Difference between revisions of "2004 AMC 12B Problems/Problem 14"
(New page: == Problem == In <math>\triangle ABC</math>, <math>AB=13</math>, <math>AC=5</math>, and <math>BC=12</math>. Points <math>M</math> and <math>N</math> lie on <math>AC</math> and <math>BC</m...) |
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− | In <math>\triangle ABC</math>, <math>AB=13</math>, <math>AC=5</math>, and <math>BC=12</math>. Points <math>M</math> and <math>N</math> lie on <math>AC</math> and <math>BC</math>, respectively, with <math>CM=CN=4</math>. Points <math>J</math> and <math>K</math> are on <math>AB</math> so that <math>MJ</math> and <math>NK</math> are perpendicular to <math>AB</math>. What is the area of pentagon <math> | + | In <math>\triangle ABC</math>, <math>AB=13</math>, <math>AC=5</math>, and <math>BC=12</math>. Points <math>M</math> and <math>N</math> lie on <math>AC</math> and <math>BC</math>, respectively, with <math>CM=CN=4</math>. Points <math>J</math> and <math>K</math> are on <math>AB</math> so that <math>MJ</math> and <math>NK</math> are perpendicular to <math>AB</math>. What is the area of pentagon <math>CMJKN</math>? |
<asy> | <asy> |
Revision as of 19:27, 1 January 2012
Problem
In , , , and . Points and lie on and , respectively, with . Points and are on so that and are perpendicular to . What is the area of pentagon ?
Solution
The triangle is clearly a right triangle, its area is . If we knew the areas of triangles and , we could subtract them to get the area of the pentagon.
Draw the height from onto . As and the area is , we get . The situation is shown in the picture below:
Now note that the triangles , , , and all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio , their areas have ratio . We will use this fact repeatedly. Below we will use to denote the area of the triangle .
We have , hence .
Also, , hence .
Now for the smaller triangles:
We know that , hence .
Similarly, , hence .
Finally, the area of the pentagon is .
See Also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |