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Difference between revisions of "1959 AHSME Problems/Problem 10"

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== Solution ==  
 
== Solution ==  
  
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Note that <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2} = 10.8</math>. Thusly, our answer is <math>\boxed{\text{(D)}}</math>, and we are done.
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Note that <math>[\triangle ABC]=[\triangle ADE]</math>, so <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}</math>.
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==See also==
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{{AHSME 50p box|year=1959|num-b=9|num-a=11}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Revision as of 12:15, 21 July 2024

Problem

In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$, a point $D$ is taken on $AB$ at a distance $1.2$ from $A$. Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$. Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$

Solution

Note that $[\triangle ABC]=[\triangle ADE]$, so $\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE$. Since $\angle BAC = \angle DAE$, we have $AB*AC = AD*AE$, so that $3.6*3.6 = 1.2*AE$. Therefore, $AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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