Q1.prove that maximum area of right triangle inscribed in a circle is isosceles
Asked by | 6th Mar, 2009, 08:46: PM
Given a circle with radius r.
Let triangle ABC be a right triangle inscribed in the circle.
Since the triangle is a right triangle, hence one of the angle is 90o.
Let angle C = 90o
Then AB is diameter of the circle, since angle in semi circle is 90o
Hence, AB= 2r, let BC=x and AC = y
Area of traingle ABC = A = 1/2 * x*y
Applying pythagoras theorem in triangle ABC, we het
y = ((2r)2-x2) = (4r2-x2)
Putting this value of y in area
A = 1/2 * x * (4r2-x2)
Differentiating A with respect to x, we get
Answered by | 20th Mar, 2009, 12:38: PM
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