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.
Despite this long proof you can easily see that each triangle is congruent to one another. 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 | |
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 |
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