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

## 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?

$[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("7",(1.25,0.2)); label("7",(2.2,0.45)); label("3",(0.45,0.35)); [/asy]$

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

## Solution 1

Label the points in the figure as shown below, and draw the segment $CF$. This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$.

$[asy] unitsize(2cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("7",(1.45,0.15)); label("7",(2.2,0.45)); label("3",(0.45,0.35)); draw( C -- F, dashed ); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",Ep,NW); label("F",F,S); label("x",(1,1)); label("y",(1.6,1)); [/asy]$

https://latex.artofproblemsolving.com/a/3/7/a373a9412edc8a46f24525404560b3b355922171.png Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$.

Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$.

Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$.

Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$.

Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\frac{15}{2}$. Then $y=x+3 = \frac{15}{2}+3 = \frac{21}{2}$, and the total area of the quadrilateral is $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}$.

## Solution 2

Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio.

Triangles $EFD$ and $FBD$ share an altitude and their bases are in a $3:7$ ratio. Therefore, their areas are in a $3:7$ ratio and the area of triangle $EFD$ is $3$.

Triangle $CED$ and $DEA$ share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases $CE$ and $EA$. The ratio is $A:(3+3) \Rightarrow A:6$ where $A$ is the area of triangle $CED$

Triangles $CEB$ and $EAB$ also share an altitude. The ratio of their areas is also equal to the ratio of bases $CE$ and $EA$. The ratio is $(A+3+7):(3+7) \Rightarrow (A+10):10$

Because the two ratios are equal, we get the equation $\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15$. We add the area of triangle $EDF$ to get that the total area of the quadrilateral is $\boxed{\textbf{(D) }18}$.

~Zeric Hang