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Grade 12th passDifferential Calculus

If AB is a diameter of a circle and C is any point on its circumference,then when isosceles, the area of triangle ABC is : (a)maximum (b)minimum. I know the answer solving it by the conventional process i.e. putting a(area)/dx=0. But that is a lengthy one. So please tell if any short trick is there to get the answer faster.

Profile image of SKS
12 Years agoGrade 12th pass
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1 Answer

Profile image of nkhlshd
12 Years ago
Plot the circle given in the problem on the xy plane as x^2+y^2=r^2 .Take A=(-r,0) B=(r,0) and C=(0,r) where r is real and positive & O=(0,0). Then length of AB=2r. For any point D (not equal to C) on the circumference , the length of the perpendicular from D to AB is strictly less than OC. Hence the claim.

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