2001 AMC 10 Problems/Problem 24
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 |
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