Difference between revisions of "1953 AHSME Problems/Problem 48"
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==Solution== | ==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== | ==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:
Solution
Let be the length of the smaller base of isosceles trapezoid , and be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is . Let point be the foot of the altitude from to .
Since the larger base of the isosceles trapezoid equals a diagonal, . Since the smaller base equals the altitude, . Since the trapezoid is isosceles, , so . Using the Pythagorean Theorem on right triangle , we have Multiplying both sides by gives Expanding the squared binomial and rearranging gives This can be factored into . Since a must be positive, , so . The ratio of the smaller base to the larger base is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 47 |
Followed by Problem 49 | |
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