<p>If anyone has it, how do you solve the math problem number 20 in section 2?</p>
<p>There is no question 20…</p>
<p>The one that starts off with “the figure above shows a square with a side of length…”</p>
<p>in the practice test that comes with the nmsqt stuff they give you when you sign up</p>
<p>Aye, the woman of delicious sandwiches is correct. Zachary must be hallucinating.</p>
<p>Yeah, that’s not a question…</p>
<p>are you talking about the question with the square of side length 6 and the weird circle shape thing in the middle?
if you are, then this is how i did it:
i just looked it as a whole square. the shaded part is the square minus the two little “triangular” circle end pieces and the two quarter-circle pieces. Since the side length of the square is 6, the area of the square is 6^2, or 36, and the radius of the circle is 3. First, I represented the two little “triangular” circle end pieces with (36-π 3^2)/2 since the 4"triangular" end pieces are made when the entire circle is subtracted from the square and it’s divided by 2 since only 2 of the 4 pieces are showing. The other 2 pieces are the quarter-circle pieces, which I represented with (π 3^2)/2 since two quarter circle pieces add up to half a circle. Then, since I found out the values of the space around the shaded area, I subtracted these values all from the square to get 36-[[(36-π 3^2)/2]+[((π 3^2)/2)]]
Once the parenthesis parts were divided by 2, I got 36-(18-π 3.5+π 3.5) and since the π3.5s canceled each other out, I then got 36-18 and thus, A.
i hope that made sense! =D</p>