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

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 $\triangle ABF = \boxed{18}$

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 ==Solution==
-{{solution}}
+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>.
+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
+then  <math>\triangle{ABF}</math> is <math>\frac{9}{64}</math> of <math>\triangle{ACE}</math> so
+<math>\triangle ABF = \boxed{18}</math>
 ==See Also==

2006 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 5"

Latest revision as of 02:20, 13 January 2019

Problem

Solution

See Also