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Difference between revisions of "2006 AMC 10B Problems/Problem 23"
(extra solution)
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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>.
Line 64: Line 64:
 
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 11:23, 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?

[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 (mass points)

We see that $EF:FB=3:7$ and $AF=FD$. We assign a mass of $7$ to $E$ and $3$ to $D$, making $F$ have mass $10$ and $A$ and $D$ each have mass 5. Now, $C$ has mass $2$. Therefore, the area of triangle $CEB$ is $10 \cdot 2.5=25$, so the area of $CEFD$ is $\boxed{(D), 18}$

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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