Difference between revisions of "2006 AMC 10B Problems/Problem 23"
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Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>C</math> and their respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio, hence <math>x:(y+7) = 3:7</math>, which gives us <math>7x = 3(y+7)</math>. | Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>C</math> and their respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio, hence <math>x:(y+7) = 3:7</math>, which gives us <math>7x = 3(y+7)</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{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>. |
== 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 23:04, 7 September 2011
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
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 .
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 |