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$AB$ is a diameter of a circle and $C$ is any point on the circle. Show that the area of $ABC$ is maximum, when it is isosceles.
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Given, AB=2r and angle ACB = 90° (since angle in the semicircle is always 90°)
Let AC=x and BC=y
Therefore, (2r)^2 = x^2 + y^2
So, y^2 = 4r^2  x^2
So, y = sqrt(4r^2  x^2) ..........(i)
Now, area of triangle ABC = 1/2*x*y
= 1/2*x*sqrt(4r^2  x^2) [using Eq. (i)]
Now, differentiating both sides with respect to x, we get
dA/dx = 1/2*[x*(1/2)*(4r^2  x^2)^(1/2)*(2x) + (4r^2  x^2)^(1/2)*1]
= 1/2*[(x^2)/(sqrt(4r^2  x^2)) + sqrt(4r^2  x^2)]
= 1/2*[(x^2 + 2r^2)/(sqrt(4r^2  x^2))]
Now, dA/dx = 0
Therefore, x^2 + 2r^2 = 0
So, r^2 + (1/2)x^2
So, r = (1/sqrt(2))*x
Therefore, x = r*sqrt(2)
Now, differentiating both sides with respect to x again, we get
d^2A/dx^2 = [2x*(sqrt(4r^2  x^2)) + (2r^2  x^2)*(2x)]/[2*(4r^2  x^2)^(3/2)]
= [16*sqrt(2)*r^3 + 8*sqrt(2)*r^3]/[2*(2r^2)^(3/2)]
= 2*sqrt(2) < 0
For x = r*sqrt(2), the area of the triangle is maximum.
For x = r*sqrt(2), y = sqrt(2)*r
Since, x = r*sqrt(2) = y
Hence, the triangle is isosceles.
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Question Text  $AB$
is a diameter of a
circle and $C$
is any point on the
circle. Show that the area of $ABC$
is maximum, when it is
isosceles.

Updated On  Jan 31, 2024 
Topic  Application of Derivatives 
Subject  Mathematics 
Class  Class 12 
Answer Type  Text solution:1 