SOLUTION: What is that largest area possible for right triangle in which the sum of the lengths of the two shorter sides is 16? [Solve by completing the square]

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Question 1019175: What is that largest area possible for right triangle in which the sum of the lengths of the two shorter sides is 16? [Solve by completing the square]
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= length of one of the shorter sides (legs) of the right triangle.
16-x= length of the other one of the shorter sides (legs) of the right triangle.
Because the shorter sides (legs) of the right triangle are perpendicular to each other,
we can consider the length of one to be the base of the triangle,
and the length of the other one to be the height.
So, the area is
area=x%2816-x%29%2F2<-->area=-x%5E2%2B16x
Completing the square:
area=%28-x%5E2%2B16x%29%2F2
area=%28-x%5E2%2B16x-64%2B64%29%2F2
area=%28-x%5E2%2B16x-64%29%2F2%2B32
area=%28-1%2F2%29%28x%5E2-16x%2B64%29%2B32
area=%28-1%2F2%29%28x%2B8%29%5E2%2B32
Since the first term is negative or zero,
%28-1%2F2%29%28x%2B8%29%5E2%3C=0 ,
area%3C=32 .
In other words, the largest area possible for that right triangle is highlight%2832%29 .