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>. | + | 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== | ||
Revision as of 12:51, 18 July 2024
Problem
Let line 
be perpendicular to line 
. Connect 
to 
, the midpoint of , and connect 
to 
,
the midpoint of . If 
and 
intersect in point 
, and 
inches,
then the area of triangle 
, in square inches, is:
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]](http://latex.artofproblemsolving.com/a/3/c/a3c67da2c2cc210104d19ff8a73b413ab9841ab7.png)
Draw 
, as seen in the diagram. From the problem, we know that 
and 
are medians of 
. Let 
be the midpoint of . Then, 
is also a median of , and it goes through 's centroid, . Because medians divide their triangle into 
smaller triangles of equal area, we know that ![$[\triangle DFE]=\frac{1}{6}[\triangle ACE]$](http://latex.artofproblemsolving.com/d/8/4/d845a47c4053dd4e1393a5a76deb6a04663a1ed3.png)
. Because ![$[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450$](http://latex.artofproblemsolving.com/2/2/8/2282e74539cd3c488ca05dd9e590c008adb6a1ae.png)
, ![$[\triangle DFE]=\frac{450}{6}=75$](http://latex.artofproblemsolving.com/4/6/c/46c99d5f81dbebb47b299ee526184231ee2807ac.png)
. Thus, our answer is 
.
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 | ||
