Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | draw((0,0)--(33,0)--(23, -9.79795897)--(2, -9.79795897)--(0,0)); | ||
+ | label("A", (0,0), NW); | ||
+ | label("B", (33,0), NE); | ||
+ | label("C", (23,-9.79795897), SE); | ||
+ | label("D", (2, -9.79795897), SW); | ||
+ | draw((2, 0)--(2, -9.79795897)); | ||
+ | draw((23, 0)--(23,-9.79795897)); | ||
+ | label("E", (2,0), N); | ||
+ | label("F", (23,0), N); | ||
+ | </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>\begin{align*}h^2 &= 100-x^2\\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> | ||
==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 17:46, 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 |
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