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Difference between revisions of "1965 AHSME Problems/Problem 16"

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Draw <math>\overline{AE}</math>, as seen in the diagram. From the problem, we know that <math>\overline{EB}</math> and <math>\overline{AD}</math> are [[median of a triangle|medians]] of <math>\triangle ACE</math>. Let <math>G</math> be the midpoint of <math>\overline{AE}</math>. Then, <math>\overline{CG}</math> is also a median of <math>\triangle ACE</math>, and it goes through <math>\triangle ACE</math>'s centroid, <math>F</math>. Because medians divide their triangle into <math>6</math> smaller triangles of equal area, we know that <math>[\triangle DFE]=\frac{1}{6}[\triangle ACE]</math>. Because <math>[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450</math>, <math>[\triangle DFE]=\frac{450}{6}=75</math>. Thus, our answer is <math>\boxed{\textbf{(C) }75}</math>.
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Draw <math>\overline{AE}</math>, as seen in the diagram. From the problem, we know that <math>\overline{EB}</math> and <math>\overline{AD}</math> are [[median of a triangle|medians]] of <math>\triangle ACE</math>. Let <math>G</math> be the midpoint of <math>\overline{AE}</math>. Then, <math>\overline{CG}</math> is also a median of <math>\triangle ACE</math>, and it goes through <math>\triangle ACE</math>'s [[centroid]], <math>F</math>. Because medians divide their triangle into <math>6</math> smaller triangles of equal area, we know that <math>[\triangle DFE]=\frac{1}{6}[\triangle ACE]</math>. Because <math>[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450</math>, <math>[\triangle DFE]=\frac{450}{6}=75</math>. Thus, our answer is <math>\boxed{\textbf{(C) }75}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1965|num-b=15|num-a=17}}
 
{{AHSME 40p box|year=1965|num-b=15|num-a=17}}
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{{MAA Notice}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Latest revision as of 15:58, 18 July 2024

Problem

Let line $AC$ be perpendicular to line $CE$. Connect $A$ to $D$, the midpoint of $CE$, and connect $E$ to $B$, the midpoint of $AC$. If $AD$ and $EB$ intersect in point $F$, and $\overline{BC} = \overline{CD} = 15$ inches, then the area of triangle $DFE$, in square inches, is:

$\textbf{(A)}\ 50 \qquad \textbf{(B) }\ 50\sqrt {2} \qquad \textbf{(C) }\ 75 \qquad \textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad \textbf{(E) }\ 100$

Solution

[asy] draw((0,0)--(30,0)); dot((0,0)); label("C", (-2,-2)); dot((30,0)); label("E", (32,-2)); dot((15,0)); label("D",(15,-2)); dot((0,15)); label("B",(-2,15)); draw((0,0)--(0,30)--(30,0)); dot((0,30)); label("A",(-2, 32)); dot((15,15)); label("G", (17,17)); draw((0,0)--(15,15)); dot((10,10)); label("F", (10,12)); draw((0,15)--(30,0)); draw((15,0)--(0,30)); markscalefactor=0.25; draw(rightanglemark((0,30),(0,0),(30,0))); label("$15$",(-2,7.5)); label("$15$",(-2,22.5)); label("$15$",(7.5,-2)); label("$15$",(22.5,-2)); [/asy]

Draw $\overline{AE}$, as seen in the diagram. From the problem, we know that $\overline{EB}$ and $\overline{AD}$ are medians of $\triangle ACE$. Let $G$ be the midpoint of $\overline{AE}$. Then, $\overline{CG}$ is also a median of $\triangle ACE$, and it goes through $\triangle ACE$'s centroid, $F$. Because medians divide their triangle into $6$ smaller triangles of equal area, we know that $[\triangle DFE]=\frac{1}{6}[\triangle ACE]$. Because $[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450$, $[\triangle DFE]=\frac{450}{6}=75$. Thus, our answer is $\boxed{\textbf{(C) }75}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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