# CB practice psat math problem

<p>the last one, of the fourth section grid-ins, i am not sure how to solve, or even begin, figuring out this problem...could someone enlighten me? its been a while since i have taken geometry, so i know the midpoint information that was given is very significant, but i dont quite remember how.</p>

<p>it was the only math problem i missed on the practice test...darn, almost had an 80...sigh</p>

<p>doesn't it suck that missing 1 pt. on raw score drops you to a 77?</p>

<p><em>sigh</em> geometry it's been a while lol, this is the way i'd do it since they tell mid point.
its not the correct way though... since i always do weird stuff in the math section cuz i have forgotten basic algebra/geometry</p>

<p>draw two lines from line QS to about mid way of PT so you have 4 triangles in one.
Parallelogram takes up 3 of the 4. Therefore 3/4
:-\ sorry if that isn't much help</p>

<p>You know that was a very hard problem. I'm sure there's a formula for solving it, but basically what I did was just assume that it was an equalaterial triangle with sides say 10, solved the area for the triangle, and solved the area of the trapezoid based on the formula (b1 + b2)(H) / 2. The area of the triangle turned out to be 10(5(3^(1/2)) / 2, or 25(3^(1/2)). Since we assumed an equilateral triangle, the top of the trapezoid would be 5. The slopy sides would also be 5, and the bottome would be 10. Solve for the height of the trapezoid with 25 - (5/2)^2 = 75 / 2. Take the square root of that which is 5(3^(1/2)) / 2. The area of the trapezoid is 75(3^(1/2)) / 4. Divide this by 25(3^(1/2)) and you got (3/4).</p>

<p>i dont get it..</p>

<p>Another way is to <em>practically</em> draw FOUR triangles, see? it works; or
midline segment theorem --> 3bh/2 / 2h 2b/2</p>

<p>are we supposed to know any formulas for the SAT, besides the stuff they give us?</p>

<p>No, but it's helpful if you do. Sequences and crap for the NEW SAT Help. Like cal and precal help me!</p>

<p>so there must be a logical and simple way to figure this math ^ out, besides using formulas...</p>

<p>Yeah, it sucks that missing one brings you down to a 77. I missed that one too, but hey, I made a valiant guess...lol, 2/3. It was 3/4. I almost had it by blindly guessing, it was either 2/3 or 3/4, grrr....</p>

<p>Solution</p>

<p>In a trapezoid, two sides are parallel, so QS is parallel to PT. That makes angle P and angle Q equal, as well as angle T and angle S. They share angle R. So, by the AAA thingy-majig, triangle QRS and triangle PRT are similar. QR = x, and PR = 2x, so the larger triangle is greater by a factor of two.</p>

<p>I'm preeeeetty sure that this means the heights are also differing by a factor of two. So, the height of the trapezoid is h, and the height of the triangle, 2h. The shorter base of the trapezoid, QS, is z, while the longer base of the trapezoid (and base of the triangle PRT) is 2z. </p>

<p>Area of triangle (obviously): half of base x height</p>

<p>Area of trapezoid (glares daggers at CB for not putting that in there, although I shoulda remembered this): height times (base1 + base2) all over two. </p>

<p>So the area of triangle PRT is 4zh over 2, or 2zh. </p>

<p>The area of the trapezoid is 3zh over 2, or 1.5zh. </p>

<p>Divide the area of the trapezoid by the area of the triangle, and you cancel out all the random variables (z is for zoogie, by the way; don't ask where that came from; h is for height) and are left with 1.5 over 2. Or, 3 over 4.</p>

<p>Damnit, if I only had the stupid trapezoid area formula at the time, I woulda gotten it. It's not like I would have run out of time thinking it out, I had around ten minutes left when I got to that problem.</p>

<p>(By the way, I just did this now...took me a bit, and I still think it's a very cruel problem. They gave you the formula for the area of a rectangle! A RECTANGLE! And not a trapezoid. Gaaaaar...)</p>