Difference between revisions of "2006 AMC 10B Problems/Problem 23"
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<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math> | <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math> | ||
− | == Solution == | + | == Solution 1== |
Label the points in the figure as shown below, and draw the segment <math>CF</math>. This segment divides the quadrilateral into two triangles, let their areas be <math>x</math> and <math>y</math>. | Label the points in the figure as shown below, and draw the segment <math>CF</math>. This segment divides the quadrilateral into two triangles, let their areas be <math>x</math> and <math>y</math>. | ||
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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 (mass points)== | ||
+ | We see that <math>EF:FB=3:7</math> and <math>AF=FD</math>. We assign a mass of <math>7</math> to <math>E</math> and <math>3</math> to <math>D</math>, making <math>F</math> have mass <math>10</math> and <math>A</math> and <math>D</math> each have mass 5. Now, <math>C</math> has mass <math>2</math>. Therefore, the area of triangle <math>CEB</math> is <math>10 \cdot 2.5=25</math>, so the area of <math>CEFD</math> is <math>\boxed{(D), 18}</math> | ||
== 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 12:23, 22 November 2015
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 (mass points)
We see that and . We assign a mass of to and to , making have mass and and each have mass 5. Now, has mass . Therefore, the area of triangle is , so the area of is
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 |
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