<p>What is the greatest possible area of a triangle with one side length 4 and another side length 9?</p>
<p>A)9
B)18
C)36
D)54
E)72</p>
<p>I am not sure of the answer, please tell of how you derived your answer. Thanks for the help.</p>
<p>The answer is 18. A right triangle makes the most use of its area. So if its legs are length 9 and 4. Then you get the area by A=.5(9)(4) which gives us 18. Hope that helps.</p>
<p>yea that is what i thought, thanks</p>
<p>If you didn’t know a right triangle maximizes its area then what you do is this:</p>
<p>area= ab/2*sinQ (Q is the angle between side A and B)
sinQ < or = 1 (think about sin graph)</p>
<p>Therefore to maximize area, sinQ must be 1
So Q=90 degrees
thus ab/2 x sin90
4x9/2=[18]</p>
<p>yea thanks a lot</p>
<p>How do you know 9 isn’t the hypotenuse?</p>
<p>
</p>
<p>The hypotenuse does not matter in THIS problem. The question is “What is the greatest possible area of a triangle with one side length 4 and another side length 9?” </p>
<p>Since the area only depends from the base and the height, there is no need to worry about the third side.</p>
<p>9 was the ‘hypotenuse,’ we would have a 9-9-4 triangle.</p>
<p>This is isosceles and there is obviously no hypotenuse in an isosceles triangle.</p>
<p>Also, the hypotenuse is irrelevant in this problem.</p>