<p>A small square region inside a larger square is shaded. The diagonal of the larger square is three times the diagonal of the small square in length. What fraction of the area of the larger square is not shaded ? </p>
<p>2/3
1/3
1/9
8/9
cannot be determined</p>
<p>help me understand it thanks</p>
<p>first find the area of the small square by using the 45-45-90 special triangle rule. then find the area of the big triangle (making sure you keep the ratio b/w the small and big triangle) using the same rule and just subtract the two areas and then divide the difference by the total area of the large square. I used x for the diagonal length of the small square so I got that the side length of the small square was x/[2^(1/2)] and the side length of the big square was 3x/[2^(1/2)]. Subsequently, the areas are (x^2)/2 (small square) and (3x^2)/2 (big square).<br>
You end up with....[(9x^2)/2]-[(x^2)/2]=4x^2. And then (4x^2)/[(9x^2)/2]=>(4x^2)[2/(9x^2)]=8/9.</p>
<p>I hope the answer is right. I wrote this instead of doing an essay lol. That's what happens when I get bored...math problems. Sorry for all the parentheses. Btw, 2^(1/2) is the square root of 2. Just in case. :)</p>
<p>I think it's 8/9</p>
<p>if the diagonals are at a ratio of 1:3, then the sides are a ratio of 1:3,
1^2:3^2 (area formula) = 1:9 or 1/9</p>
<p>subtract 1/9 from 1 and you get 8/9</p>
<p>Yep, I got 8/9 too...</p>
<p>Make a square grid 3x3; shade a corner square.
8 out of 9 small squares are unshaded; it's 8/9 of the larger square.</p>
<p>"Make a square grid 3x3; shade a corner square.
8 out of 9 small squares are unshaded; it's 8/9 of the larger square."</p>
<p>That's a clever way of thinking of it, especially to save time. I always tend to look at these questions in terms of algebra, when sometimes they can be pictured much more simply...</p>
<p>All the above stategies are great, but this one can be applied to almost all of the math questions involving variables. </p>
<p>PICK your own #s
Side of large square=3, so diagonal=3<em>Square root of 2, so then the diagonal of the smallar square=Square root of 2(with sides of 1). So 3</em>3=9. 9-(1*1)=8 out of 9. Again, sometimes the above method are better and faster, but you have more chances to make errors with algebra. Just pick your own #s because then you have full control of the problem.</p>
<p>POE lol...no way could it be 2/3, and of course there was enough info, so 8/9 was the obvious choice</p>
<p>If you can't understand the algebra or can't think of it quickly on TestDay, just pick #s. Diagonal of large square is 9<em>SQ2 and the smaller square's diagonal is 3</em>SQ2. So then the sides are 3 and 9, and the areas are 9 and 81. So 81-9=73. 73/81=8/9. Notice how I picked really easy to use #s? If you picked 9 and 3 for the diagonals, you would have to divide them by the SQ2, and that's not fun to do at all! Pick #s that help you out.</p>