Difference between revisions of "1960 AHSME Problems/Problem 13"

(Created page with "==Problem== The polygon(s) formed by <math>y=3x+2, y=-3x+2</math>, and <math>y=-2</math>, is (are): <math>\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text...")
 
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==Solution==
 
==Solution==
 
Graph the equations on the coordinate grid.
 
  
 
[[Image:B9CCCDD9-3865-4F51-8308-230C4C1796F4.jpeg|350px]]
 
[[Image:B9CCCDD9-3865-4F51-8308-230C4C1796F4.jpeg|350px]]
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Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>.  That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>.
 
Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>.  That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1960|num-b=12|num-a=14}}
 
{{AHSME box|year=1960|num-b=12|num-a=14}}

Revision as of 19:52, 8 May 2018

Problem

The polygon(s) formed by $y=3x+2, y=-3x+2$, and $y=-2$, is (are):

$\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$

Solution

B9CCCDD9-3865-4F51-8308-230C4C1796F4.jpeg

The points of intersection of two of the lines are $(0,2)$ and $(\pm \frac{4}{3} , -2)$, so use the Distance Formula to find the sidelengths.

Two of the side lengths are $\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}$ while one of the side lengths is $4$. That makes the triangle isosceles, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1960 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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