<p>what was the answer to it? it looked something like this:
<a href="http://img476.imageshack.us/img476/959/psatproblemma6.png%5B/url%5D">http://img476.imageshack.us/img476/959/psatproblemma6.png</a></p>
<p>and asked for the area. I think the choices were between 2 and 5 or something. anyone know?</p>
<p>The same question stumped me (when I took the test). It was pretty easy though...</p>
<p>Basically, draw lines around the figure to form a parallelogram. Then using the points on the graph and the formula for the area of a triangle. Area of the parallelogram - area of the triangles = area of the figure.</p>
<p>Edit: I do not have my scores back yet. So I can't tell you the answer, only how to get there.</p>
<p>Yep, that's what I found out after the test too...but I'm still hoping against hope that somehow the answer was 4</p>
<p>Just divide it into shapes. The top half is just a triangle (use the normal base x height formula, even though it's skewed), the bottom half was a square and two triangles. 'Twas how I did it anyways.
(I want my scores!!)</p>
<p>"Actually..for the quadrilateral one I got 4.5..what I did was to solve for the equations of both the diagonal lines, then graph them and find the intersection point. From the intersection point I was able to find the distance from vertex to vertex and use 1/2bh to solve for the area of the triangle..that gave me the answer of 4.5"</p>
<p>That's from the october 18 discussion board - i posted that. However, most people on there got 3.5, not 4.5.</p>
<p>I haven't seen the problem, only that picture, but I think Krofh's solution is the quickest and neatest one you'll find. You could also combine the bottom part into a trapezoid if you prefer.</p>
<p>Oh, and if I understand the picture correctly, the area is in fact 7/2.</p>
<p>when did u guys get ur scores back</p>
<p>Those who said 3.5 or, as it was on the test, 3 1/2 are right.</p>
<p>I just checked it out ont he answer manual. The letter answer is B by the way.</p>
<p>Ohio score received today.</p>
