Difference between revisions of "2001 AMC 10 Problems/Problem 24"
Ironicninja (talk | contribs) (→Solution 2) |
Phoenixfire (talk | contribs) (→Solution 2) |
||
Line 43: | Line 43: | ||
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> | + | 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 by IronicNinja |
Revision as of 03:36, 9 November 2019
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 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.