Difference between revisions of "1967 AHSME Problems/Problem 9"
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\textbf{(B)}\ K \; \text{must be a rational fraction} \\ | \textbf{(B)}\ K \; \text{must be a rational fraction} \\ | ||
\textbf{(C)}\ K \; \text{must be an irrational number} \qquad | \textbf{(C)}\ K \; \text{must be an irrational number} \qquad | ||
| − | \textbf{(D)}\ K\; \text{must be an integer or a rational fraction} \qquad</math> | + | \textbf{(D)}\ K \; \text{must be an integer or a rational fraction} \qquad</math> |
<math>\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}</math> | <math>\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}</math> | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | From the problem we can set the altitude equal to <math>a</math>, the shorter base equal to <math>a-d</math>, and the longer base equal to <math>a+d</math>. By the formula for the area of a trapezoid, we have <math>K=a^2</math>. However, since <math>a</math> can equal any real number <math>(3, 2.7, \pi)</math>, none of the statements <math>A, B, C, D</math> need to be true, so the answer is <math>\fbox{E}</math>. |
== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=8|num-a=10}} | + | {{AHSME 40p box|year=1967|num-b=8|num-a=10}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:36, 16 August 2023
Problem
Let 
, in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:
Solution
From the problem we can set the altitude equal to 
, the shorter base equal to 
, and the longer base equal to 
. By the formula for the area of a trapezoid, we have 
. However, since can equal any real number 
, none of the statements 
need to be true, so the answer is 
.
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
