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Difference between revisions of "1995 AHSME Problems/Problem 8"

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size(120);
 
size(120);
 
defaultpen(0.7);
 
defaultpen(0.7);
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pair  A = (0,6), B = (8,0), C= (0,0), D = (56/15,3.2), E = (4/3,0), F = (0, 3), G = (38/15,1.6);
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pair  A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6);
 
draw(A--B--E--D--E--B--C--A--B--cycle);
 
draw(A--B--E--D--E--B--C--A--B--cycle);
 
label("\(A\)",A,W);
 
label("\(A\)",A,W);
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</asy>
 
</asy>
  
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Since <math>\triangle BED</math> is similar to <math>\triangle ABC</math>, it is a 3-4-5 triangle, and thus <math>BD=4\cdot \dfrac{4}{3}=\dfrac{16}{3}\Rightarrow \boxed{\mathrm{(B)}}</math>.
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<math>\triangle BAC</math> is a <math>6-8-10</math> right triangle with hypotenuse <math>AB = 10</math>.
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<math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 90^\circ</math> and <math>B=B</math>, and thus <math>\frac{BD}{BA} = \frac{DE}{AC}</math>.
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 +
Solving the above for <math>BD</math>, we get <math>BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>.
  
 
==See also==
 
==See also==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:58, 5 July 2013

Problem

In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }$

Solution

[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6); draw(A--B--E--D--E--B--C--A--B--cycle); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,SW); label("\(D\)",D,NE); label("\(E\)",E,S); label("\(6\)",F,W); label("\(4\)",G,NW); [/asy]

$\triangle BAC$ is a $6-8-10$ right triangle with hypotenuse $AB = 10$.

$\triangle BDE$ is similar to $\triangle BAC$ by angle-angle similarity since $E=C = 90^\circ$ and $B=B$, and thus $\frac{BD}{BA} = \frac{DE}{AC}$.

Solving the above for $BD$, we get $BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
All AHSME Problems and Solutions

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