1964 AHSME Problems/Problem 29

Revision as of 16:43, 5 March 2014 by Bobthesmartypants (talk | contribs) (Created page with "==Problem== In this figure <math>\angle RFS=\angle FDR</math>, <math>FD=4</math> inches, <math>DR=6</math> inches, <math>FR=5</math> inches, <math>FS=7\dfrac{1}{2}</math> inches...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The lenth of $RS$, in inches, is:

$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$

[asy] import olympiad;  draw((0,0)--7.5*dir(0)); draw((0,0)--5*dir(55)); draw((0,0)--4*dir(140)); draw(4*dir(140)--5*dir(55)--7.5*dir(0)); markscalefactor=0.1; draw(anglemark((0,0),4*dir(140),5*dir(55))); draw(anglemark(7.5*dir(0),(0,0),5*dir(55))); label("$F$",(0,0),dir(240)); label("$D$",4*dir(140),dir(180)); label("$R$",5*dir(55),dir(80)); label("$S$",7.5*dir(0),dir(-25)); label("$4$",2*dir(140),dir(230)); label("$5$",2.5*dir(55),dir(155)); label("$6$",(4*dir(140)+5*dir(55))/2,dir(100)); label("$7\frac{1}{2}$",3.75,dir(-90)); [/asy]