Difference between revisions of "2006 AMC 10B Problems/Problem 23"

Revision as of 14:05, 2 August 2006

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?

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$

Solution

To help in finding the area of the quadrilateral, draw an auxillary line from the top vertex to the intersection of the two lines.

Doing so and labling the points yeilds this as the resulting diagram:

Since triangles $AFC$ and $EFC$ share an altitude from $C$ and have equal area, their bases must be equal.

So $AF=EF$ .

Since triangles $AFB$ and $EFB$ share an altitude from $B$ and their respective bases are equal, their areas must be equal.

So $x+3=y$ .

Since triangles $DFA$ and $CFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$ , their bases must be in the same ratio.

So $\frac{DF}{3}=\frac{CF}{7}$ .

Since triangles $DFB$ and $CFB$ share an altitude from $B$ and their respective bases are in the ratio $3:7$ , their areas must be in the same ratio.

So $\frac{x}{3}=\frac{y+7}{7}$ .

Substituting $x+3$ for $y$ :

$\frac{x}{3}=\frac{x+3+7}{7}$

$\frac{x}{3}=\frac{x+10}{7}$

$7x=3x+30$

$4x=30$

$x=\frac{15}{2}$

$y=\frac{21}{2}$

Therefore, the shaded area is $x+y = \frac{15}{2} + \frac{21}{2} = 18 \Rightarrow D$

@@ Line 47: / Line 47: @@
 == See Also ==
 *[[2006 AMC 10B Problems]]
+*[[2006 AMC 10B Problems/Problem 22|Previous Problem]]
+*[[2006 AMC 10B Problems/Problem 24|Next Problem]]

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Difference between revisions of "2006 AMC 10B Problems/Problem 23"

Revision as of 14:05, 2 August 2006

Problem

Solution

See Also