Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (added category)
Line 2: Line 2:
 
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?
 
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?
  
−
[[Image:2006amc10b23.gif]]
+
<asy>
 +
unitsize(1.5cm);
 +
defaultpen(.8);
  
−
<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>
+
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 );
  
−
== Solution ==
+
draw( A -- B -- C -- cycle );
−
To help in finding the [[area]] of the [[quadrilateral]], draw an auxillary line from the top [[vertex]] to the intersection of the two lines.
+
draw( A -- D );
 +
draw( B -- Ep );
 +
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );
  
−
Doing so and labling the [[point]]s yeilds this as the resulting diagram:
+
label("$7$",(1.25,0.2));
 +
label("$7$",(2.2,0.45));
 +
label("$3$",(0.45,0.35));
 +
</asy>
  
−
[[Image:2006amc10b23solution.gif]]
+
<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>
  
−
Since [[triangle]]s <math>AFC</math> and <math>EFC</math> share an [[altitude]] from <math>C</math> and have equal area, their bases must be equal.
+
== Solution ==
  
−
So <math>AF=EF</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>.
  
−
Since triangles <math>AFB</math> and <math>EFB</math> share an altitude from <math>B</math> and their respective bases are equal, their areas must be equal.  
+
<asy>
 +
unitsize(2cm);
 +
defaultpen(.8);
  
−
So <math>x+3=y</math>.
+
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 );
  
−
Since triangles <math>DFA</math> and <math>CFA</math> share an altitude from <math>A</math> and their  respective areas are in the ratio <math>3:7</math>, their bases must be in the same ratio.
+
draw( A -- B -- C -- cycle );
 +
draw( A -- D );
 +
draw( B -- Ep );
 +
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );
  
−
So <math>\frac{DF}{3}=\frac{CF}{7}</math>.
+
label("$7$",(1.45,0.15));
 +
label("$7$",(2.2,0.45));
 +
label("$3$",(0.45,0.35));
  
−
Since triangles <math>DFB</math> and <math>CFB</math> share an altitude from <math>B</math> and their  respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio.
+
draw( C -- F, dashed );
  
−
So <math>\frac{x}{3}=\frac{y+7}{7}</math>.
+
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,N);
 +
label("$D$",D,NE);
 +
label("$E$",Ep,NW);
 +
label("$F$",F,S);
  
−
Substituting <math>x+3</math> for <math>y</math>:
+
label("$x$",(1,1));
 +
label("$y$",(1.6,1));
 +
</asy>
  
−
<math>\frac{x}{3}=\frac{x+3+7}{7}</math>
+
Since [[triangle]]s <math>AFB</math> and <math>DFB</math> share an [[altitude]] from <math>B</math> and have equal area, their bases must be equal, hence <math>AF=DF</math>.
  
−
<math>\frac{x}{3}=\frac{x+10}{7}</math>
+
Since triangles <math>AFC</math> and <math>DFC</math> share an altitude from <math>C</math> and their respective bases are equal, their areas must be equal, hence <math>x+3=y</math>.
  
−
<math>7x=3x+30</math>
+
Since triangles <math>EFA</math> and <math>BFA</math> share an altitude from <math>A</math> and their respective areas are in the ratio <math>3:7</math>, their bases must be in the same ratio, hence <math>DF:CF = 3:7</math>.
  
−
<math>4x=30</math>
+
Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>B</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>.
  
−
<math>x=\frac{15}{2}</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>.
  
−
<math>y=\frac{21}{2}</math>
+
== See Also ==
  
−
Therefore, the shaded area is <math> x+y = \frac{15}{2} + \frac{21}{2} = 18 \Rightarrow D </math>
 
−
 
−
== See Also ==
 
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]
  

Revision as of 16:44, 15 March 2009

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

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 $DF:CF = 3:7$.

Since triangles $EFC$ and $BFC$ share an altitude from $B$ 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 $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{18}$.

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&oldid=30730"