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 . 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, the area of the shaded region is .
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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