2002 AMC 8 Problems/Problem 20
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 1
The shaded region is a right trapezoid. Assume WLOG that . Then because the area of is equal to 8, the height of the triangle . Because the line is a midsegment, the top base of the trapezoid is . Also, divides in two, so the height of the trapezoid is . The bottom base is . The area of the shaded region is .
Solution 2
Since and are the midpoints of and , respectively, . Draw segments and . Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that . is the line that connects the midpoints of two sides of a triangle together, which means that is parallel to and half in length of . Then . Since is parallel to , and is the transversal, Similarly, Then, by SAS, . Since corresponding parts of congruent triangles are congruent,. Since ACY and BCZ are now isosceles triangles, and Using the fact that is parallel to , and . Now . Draw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3.
Basically the proof is to show . If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area 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
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.