Hard math question for all you math gurus

<p>how can you figure this question out and quickly ? ironically its a level 3 question and the only one I got wrong on a pre 2005 official practice test</p>

<p><a href="http://www.aplusclick.com/image/png/1055.png%5B/url%5D"&gt;http://www.aplusclick.com/image/png/1055.png&lt;/a&gt;&lt;/p>

<p>There is no question, just a picture.</p>

<p>oh yes sorry. What is the area of the red/divided by the area of the triangle.</p>

<p>…(E) Not enough information?
Where are the given lengths?</p>

<p>Well, you wouldn’t need the actual lengths, but what’s “the triangle”?</p>

<p>area of red/area of entire rectangle =1/4?</p>

<p>It’s 1/4 because the left and right triangles take up 1/2 of the rectangles area since they are formed by the diagonals and then the other three triangle on the top must be equivalent to the three on the bottom due to the vertical angles theorem and since one of each size is shaded green, that makes 1/2 of the remaining area green as well totaling 3/4 green and thus 1/4 red. Hope you could understand what I was saying because I had trouble putting it into words haha.</p>

<p>to look at it visually, imagine the green triangle on the top switching with the the only red triangle on the bottom half. the since you have 2 diagonals thay divide the rectangle into 4 equal pieces, you can see that only 1/4 of it is left red.</p>

<p>Well the lengths could be in terms of x or some other variable. The question is pretty broad and ambiguous but I agree the area of the red region compared to the green is 1/4 just by winging it.</p>

<p>It really depends on the length. It can be 1 /100 or 1/2 or anything. Was it in the official test? Because I didn’t see anything like this.</p>

<p>Here is the complete problem statement. This is a level 3 problem according to the test writers. I will reveal the answer after everyone has looked at the complete question.</p>

<p>[SAT</a> Question 2004-05test Q13 - Dabral’s library](<a href=“http://www.screencast.com/t/K95uvlkxh9aQ]SAT”>http://www.screencast.com/t/K95uvlkxh9aQ)</p>

<p>I think it is 1/8</p>

<p>I’d say its 1/4 but</p>

<p>As other people have said, it should be 1/4.</p>

<p>First, the triangles created by the diagonals of a rectangle all have equal area. So the left and rightmost triangles are congruent, and they account for 1/2 of the total area. Now, of the remaining section, it is easy to see that half of it is shaded and half of it is not. You could formally prove that each triangle is congruent with its partner on the bottom, but I don’t think it’s necessary (also, it would be hard to explain without labeled angles/sides). So the shaded area makes up one half of the the two big triangles on the top and bottom. Since those two triangles only account for half of the area of the rectangle, the shaded are is 1/2 * 1/2 = 1/4.</p>

<p>The picture given by SATquantum is so much better its obviously 1/4.</p>

<p>How do you know that the left and right triangle is 1/2 the area</p>

<p><a href=“https://www.khanacademy.org/math/geometry/basic-geometry/perimeter_area_tutorial/v/area-of-diagonal-generated-triangles-of-rectangle-are-equal[/url]”>https://www.khanacademy.org/math/geometry/basic-geometry/perimeter_area_tutorial/v/area-of-diagonal-generated-triangles-of-rectangle-are-equal&lt;/a&gt;&lt;/p&gt;

<p>He explains it better than I would :P</p>

<p>The answer is indeed (B) 1/4, which many pointed out correctly.</p>

<p>Here is one possible solution. </p>

<p>[Solution</a> to Q13 SAT 2004-05 - Dabral’s library](<a href=“http://www.screencast.com/t/1uHlJpSGHt3]Solution”>http://www.screencast.com/t/1uHlJpSGHt3)</p>

<p>One can also use a symmetry argument without resorting to any algebra. Other approaches proposed by Hittce and Extractum11 also work.</p>

<p>Here is another similar shaded area question from the Official SAT test. I don’t remember which year this one is from. I will post the answer later.</p>

<p>[SAT</a> Question Online Tests Q16 - Dabral’s library](<a href=“http://www.screencast.com/t/oztUjgdG3Wee]SAT”>http://www.screencast.com/t/oztUjgdG3Wee)</p>

<p>nvm…i understood</p>

<p>Nope there is no M level III .</p>

<p>Btw, answer to the second question is 5/8</p>