2009 AMC 10B Problems/Problem 18
Contents
Problem
Rectangle has and . Point is the midpoint of diagonal , and is on with . What is the area of ?
Solution 1 (Coordinate Geo)
Set to . Since is the midpoint of the diagonal, it would be . The diagonal would be the line . Since is perpendicular to , its line would be in the form . Plugging in and for and would give . To find the x-intercept of we plug in for and get . Then, using the Shoelace Formula for , , and , we find the area is .
Solution 2
By the Pythagorean theorem we have , hence .
The triangles and have the same angle at and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is , hence the ratio of their areas is .
And as the area of triangle is , the area of triangle is .
Solution 3 (Only Pythagorean Theorem)
Draw as shown from the diagram. Since is of length , we have that is of length , because of the midpoint . Through the Pythagorean theorem, we know that , which means . Define to be for the sake of clarity. We know that . From here, we know that . From here, we can write the expression . Now, remember . , since we set in the start of the solution. Now to find the area
Video Solution
https://www.youtube.com/watch?v=VNJrZ-ABtS4
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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