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]$

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

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
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2006 AMC 10B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See also