Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
− | We know it is a trapezoid and that AB and CD are perpendicular to AD. If they are perpendicular to AD 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 AD is 7. We can then set the length of AB to be x and the length of DC to be y. BC would then be x+y. Let's draw a straight line down from point B which is perpendicular to DC and parallel to AD. Let's name this line M. Then let's name the point at which line M intersects DC point E. Line M partitions the trapezoid into rectangle ADEB and triangle BEC. We will use the triangle to solve for x*y using the Pythagorean theorem. The line segment EC would be y-x because DC is y and DE is x. DE is x because it is parallel to AB and both are of equal length. Because of the Pythagorean theorem, we know that (EC)^2+(BE)^2=(BC)^2. Substituting the values we have we get (y-x)^2+(7)^2=(x+y)^2. Simplifying this we get (y^2-2xy+x^2)+(49)=(x^2+2xy+y^2). Now we get rid of the x^2 and y^2 terms from both sides to get (-2xy)+(49)=(2xy). Combining like terms we get (49)=(4xy). Then we divide by 4 to get (12.25)=(xy). Now we know that x*y (same thing as xy) is equal to 12.25 which is answer choice <math>\boxed{\textbf{(B)}\ 12.25} </math>. | + | 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*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*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 | Solution By: MATHCOUNTSCMS25 | ||
P.S. I Don't Know How To Format It Properly Using Latex So Could Someone Please Fix It | P.S. I Don't Know How To Format It Properly Using Latex So Could Someone Please Fix It | ||
+ | |||
+ | EDIT: Fixed! (As much as my ability can)-Mliu630XYZ | ||
==See Also== | ==See Also== |
Revision as of 18:40, 10 January 2021
Contents
[hide]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 Latex So Could Someone Please Fix It
EDIT: Fixed! (As much as my ability can)-Mliu630XYZ
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.