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

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 $BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$ .

1995 AHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1995 AHSME Problems/Problem 8"

Latest revision as of 12:58, 5 July 2013

Problem

Solution

See also