1953 AHSME Problems/Problem 48
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
![[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]](http://latex.artofproblemsolving.com/a/e/7/ae7897dd2f5940ecbbd0857b8c6dd200a7e12355.png)
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
![\[a^2+\left(\frac{a+1}{2}\right)^2=1\]](http://latex.artofproblemsolving.com/c/f/a/cfa93f0905263b79480af39c0c867c301e3ec199.png)
Multiplying both sides by 
gives
![\[4a^2+(a+1)^2=4\]](http://latex.artofproblemsolving.com/3/1/1/3118d14c42c205e5e91012a8754b88c8d7869be3.png)
Expanding the squared binomial and rearranging gives
![\[5a^2+2a-3=0\]](http://latex.artofproblemsolving.com/3/9/2/392b6c08f8a85bb3ec23687c8d63b2ef803cbe0e.png)
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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