<p>Two sides of a triangle are 6 and 7.</p>
<p>Can the area be:</p>
<li>13</li>
<li>21</li>
<li>24</li>
</ol>
<p>…or perhaps a combination of the above?</p>
<p>I think I should mention that this is from the Classic Red Book; I do have the answer.</p>
<p>I am looking for a solution.</p>
<p>Well it can def. be 21, if 6 and 7 are the two legs of a right triangle.</p>
<p>(I know, that was really obvious, but I have no idea about the other two choices.)</p>
<p>As you can CLEARLY see, the SAT has improved since 1994. Lol.</p>
<p>Use heron's formula.</p>
<p>21 is the max area, because any other leg will give less than optimum area (a right triangle yields biggest area). If the other side is 1 (the smallest it can be), the area is...well u use that rule with the rt s (s-A)(s-b)(s-c)/2 crap, but i dont wanna solve. I think it's only 21 tho, i think 13 is too small.</p>
<p>I believe 24 is impossible. If 7 is the longest side, the others must be shorter and can't have an area of 24. If 7 isn't the longest side, that means it can't be the height or base (if it was, it would be remaining side would be shorter).</p>
<p>agree with NJPitcher, but I'll add that 13 is a possibility, though the third side may not necessarily be an integer. Why? Because the smallest area is 6. Use this interval, between 6 and 21, and both 1 and 2 are right answers.</p>
<p>*correction, smallest area is not 6, it is actually slightly bigger than 3.</p>
<p>13 = 26 without dividing by half
21 = 42 without dividing by half
24 = 48 without dividing by half</p>
<p>Now 6,7 is given so the other side can range greater than 1 and less than 13.</p>
<p>6<em>7 = 42/2 = 21 WORKS
The 24 is a trap, because people will think 6</em>8 = 48 and then end up with 24, but this is not possible, since you can't multiply the hyp by a leg.
The 13 is another trap in that someone might do 6+7.</p>
<p>*also if you were thinking of numbers other than integers on this problem, think again, because rmbr you can solve ALL SAT problems without a calc and so think PROPERTIES not crazy calculations or formulas (herons) even though us pre-calc and calc students immediatly think straight to that...go back to 8th grade mode using the wisdom you have gained over the years lol.</p>
<p>by the way, was this a medium question?</p>
<p>merudh123, why do you even NEED to calculate the areas? just assume an interval, and it's a 10 second problem.</p>
<p>lol I know...I just felt I would explain it.</p>
<p>No one said that they have to be integers.</p>
<p>it doesn't matter. you just gotta know the properties. I mentioned the integer thing in order to show that you don't need to think too complex.</p>
<p>The shortest side can't be less than one. If we say it's 1.1 (it can't be one exactly), then using herons u get rt (7.05)(.05)(1.05)(5.95), which is...1.484, which is well under 13. So yea, 21 and 13 work.</p>
<p>The other side must be between 1 and 13, and not equal to any of those</p>
<p>13 and 21 is the correct answer.</p>
<p>This was question number 25 on the first math section of a test administered on March 1994. It's the first test in The Classic Red Book, 2nd ed.</p>
<p>It's labeled as a hard question.</p>
<p>The sad thing is that by new SAT standards, this is a "hard" question.</p>
<p>Why do they insist on putting out such easy tests and then murdering people for silly mistakes with terrible curves (which seem necessary to disperse the scores at the top)?</p>
<p>Agreede random. I missed what will no doubt be an "easy" question because I misread horribly. I saw a graph, it asked for a point if the graph were moved up 6 points, but i only saw half the graph). It was insanely easy, but there goes a top score. Oh well. I'd rather that half the questions be ridiculously hard; it'd be much more meritorious that way, IMO. Oh well, I'd probably be saying something (the contrary) else had i gotten that one question, this is purely emotional talk. Blah.</p>
<p>Random...this was considered "hard" by the old SAT standards.</p>