<p>ok, for number 10 the answer is 35. You know that 7 and 10 are two of the sides. If any of these were to be the hypotenuse, the other side would be smaller than 7/10. (I.e. If the hypotenuse was 10, the side would be the square root of 10^2-7^2 which is less than having a side of ten). Now, if u can recall, the area of a right triangle is one side times another divided by 2. This gives u the max area.
As for 15, I would just guess and check. Though a tedious way to solve the problem, it really isnt all that bad. You know that p and q need to be integers, so plug in random numbers. p could be 4 and q could be 2, so 4-2=2… II works. none of the others work. The answer is B</p>
<p>tyty i forgot that n>0 and included -2. </p>
<p>For 10, I tried to use 10, 7 and 16. I drew a midpoint on 16, but didn’t realize untill you explained it that I can’t draw a midpoint mathematically correct on a non isosceles triangle.</p>
<p>no problem! Good luck if u r taking the SAT tomorrow</p>
<p>I have a problem with number 10. It never specifies that it is a right triangle…</p>
<p>^It asks for greatest possible area; If it wasn’t a right triangle the area would be smaller than 35.</p>