Difference between revisions of "1953 AHSME Problems/Problem 48"

(Created page with "==Problem== If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is...")
 
 
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==Solution==
 
==Solution==
  
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<asy>
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draw((0,0)--(1,3)--(4,3)--(5,0)--cycle);
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draw((0,0)--(4,3));
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draw((4,3)--(4,0));
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draw((3.8,0)--(3.8,0.2)--(4,0.2));
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label("$A$",(0,0),W);
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label("$B$",(1,3),NW);
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label("$C$",(4,3),NE);
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label("$D$",(5,0),E);
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label("$E$",(4,0),S);
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label("1",(2,1.5),NW);
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</asy>
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Let <math>a</math> be the length of the smaller base of isosceles trapezoid <math>ABCD</math>, and <math>1</math> be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is <math>\frac a1=a</math>. Let point <math>E</math> be the foot of the altitude from <math>C</math> to <math>\overline{AD}</math>.
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Since the larger base of the isosceles trapezoid equals a diagonal, <math>AC=AD=1</math>. Since the smaller base equals the altitude, <math>BC=CE=a</math>. Since the trapezoid is isosceles, <math>DE=\frac{1-a}{2}</math>, so <math>AE = 1-\frac{1-a}{2} = \frac{a+1}{2}</math>. Using the [[Pythagorean Theorem]] on right triangle <math>ACE</math>, we have
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<cmath>a^2+\left(\frac{a+1}{2}\right)^2=1</cmath>
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Multiplying both sides by <math>4</math> gives
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<cmath>4a^2+(a+1)^2=4</cmath>
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Expanding the squared binomial and rearranging gives
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<cmath>5a^2+2a-3=0</cmath>
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This can be factored into <math>(5a-3)(a+1)=0</math>. Since a must be positive, <math>5a+3=0</math>, so <math>a=\frac 35</math>. The ratio of the smaller base to the larger base is <math>\boxed{\textbf{(D) } \frac 35}</math>.
 
==See Also==
 
==See Also==
  

Latest revision as of 23:37, 14 February 2020

Problem

If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution

[asy] draw((0,0)--(1,3)--(4,3)--(5,0)--cycle); draw((0,0)--(4,3)); draw((4,3)--(4,0)); draw((3.8,0)--(3.8,0.2)--(4,0.2)); label("$A$",(0,0),W); label("$B$",(1,3),NW); label("$C$",(4,3),NE); label("$D$",(5,0),E); label("$E$",(4,0),S); label("1",(2,1.5),NW); [/asy]

Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$, and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac a1=a$. Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$.

Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$. Since the smaller base equals the altitude, $BC=CE=a$. Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$, so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$. Using the Pythagorean Theorem on right triangle $ACE$, we have \[a^2+\left(\frac{a+1}{2}\right)^2=1\] Multiplying both sides by $4$ gives \[4a^2+(a+1)^2=4\] Expanding the squared binomial and rearranging gives \[5a^2+2a-3=0\] This can be factored into $(5a-3)(a+1)=0$. Since a must be positive, $5a+3=0$, so $a=\frac 35$. The ratio of the smaller base to the larger base is $\boxed{\textbf{(D) } \frac 35}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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All AHSME Problems and Solutions


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