Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | ||
− | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. | + | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up |
Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. | Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. |
Revision as of 18:50, 19 September 2015
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
. (An alternate way is by seeing that the set-up
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.