revision as of 08:18, 23 September 2021

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 $\bigtriangleup AFC$ 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 $\boxed{\textbf{(B) }25}$.


Solution 3

Using the same way as Solution 1, obtain that AE=2. Let us assume that point G is point A moving in a straight line down until it touches DC if DC was extended both ways by infinity. We know that GC=23. Thus the answer would be the square root of (23 square and (4*square root of 6) square). Thus we get 625. Square root of 625 would be 25. -Reality Writes

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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