Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. | + | <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> | ||
+ | |||
+ | If <math> AB=x </math> and <math> CD=y </math>, then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49. </math> | ||
+ | Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Simpler is just drawing the trapezoid and then using what is given to solve. | ||
+ | Draw a line parallel to <math>\overline{AD}</math> that connects the longer side to the corner of the shorter side. Name the bottom part <math>x</math> and top part <math>a</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{\textbf{(B)}\ 12.25} </math>. | ||
+ | |||
+ | Solution by IronicNinja | ||
+ | |||
+ | ==Solution 3== | ||
+ | We know it is a trapezoid and that <math>\overline{AB}</math> and <math>\overline{CD}</math> are perpendicular to <math>\overline{AD}</math>. If they are perpendicular to <math>\overline{AD}</math> that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know <math>\overline{AD}</math> is <math>7</math>. We can then set the length of <math>\overline{AB}</math> to be <math>x</math> and the length of <math>\overline{DC}</math> to be <math>y</math>. <math>\overline{BC}</math> would then be <math>x+y</math>. Let's draw a straight line down from point <math>B</math> which is perpendicular to <math>\overline{DC}</math> and parallel to <math>\overline{AD}</math>. Let's name this line <math>M</math>. Then let's name the point at which line <math>M</math> intersects <math>\overline{DC}</math> point <math>E</math>. Line <math>M</math> partitions the trapezoid into rectangle <math>ADEB</math> and triangle <math>BEC</math>. We will use the triangle to solve for <math>x \times y</math> using the Pythagorean theorem. The line segment <math>\overline{EC}</math> would be <math>y-x</math> because <math>\overline{DC}</math> is <math>y</math> and <math>\overline{DE}</math> is <math>x</math>. <math>\overline{DE}</math> is <math>x</math> because it is parallel to <math>\overline{AB}</math> and both are of equal length. Because of the Pythagorean theorem, we know that <math>(EC)^2+(BE)^2=(BC)^2</math>. Substituting the values we have we get <math>(y-x)^2+(7)^2=(x+y)^2</math>. Simplifying this we get <math>(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)</math>. Now we get rid of the <math>x^2</math> and <math>y^2</math> terms from both sides to get <math>(-2xy)+(49)=(2xy)</math>. Combining like terms we get <math>(49)=(4xy)</math>. Then we divide by <math>4</math> to get <math>(12.25)=(xy)</math>. Now we know that <math>x \times y</math> (same thing as <math>xy</math>) is equal to <math>12.25</math> which is answer choice <math>\boxed{\textbf{(B)}\ 12.25} </math>. | ||
+ | |||
+ | Solution By: MATHCOUNTSCMS25 | ||
+ | |||
+ | P.S. I Don't Know How to Format It Properly Using <math>\LaTeX</math> So Could Someone Please Fix It | ||
+ | |||
+ | EDIT: Fixed! (As much as my ability can) - Mliu630XYZ | ||
+ | |||
+ | EDIT: Fixed Completely! - palaashgang | ||
==See Also== | ==See Also== |
Latest revision as of 00:56, 1 June 2021
Problem
In trapezoid , and are perpendicular to , with , , and . What is ?
Solution
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 parallel to that connects the longer side to the corner of the shorter side. Name the bottom part and top part . 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
Solution 3
We know it is a trapezoid and that and are perpendicular to . If they are perpendicular to that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know is . We can then set the length of to be and the length of to be . would then be . Let's draw a straight line down from point which is perpendicular to and parallel to . Let's name this line . Then let's name the point at which line intersects point . Line partitions the trapezoid into rectangle and triangle . We will use the triangle to solve for using the Pythagorean theorem. The line segment would be because is and is . is because it is parallel to and both are of equal length. Because of the Pythagorean theorem, we know that . Substituting the values we have we get . Simplifying this we get . Now we get rid of the and terms from both sides to get . Combining like terms we get . Then we divide by to get . Now we know that (same thing as ) is equal to which is answer choice .
Solution By: MATHCOUNTSCMS25
P.S. I Don't Know How to Format It Properly Using So Could Someone Please Fix It
EDIT: Fixed! (As much as my ability can) - Mliu630XYZ
EDIT: Fixed Completely! - palaashgang
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.