Difference between revisions of "1967 AHSME Problems/Problem 9"

Latest revision as of 00:36, 16 August 2023

Problem

Let $K$ , 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:

$\textbf{(A)}\ K \; \text{must be an integer} \qquad \textbf{(B)}\ K \; \text{must be a rational fraction} \\ \textbf{(C)}\ K \; \text{must be an irrational number} \qquad \textbf{(D)}\ K \; \text{must be an integer or a rational fraction} \qquad$ $\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}$

Solution

From the problem we can set the altitude equal to $a$ , the shorter base equal to $a-d$ , and the longer base equal to $a+d$ . By the formula for the area of a trapezoid, we have $K=a^2$ . However, since $a$ can equal any real number $(3, 2.7, \pi)$ , none of the statements $A, B, C, D$ need to be true, so the answer is $\fbox{E}$ .

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.

@@ Line 13: / Line 13: @@
 == 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]]
 {{MAA Notice}}

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Difference between revisions of "1967 AHSME Problems/Problem 9"

Latest revision as of 00:36, 16 August 2023

Problem

Solution

See also