Difference between revisions of "2012 AMC 10B Problems/Problem 19"
(add coordbash solution) |
|||
Line 55: | Line 55: | ||
Let <math>D=(0,0)</math>. | Let <math>D=(0,0)</math>. | ||
We know these points from the problem statement: | We know these points from the problem statement: | ||
− | <math>C=(6,0) | + | <math>C=(6,0), B=(6,30), A=(0,30), G=(0,15), E=(8,30)</math> |
− | + | ||
− | |||
− | |||
− | |||
We can use the Shoelace Formula to find the area of quadrilateral <math>BFDG</math>. We know the coordinates of all of the points in <math>BFDG</math> except <math>F</math>. Since <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>, we can use a system of equations to solve for the coordinates of <math>F</math>. The line for <math>\overline{BC}</math> is simply <math>x = 6</math>. The line for <math>\overline{ED}</math> passes through the origin so it has a y-intercept of <math>0</math>, and a slope of <math>\frac{30-0}{8-0}=\frac{30}{8}=\frac{15}{4}</math>. Therefore, the line for <math>\overline{ED}</math> is <math>y=\frac{15}{4} x</math>. Substituting <math>6</math> for <math>x</math>, we find that <math>y = \frac{45}{2}</math>. Therefore, <math>F=(6, \frac{45}{2})</math>. Applying Shoelace on these points gives us that <math>[BFDG] = \boxed{\frac{135}{2}}</math>. | We can use the Shoelace Formula to find the area of quadrilateral <math>BFDG</math>. We know the coordinates of all of the points in <math>BFDG</math> except <math>F</math>. Since <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>, we can use a system of equations to solve for the coordinates of <math>F</math>. The line for <math>\overline{BC}</math> is simply <math>x = 6</math>. The line for <math>\overline{ED}</math> passes through the origin so it has a y-intercept of <math>0</math>, and a slope of <math>\frac{30-0}{8-0}=\frac{30}{8}=\frac{15}{4}</math>. Therefore, the line for <math>\overline{ED}</math> is <math>y=\frac{15}{4} x</math>. Substituting <math>6</math> for <math>x</math>, we find that <math>y = \frac{45}{2}</math>. Therefore, <math>F=(6, \frac{45}{2})</math>. Applying Shoelace on these points gives us that <math>[BFDG] = \boxed{\frac{135}{2}}</math>. | ||
Latest revision as of 14:23, 23 August 2022
Problem
In rectangle ,
,
, and
is the midpoint of
. Segment
is extended 2 units beyond
to point
, and
is the intersection of
and
. What is the area of quadrilateral
?
Solution
Note that the area of equals the area of
.
Since
. Now,
, so
and
so
Therefore,
hence our answer is
Solution 2
Notice that is a trapezoid with height
, so we need to find
.
, so
. Since
,
. The area of
is
Solution 3 (Coordinate Bash)
Let .
We know these points from the problem statement:
We can use the Shoelace Formula to find the area of quadrilateral . We know the coordinates of all of the points in
except
. Since
is the intersection of
and
, we can use a system of equations to solve for the coordinates of
. The line for
is simply
. The line for
passes through the origin so it has a y-intercept of
, and a slope of
. Therefore, the line for
is
. Substituting
for
, we find that
. Therefore,
. Applying Shoelace on these points gives us that
.
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.