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

(See also)
(Solution)
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size(120);
 
size(120);
 
defaultpen(0.7);
 
defaultpen(0.7);
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);
+
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>
  
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>.
+
Since <math>\triangle BDE</math> is similar to <math>\triangle ABC</math>, and thus <math>BD=10\cdot \dfrac{2}{3}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>.
  
 
==See also==
 
==See also==

Revision as of 21:34, 4 August 2008

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]

Since $\triangle BDE$ is similar to $\triangle ABC$, and thus $BD=10\cdot \dfrac{2}{3}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AHSME Problems and Solutions