Difference between revisions of "2006 AMC 10B Problems/Problem 23"
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Substituting <math>y=x+3</math> into the second equation we get <math>7x = 3(x+10)</math>, which solves to <math>x=\frac{15}{2}</math>. Then <math>y=x+3 = \frac{15}{2}+3 = \frac{21}{2}</math>, and the total area of the quadrilateral is <math>x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}</math>. | Substituting <math>y=x+3</math> into the second equation we get <math>7x = 3(x+10)</math>, which solves to <math>x=\frac{15}{2}</math>. Then <math>y=x+3 = \frac{15}{2}+3 = \frac{21}{2}</math>, and the total area of the quadrilateral is <math>x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}</math>. | ||
+ | ==Solution 2== | ||
+ | |||
+ | Connect points <math>E</math> and <math>D</math>. Triangles <math>EFA</math> and <math>FAB</math> share an altitude and their areas are in the ration <math>3:7</math>. Their bases, <math>EF</math> and <math>FB</math>, must be in the same <math>3:7</math> ratio. | ||
+ | |||
+ | Triangles <math>EFD</math> and <math>FBD</math> share an altitude and their bases are in a <math>3:7</math> ratio. Therefore, their areas are in a <math>3:7</math> ratio and the area of triangle <math>EFD</math> is <math>3</math>. | ||
+ | |||
+ | Triangle <math>CED</math> and <math>DEA</math> share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>A:(3+3) \Rightarrow A:6</math> where <math>A</math> is the area of triangle <math>CED</math> | ||
+ | |||
+ | Triangles <math>CEB</math> and <math>EAB</math> also share an altitude. The ratio of their areas is also equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>(A+3+7):(3+7) \Rightarrow (A+10):10</math> | ||
+ | |||
+ | Because the two ratios are equal, we get the equation <math>\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15</math>. We add the area of triangle <math>EDF</math> to get that the total area of the quadrilateral is <math>\boxed{\textbf{(D) }18}</math>. | ||
+ | |||
+ | ~Zeric Hang | ||
== See also == | == See also == | ||
{{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}} |
Revision as of 14:13, 19 August 2018
Contents
Problem
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?
Solution 1
Label the points in the figure as shown below, and draw the segment . This segment divides the quadrilateral into two triangles, let their areas be and .
Since triangles and share an altitude from and have equal area, their bases must be equal, hence .
Since triangles and share an altitude from and their respective bases are equal, their areas must be equal, hence .
Since triangles and share an altitude from and their respective areas are in the ratio , their bases must be in the same ratio, hence .
Since triangles and share an altitude from and their respective bases are in the ratio , their areas must be in the same ratio, hence , which gives us .
Substituting into the second equation we get , which solves to . Then , and the total area of the quadrilateral is .
Solution 2
Connect points and . Triangles and share an altitude and their areas are in the ration . Their bases, and , must be in the same ratio.
Triangles and share an altitude and their bases are in a ratio. Therefore, their areas are in a ratio and the area of triangle is .
Triangle and share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases and . The ratio is where is the area of triangle
Triangles and also share an altitude. The ratio of their areas is also equal to the ratio of bases and . The ratio is
Because the two ratios are equal, we get the equation . We add the area of triangle to get that the total area of the quadrilateral is .
~Zeric Hang
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.