Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
+ | ==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, <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 = 4\sqrt{6}</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("$4\sqrt{6}$",(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 <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>\bigtriangleup AFC</math> has a shorter base than <math>\bigtriangleup BED</math>. | ||
+ | |||
+ | ==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 <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>. | ||
==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 19:59, 27 December 2019
Contents
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 1
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 and
.
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 . Since , is the shorter length*, so the answer is .
- Or, alternatively, one can notice that the two triangles have the same height but has a shorter base than .
Solution 2
The area of is by Heron's, . This makes the length of the altitude from onto equal to . One may now proceed as in Solution to obtain an answer of .
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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