2002 AMC 8 Problems/Problem 20
Contents
Problem
The area of triangle is 8 square inches. Points and are midpoints of congruent segments and . Altitude bisects . The area (in square inches) of the shaded region is
Solution 3
We know the area of triangle is square inches. The area of a triangle can also be represented as or in this problem . By solving, we have
With SAS congruence, triangles and are congruent. Hence, triangle . (Let's say point is the intersection between line segments and .) We can find the area of the trapezoid by subtracting the area of triangle from .
We find the area of triangle by the formula- . is of from solution 1. The area of is .
Therefore, the area of the shaded area- trapezoid has area .
- sarah07
Video Solution
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AJHSME/AMC 8 Problems and Solutions |
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