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Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 5"

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==Solution==
 
==Solution==
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Since we know <math>BF</math> is equal to the short side of both triangles and that <math>DB</math> is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line <math>DF</math> (as seen below), all four triangles in <math>\triangle ACE</math> are congruent.
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In general, if <math>G</math> and <math>H</math> are similar triangles with sides <math>s_1</math> and <math>s_2</math> , then <math>\frac{Area G}{Area H}=\frac{(s_1)^2}{(s_2)^2}</math>.
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Here, <math>\triangle{BCD}</math> is <math>\frac{50}{128}=\frac{25}{64}</math> of <math>\triangle{ACE}</math>, so BC is <math>\frac{5}{8}</math> of AC or AB is <math>\frac{3}{8}</math> of AC
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<asy>
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then  <math>\triangle{ABF}</math> is <math>\frac{9}{64}</math> of <math>\triangle{ACE}</math> so
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draw((0,0)--(1,2)--(4,0)--cycle,black);
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<math>\triangle ABF = \boxed{18}</math>
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draw((1/2,1)--(2.5,1)--(2,0),black);
 
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draw((1/2,1)--(2,0),black);
 
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MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW);
 
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MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S);
 
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</asy>
 
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Then, since we know the sum of all four triangles is <math>128</math>, we can solve the easy division problem <math>128/4</math> and see that the area of <math>\triangle ABF = \boxed{32}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 02:20, 13 January 2019

Problem

In the figure $BD$ is parallel to $AE$ and also $BF$ is parallel to $DE$. The area of the larger triangle $ACE$ is $128$. The area of the trapezoid $BDEA$ is $78$. Determine the area of triangle $ABF$.

[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW); MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S); [/asy]

Solution

In general, if $G$ and $H$ are similar triangles with sides $s_1$ and $s_2$ , then $\frac{Area G}{Area H}=\frac{(s_1)^2}{(s_2)^2}$. Here, $\triangle{BCD}$ is $\frac{50}{128}=\frac{25}{64}$ of $\triangle{ACE}$, so BC is $\frac{5}{8}$ of AC or AB is $\frac{3}{8}$ of AC then $\triangle{ABF}$ is $\frac{9}{64}$ of $\triangle{ACE}$ so $\triangle ABF = \boxed{18}$

See Also

2006 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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