2007 AMC 10A Problems/Problem 18
Problem
Consider the -sided polygon
, as shown. Each of its sides has length
, and each two consecutive sides form a right angle. Suppose that
and
meet at
. What is the area of quadrilateral
?
Solution
Solution 1
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles
and
.
We know that triangles and
are similar because
. Also, since
, the ratio of the distance from
to
to the distance from
to
is also
. Solving with the fact that the distance from
to
is 4, we see that the distance from
to
is
.
The area of is simply
, the area of
is
, and the area of rectangle
is
.
Taking the area of rectangle and subtracting the combined area of
and
yields
.
Solution 2
Extend and
and call their intersection
.
The triangles and
are clearly similar with ratio
, hence
and thus
. The area of the triangle
is
.
The triangles and
are similar as well, and we now know that the ratio of their dimensions is
.
Draw altitudes from onto
and
, let their feet be
and
. We get that
. Hence
.
Then the area of is
, and the area of
can be obtained by subtracting the area of
, which is
. Hence the answer is
.
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AMC 10 Problems and Solutions |