<p>Hey guys I came across something unusual on an SAT math question. This following question was from a previous collegeboard SAT practice test:
If a triangle has sides of 7 and 10, then what is the greatest possible area?
a. 17
b. 34
c.35
d. 70
e. 140
The collegeboard says the answer is 35 as c. but when I use heron's law and try to calculate area by putting in a side between 3<x<17 I get numbers greater then 35. For example, I plugged in 15 as the mystery side, used heron's law and got an area of 54. I suppose the collegeboard wants you to assume it is right but how can you? Anyone want to explain?</p>
<p>Take a look…</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/667622-need-help-math-q.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/667622-need-help-math-q.html</a></p>
<p>Let theta be the angle between the sides of length 7 and 10. Then the area can be written as</p>
<p>A = 7<em>10</em>sin(theta)/2 = 35 sin(theta)</p>
<p>sin(theta) is maximized when theta = pi/2, so the answer is 35.</p>
<p>Also, Heron’s formula is quite long and tedious. If m is the unknown side, then semiperimeter s = (17+m)/2. Then you are trying to maximize</p>
<p>sqrt(s(s-10)(s-7)(s-m)), equivalent to maximizing s(s-10)(s-7)(s-m), which is a quartic equation. You’ll probably need calculus to maximize the area. Then you’ll get something like m = sqrt(149), only to realize that the triangle is a right triangle.</p>
<p>If OP has a graphing calculator, s/he can graph that thing on the calculator and find the maximum that way. It’s still very tedious and you run a high risk of punching in some numbers wrong, or forgetting some parentheses.</p>
<p>Use the area formula that rspence posted. A = a<em>b</em>sin(c). Remember that sine = 1 when the angle is 90 degrees, so this will give you the maximum area of the triangle.</p>
<p>Heron’s is only useful for when you know all three sides and want the semiperimeter, because solving for one of the sides if you know the area is a pain in the ass.</p>
<p>Also, no idea how you got 54 for the area with side lengths of 7, 10, 15. Check your work - [area</a> of triangle with sides 7, 15, 10 - Wolfram|Alpha](<a href=“area of triangle with sides 7, 15, 10 - Wolfram|Alpha”>area of triangle with sides 7, 15, 10 - Wolfram|Alpha)</p>
<p>s = (15+10+7)/2 = 32/2 = 16
A = sqrt((16)(16-15)(16-10)(16-7))
A = sqrt(16<em>1</em>6*9)
A = 12sqrt(6) which is about 30 (mental math, that number is just an estimate)</p>
<p>This is a really simple problem and there’s no need to complicate it with calculus. </p>
<p>A= 1/2(10)(7) = 35 :)</p>
<p>@aares1, 35 is correct, but when explaining solutions you should at least prove that 35 is optimal.</p>
<p>The solution is simple. To maximize the area, set the two sides next to each other at 90 degrees because this will maximize area for any given dimensions. We don’t need to know the 3rd side’s dimension. Just solve for 1/2 x 7 x 10. That’s 35.</p>