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 . What is the area (in square inches) of the shaded region?
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 triangle 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 . Since , it means that is on the perpendicular bisector of YZ. Then . 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,. Using the fact that is parallel to , and . Also, because is isosceles. 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- . Line segment is half of segment using AA similarity between triangles and . is of . 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 | |
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