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Difference between revisions of "1964 AHSME Problems/Problem 29"

(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...")
 
(→‎Solution)
 
(3 intermediate revisions by one other user not shown)
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==Problem==
 
==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. The lenth of <math>RS</math>, in inches, is:
+
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. The length of <math>RS</math>, in inches, is:
  
 
<math>\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</math>
 
<math>\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</math>
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label("$7\frac{1}{2}$",3.75,dir(-90));
 
label("$7\frac{1}{2}$",3.75,dir(-90));
 
</asy>
 
</asy>
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 +
==Solution==
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 +
We examine <math>\triangle RDF</math> and <math>\triangle SFR</math>.  We are given <math>\angle RDF \cong \angle SFR</math>. Also note that <math>\frac{SF}{RD} = \frac{7.5}{6} = 1.25</math> and <math>\frac{FR}{DF} = \frac{5}{4} = 1.25</math>, so <math>\frac{SF}{RD} = \frac{FR}{DF}</math>.
 +
 +
If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence.  Therefore, the third pair of sides must also be in the same proportion, so
 +
 +
<math>\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25</math>
 +
 +
<math>\frac{RS}{FR} = 1.25</math>
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<math>\frac{RS}{5} = 1.25</math>
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 +
<math>RS = 6.25</math>, which is answer <math>\boxed{\textbf{(E) }}</math>
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 +
==Solution 2==
 +
 +
Let <math>\angle RFS = \angle FDR = \theta</math>. By Cosine Law, we have:
 +
<cmath>
 +
\begin{align*}
 +
RF^2 &= DR^2 + DF^2 - 2(DR)(DF)cos(\theta) \\
 +
5^2 &= 6^2 + 4^2 - 2(6)(4)cos(\theta) \\
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cos(\theta) &= \frac{6^2+4^2-5^2}{2(6)(4)} = \frac{9}{16}
 +
\end{align*}
 +
</cmath>
 +
Applying Cosine Law again in <math>\triangle RFS</math>, we have:
 +
<cmath>
 +
\begin{align*}
 +
RS^2 &= RF^2 + FS^2 - 2(RF)(FS)cos(\theta) \\
 +
RS^2 &= 25 + (\frac{15}{2})^2 - 2(5)(\frac{15}{2})(\frac{9}{16}) \\
 +
RS^2 &= \frac{625}{16}
 +
\end{align*}
 +
</cmath>
 +
giving us <math>RS = \frac{25}{4}</math>, which is the answer <math>\boxed{\textbf{(E)}}</math>. -nullptr07
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 +
==See Also==
 +
{{AHSME 40p box|year=1964|num-b=28|num-a=30}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
 +
{{MAA Notice}}

Latest revision as of 21:59, 29 June 2023

Problem

In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length 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]

Solution

We examine $\triangle RDF$ and $\triangle SFR$. We are given $\angle RDF \cong \angle SFR$. Also note that $\frac{SF}{RD} = \frac{7.5}{6} = 1.25$ and $\frac{FR}{DF} = \frac{5}{4} = 1.25$, so $\frac{SF}{RD} = \frac{FR}{DF}$.

If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so

$\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25$

$\frac{RS}{FR} = 1.25$

$\frac{RS}{5} = 1.25$

$RS = 6.25$, which is answer $\boxed{\textbf{(E) }}$

Solution 2

Let $\angle RFS = \angle FDR = \theta$. By Cosine Law, we have: \begin{align*} RF^2 &= DR^2 + DF^2 - 2(DR)(DF)cos(\theta) \\ 5^2 &= 6^2 + 4^2 - 2(6)(4)cos(\theta) \\ cos(\theta) &= \frac{6^2+4^2-5^2}{2(6)(4)} = \frac{9}{16} \end{align*} Applying Cosine Law again in $\triangle RFS$, we have: \begin{align*} RS^2 &= RF^2 + FS^2 - 2(RF)(FS)cos(\theta) \\ RS^2 &= 25 + (\frac{15}{2})^2 - 2(5)(\frac{15}{2})(\frac{9}{16}) \\ RS^2 &= \frac{625}{16} \end{align*} giving us $RS = \frac{25}{4}$, which is the answer $\boxed{\textbf{(E)}}$. -nullptr07

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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
All AHSME Problems and Solutions

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