Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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Simpler is just drawing the trapezoid and then using what is given to solve. | Simpler is just drawing the trapezoid and then using what is given to solve. | ||
Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a. | Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a. | ||
| − | By the Pythagorean theorem, it is obvious that <math>a^{2} + 49 = (2x+a)^{2}</math> (the RHS is the fact the two sides added together equals that). Then, we get <math>a^2 + 49 = 4x^2 + 4ax + a^2</math>, cancel out and factor and we get <math>49 = 4x(x+a)</math>. Notice that <math>x(x+a)</math> is what the question asks, so the answer is <math>\boxed{\frac{49}{4}}</math> | + | By the Pythagorean theorem, it is obvious that <math>a^{2} + 49 = (2x+a)^{2}</math> (the RHS is the fact the two sides added together equals that). Then, we get <math>a^2 + 49 = 4x^2 + 4ax + a^2</math>, cancel out and factor and we get <math>49 = 4x(x+a)</math>. Notice that <math>x(x+a)</math> is what the question asks, so the answer is <math>\boxed{\frac{49}{4}}</math> |
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| + | Solution by IronicNinja | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2001|num-b=23|num-a=25}} | {{AMC10 box|year=2001|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:13, 8 January 2019
Contents
Problem
In trapezoid 
, 
and 
are perpendicular to 
, with 
, 
, and 
. What is 
?
Solution
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((0.2,4.92), 1.3)); draw(circle((1.04,1.58), 2.14)); draw((-1.1,4.92)--(0.2,4.92)); draw((0.2,4.92)--(1.04,1.58)); draw((1.04,1.58)--(-1.1,1.58)); draw((-1.1,1.58)--(-1.1,4.92)); /* dots and labels */ dot((-1.1,4.92),dotstyle); label("$A$", (-1.02,5.12), NE * labelscalefactor); dot((0.2,4.92),dotstyle); label("$B$", (0.28,5.12), NE * labelscalefactor); dot((-1.1,1.58),dotstyle); label("$D$", (-1.02,1.78), NE * labelscalefactor); dot((1.04,1.58),dotstyle); label("$C$", (1.12,1.78), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]](http://latex.artofproblemsolving.com/d/6/b/d6b496ebb456bc614e98bb77b915a46f0cf146fa.png)
If 
and 
, then 
. By the Pythagorean theorem, we have 
Solving the equation, we get 
.
Solution 2
Simpler is just drawing the trapezoid and then using what is given to solve.
Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a.
By the Pythagorean theorem, it is obvious that 
(the RHS is the fact the two sides added together equals that). Then, we get 
, cancel out and factor and we get 
. Notice that 
is what the question asks, so the answer is 
Solution by IronicNinja
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
