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

Latest revision as of 22:37, 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 $\boxed{\textbf{(D) } \frac 35}$ .

1953 AHSC (Problems • Answer Key • Resources)
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Difference between revisions of "1953 AHSME Problems/Problem 48"

Latest revision as of 22:37, 14 February 2020

Problem

Solution

See Also

@@ Line 12: / Line 12: @@
 ==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 <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>.
+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
+<cmath>a^2+\left(\frac{a+1}{2}\right)^2=1</cmath>
+Multiplying both sides by <math>4</math> gives
+<cmath>4a^2+(a+1)^2=4</cmath>
+Expanding the squared binomial and rearranging gives
+<cmath>5a^2+2a-3=0</cmath>
+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==