Difficult math problem

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<p>Picture: <a href="http://img4.imageshack.us/img4/4727/31367426.jpg%5B/url%5D"&gt;http://img4.imageshack.us/img4/4727/31367426.jpg&lt;/a&gt;&lt;/p>

<p>Problem: 18) In the figure above, the length of FJ is 12 and the length of each side of paralellogram FGHI is 13. What is the area of the quadrilateral FGHJ?</p>

<p>been stuck at this for like 20 mins, its a diagram one so i drew a quick sketch of it on windows paint lol, i can't seem to figure it out the only thing that i can get that might be useful is that JI is 5.</p>

<p>A - 72.5
B - 186
C - 197
D - 200
E - 216</p>

<p>the correct answer is B - 186
can someone please explain how to get this?</p>

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<p>Well, you can use Pythagorean theorem to get the side JI as 5, and then use one half base x height to get the area of the triangle, which is 30. Looking at the drawing, I couldn’t tell much of this, but it gave the correct answer, so I did this. Assume that the base of the parallelogram is also 13. Then the area of a parallelogram is simply base x height, or 12 x 13, which is 156. That plus the area of the triangle, 30, gives 186. Does that help?</p>

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<p>hm yea but how come you use 12 x 13 and not 13 x 13 because i figured when you get teh area of the triangle the 12 is covered , so why isn’t it 13 x 13 + 30?</p>

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<p>because 13 is not actually the height of the parallelogram. 13 is the slanted side, and 12 is the direct height of both the triangle and the parallelogram.</p>

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<p>ohhh alright makes sense, thank you very much.</p>

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<p>Because you want the height of the parallelogram, and not just multiplying the sides. The twelve comes in simply because (I hope you can see what I’m trying to explain) you need the height of the parallelogram, and if you look at the picture of the diagram in your SAT book, can you kind of see how “dragging” the twelve side to the middle of the parallelogram give you the height? </p>

<p>[Area</a> of a Parallelogram](<a href=“http://www.mathgoodies.com/lessons/vol1/area_parallelogram.html]Area”>http://www.mathgoodies.com/lessons/vol1/area_parallelogram.html)</p>

<p>Check this website out, or at least the red pictures with the height labeled. Now if you put your triangle next to it, the twelve side will be the height. </p>

<p>Can you see that? I hope I explained it properly.</p>

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<p>Ah, that came up while I was writing my response!!</p>

<p>Glad to see you figured it out, doctaJ. Good luck on your SAT!!!</p>

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<p>appreciate it biggie
ah might as well post it in this thread i just had one more question i couldn’t figure out from a practice test </p>

<p>theres no diagram (well you just make your own) but the question goes</p>

<p>What is the perimeter of an equilateral triangle inscribed in a circle of radius 2?
A - 6root3
B - 6
C - 2root3
D - 3
E - root3</p>

<p>answer is A.</p>

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<p>you use 30,60,90 triangle rule</p>

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<p>oops, change root2 to root3 in my drawing.
I made a mistake :stuck_out_tongue:
but the explanation is correct</p>

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<p>^^^A diagram is really needed here, imo. </p>

<p>So, you inscribe your eq. triangle within the circle. What I did from here is bisect all the angles of the triangle, and these bisectors all intersect at what should be the center of the circle (just something I assumed, it seems to make sense, but I didn’t want to prove it). So, the distance from the vertex of the triangle to this intersection thingy is 2 (the radius length). You should also see a right triangle formed, with the hypotenuse equal to this segment, the height equal to a littler segment, the base equal to 1/2 of the side length (the entire side length I’ll call x), and the angle between the base and hypotenuse of 60/2=30 degrees. So, the base of said triangle is x/2. The hypotenuse is 2. The angle is 30. So, the cos(30)=(x/2)/2=x/4. Remembering that cos(30) is root(3)/2, which you can use the unit circle to confirm, we have root(3)/2=x/4, and therefor x=2root(3) using some algebra. Since x is a side length, and we are looking for the perimeter of an equalateral triangle where all side lengths are the same, we know the perimeter is 3x=3*2root(3)=6root(3).</p>

<p>Edit: howon your diagram is hard to see fyi, at least on my computer.</p>