<p>What is the greatest possible area of a triangle with
one side of length 7 and another side of length 10?
(A) 17
(B) 34
(C) 35
(D) 70
(E) 140 </p>
<p>The answer is choice C, and I know how to get it, but I'm wondering why the answer can't be 80? It can be proposed that a third side is 16 (a + b > c, 10 + 17 > 16) and then the largest side would be .5 x (5) x 16 or .5 x (7) x 16, which are both 80.</p>
<p>take the side of length 7. If you ‘connect’ that side to the side with length 10, you can add another side to make a triangle, right?
Well if you think about it, the maximum area will occur when the 7 side and the 10 side are at right-angles.</p>
<p>So using the formula for area of a triangle:
A = 1/2bh</p>
<p>The area would be:
A = 1/2(7)(10)
A = 35</p>
<p>So C.</p>
<p>How come the maximum area will occur when it’s a right triangle?</p>
<p>The simple answer is that when the sides are perpendicular, the height of the triangle is greatest.</p>
<p>If you know a little trig, and you think of the area of the triangle as (1/2)bc*sin(A), b and c are fixed, and sin(A) is at its greatest when angle A measures 90°.</p>
<p>“The simple answer is that when the sides are perpendicular, the height of the triangle is greatest.”</p>
<p>Never knew that</p>
<p>Is this common SAT-taker knowledge?</p>
<p>It’s not, but an average Joe would just take those 2 numbers and do the area of a triangle formula and luckily get it right</p>
<p>The area of triangle ABC is ( AB x AC x sin A ) / 2 or the square root of p(p-AB)(p-BC)(p-CA) where p = (AB+BC+CA) / 2</p>
<p>If you use the first formula you can immediate see that the area is =< 35 since sinA =< 1.
I don’t know what formula you use when assuming that the third size is 16.</p>