Difference between revisions of "2014 AMC 10B Problems/Problem 21"
(→Solution: removed solution by -----) |
(→Solution: fixed diagrams) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
+ | |||
<asy> | <asy> | ||
− | + | size(7cm); | |
− | label(" | + | pair A,B,C,D,CC,DD; |
− | label(" | + | A = (-2,7); |
− | label(" | + | B = (14,7); |
− | label(" | + | C = (10,0); |
− | + | D = (0,0); | |
− | + | CC = (10,7); | |
− | label(" | + | DD = (0,7); |
− | label(" | + | 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); | ||
+ | draw(C--CC); draw(D--DD); | ||
</asy> | </asy> | ||
Line 24: | Line 34: | ||
\\h^2 &= 196-(12-x)^2\end{align}</math>. | \\h^2 &= 196-(12-x)^2\end{align}</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>. 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>. | + | 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); | ||
+ | draw(C--CC); draw(D--DD); | ||
+ | label("21",(CC+DD)/2,N); | ||
+ | label("$2$",(A+DD)/2,N); | ||
+ | label("$8$",(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>. Obviously, <math>25</math> is the shorter length, and thus 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 19:17, 20 February 2014
Problem
Trapezoid 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 ?
Solution
In the diagram, . Denote and . In right triangle , we have from the Pythagorean theorem: . Note that since , we have . Using the Pythagorean theorem in right triangle , we have .
We isolate the term in both equations, getting $\begin{align*}h^2 &= 100-x^2
\\h^2 &= 196-(12-x)^2\end{align}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.).
Setting these equal, we have . Now, we can determine that .
The two diagonals are and . Using the Pythagorean theorem again on and , we can find these lengths to be and . Obviously, is the shorter length, and thus the answer is .
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.