Difference between revisions of "1972 AHSME Problems/Problem 1"

(Created page with "Let's test out the four triangles: I: <math>3^2 + 4^2 = 5^2</math> (true) II: <math>4^2 + 7.5^2 = 8.5^2</math> (true) III: <math>7^2 + 24^2 = 25^2</math> The only answer c...")
 
(Add statement)
 
Line 1: Line 1:
 +
== Problem ==
 +
 +
The lengths in inches of the three sides of each of four triangles <math>I, II, III</math>, and <math>IV</math> are as follows:
 +
 +
<math>\begin{array}{rlrl}
 +
\hbox{I}& 3,\ 4,\ \hbox{and}\ 5\qquad &
 +
\hbox{III}& 7,\ 24,\ \hbox{and}\ 25\\
 +
\hbox{II}& 4,\ 7\frac{1}{2},\ \hbox{and}\ 8\frac{1}{2}\qquad &
 +
\hbox{IV}& 3\frac{1}{2},\ 4\frac{1}{2},\ \hbox{and}\ 5\frac{1}{2}.\end{array}</math>
 +
 +
Of these four given triangles, the only right triangles are
 +
 +
<math>\text{(A)  I and II}\qquad
 +
\text{(B)  I and III}\qquad
 +
\text{(C)  I and IV} \quad \\
 +
\text{(D)  I, II, and III} \qquad
 +
\text{(E)  I, II, and IV }</math>
 +
 +
== Solution ==
 +
 
Let's test out the four triangles:
 
Let's test out the four triangles:
  

Latest revision as of 12:37, 16 July 2024

Problem

The lengths in inches of the three sides of each of four triangles $I, II, III$, and $IV$ are as follows:

$\begin{array}{rlrl} \hbox{I}& 3,\ 4,\ \hbox{and}\ 5\qquad & \hbox{III}& 7,\ 24,\ \hbox{and}\ 25\\  \hbox{II}& 4,\ 7\frac{1}{2},\ \hbox{and}\ 8\frac{1}{2}\qquad & \hbox{IV}& 3\frac{1}{2},\ 4\frac{1}{2},\ \hbox{and}\ 5\frac{1}{2}.\end{array}$

Of these four given triangles, the only right triangles are

$\text{(A)  I and II}\qquad  \text{(B)  I and III}\qquad \text{(C)  I and IV} \quad \\ \text{(D)  I, II, and III} \qquad \text{(E)  I, II, and IV }$

Solution

Let's test out the four triangles:

I: $3^2 + 4^2 = 5^2$ (true)

II: $4^2 + 7.5^2 = 8.5^2$ (true)

III: $7^2 + 24^2 = 25^2$

The only answer choice with I, II, and III in the answer is $\fbox{D} \rightarrow$ I, II, and III.