Determine the maximum area of a triangle that can be inscribed in the first quadrant of a circle with a radius of 10root2 where one of the sides must be on the x-axis and one vertex at the circumference of the circle. This question i have no clue how to do.
Intuitively the maximum area of this triangle will occur when the 2 shorter sides are equal.
If this isn't obvious then you should be able to see this by doing some simple substitutions and as the length of the 2 shorter sides converge the area will be a max.
Given this the base (and height) length can be found by using pythagora's theorem using x as the base
x^2 + x^2 = (10 sqrt(2))^2
....
yields x = sqrt(10)
Then Area = 0.5 *sqrt(10)* sqrt(10) = 50 units squared.
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