Difference between revisions of "2012 AMC 10B Problems/Problem 19"
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− | + | ==Problem== | |
+ | In rectangle <math>ABCD</math>, <math>AB=6</math>, <math>AD=30</math>, and <math>G</math> is the midpoint of <math>\overline{AD}</math>. Segment <math>AB</math> is extended 2 units beyond <math>B</math> to point <math>E</math>, and <math>F</math> is the intersection of <math>\overline{ED}</math> and <math>\overline{BC}</math>. What is the area of quadrilateral <math>BFDG</math>? | ||
+ | |||
+ | <math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Solution== | ||
+ | <asy> | ||
+ | unitsize(10); | ||
+ | pair B=(0,0); | ||
+ | pair A=(0,6); | ||
+ | pair C=(30,0); | ||
+ | pair D=(30,6); | ||
+ | pair G=(15,6); | ||
+ | pair E=(0,-2); | ||
+ | pair F=(15/2,0); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(G); | ||
+ | dot(E); | ||
+ | dot(F); | ||
+ | label("A",(0,6),NW); | ||
+ | label("B",(0,0),W); | ||
+ | label("C",(30,0),E); | ||
+ | label("D",(30,6),NE); | ||
+ | label("G",(15,6),N); | ||
+ | label("E",(0,-2),SW); | ||
+ | label("F",(15/2,0),N); | ||
+ | label("15",(A--G),N); | ||
+ | label("15",(G--D),N); | ||
+ | label("6",(A--B),W); | ||
+ | label("2",(B--E),W); | ||
+ | label("6",(D--C),E); | ||
+ | draw(A--B); | ||
+ | draw(B--C); | ||
+ | draw(C--D); | ||
+ | draw(D--A); | ||
+ | draw(E--D); | ||
+ | draw(B--E); | ||
+ | draw(B--G); | ||
+ | </asy> | ||
+ | |||
+ | Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. | ||
+ | Since <math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5</math> and <math>FC=22.5,</math> so <math>\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}. </math> | ||
+ | |||
+ | Therefore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath> | ||
+ | hence our answer is <math>\fbox{C}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Notice that <math>BFDG</math> is a trapezoid with height <math>6</math>, so we need to find <math>BF</math>. <math>\triangle BFE\sim \triangle ADE</math>, so <math>4\cdot BF = AD</math>. Since <math>AD = 30</math>, <math>BF = \frac{15}{2}</math>. The area of <math>BFDG</math> is <math>6\cdot \frac{15+\frac{15}{2}}{2}=3\cdot \frac{45}{2}=\boxed{\textbf{(C) }\frac{135}{2}}</math> | ||
+ | ==Solution 3 (Coordinate Bash)== | ||
+ | Let <math>D=(0,0)</math>. | ||
+ | We know these points from the problem statement: | ||
+ | <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>. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | |||
+ | {{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}} | ||
+ | |||
+ | {{MAA Notice}} |
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.