Difference between revisions of "2002 AMC 8 Problems/Problem 20"
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draw((2.5,2)--(7.5,2)); | draw((2.5,2)--(7.5,2)); | ||
draw((5,4)--(5,0)); | draw((5,4)--(5,0)); | ||
− | fill((0,0)--(2.5,2)--( | + | fill((0,0)--(2.5,2)--(4,2)--(5,0)--cycle, mediumgrey); |
− | label(scale(0.8)*"$X$", (5, | + | label(scale(0.8)*"$X$", (5,6), N); |
label(scale(0.8)*"$Y$", (0,0), W); | label(scale(0.8)*"$Y$", (0,0), W); | ||
label(scale(0.8)*"$Z$", (10,0), E); | label(scale(0.8)*"$Z$", (10,0), E); |
Revision as of 14:30, 26 October 2016
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, 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
.
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.