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
. 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
Solution 4 (Dummed down)
The area of triangle is
square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being
and another triangle on the right side, with its hypotenuse being
. After drawing the square, you can cut it into
squares. The area of the rectangle is
square inches because the triangle on the left is half of
and there's another triangle on the other side, equaling them to
square inches. We will be focusing on the left side of the rectangle of
because it includes the shaded region. It's split into
equal squares. The area of this triangle is
square inches because the total area of the rectangle is
square inches and
divided by
is
. There are
sections, so you would do
divided by
in order to find the area of one square. That means that the area of the top right square is
inches and because it's not needed, we will subtract
from
to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is
should then be divided by
in order to find the shaded region since the shaded region is equal to the other square and half a square.
divided by
is
, so the answer is
.
Solution 5 (Dividing into Congruent Triangles)
/* Super fast after convincing yourself they are congruent */
We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates °. (This is not a random idea - the two original two triangles on the top inspires us to do so.)
Note that is perpendicular to
;
is perpendicular to
.
is perpendicular to
, so because of corresponding angles theorem,
. We are also told
. Both
and
have a
° angle. Because of AAS Congruency (Angle-Angle-Side;
°,
, and side
),
is congruent to
. Because of symmetry, the same goes for
and
.
is congruent to
, which is congruent to
because of symmetry. So there are
congruent triangles on the left side. There is also
congruent triangles on the right side because of symmetry.
There are congruent triangles in all. We are told the area of the triangle
is 8.
, so each of the small congruent triangles is
. The shaded area contains
small triangles, so the area of the shaded section is
.
- JoyfulSapling
Video Solution
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David
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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