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5. AB is a diameter of a circle and C is any point on the circumference of the circle. Then (a) The area of D ABC is maximum when it is isosceles (b) The area of D ABC is minimum when it is isosceles (c) The perimeter of D ABC if maximum when it is isosceles (d) None of these


5.     AB is a diameter of a circle and C is any point on the circumference of the circle. Then


        (a)    The area of DABC is maximum when it is isosceles


(b)    The area of DABC is minimum when it is isosceles


(c)    The perimeter of DABC if maximum when it is isosceles


(d)    None of these


Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
10 years ago
Ans: (a)
Sol:
Let the circle be
x^{2}+y^{2}=(\frac{AB}{2})^{2}
Let AB is the the diameter on the x-axis. Then coordinates of C will be:
(\frac{AB}{2}cos\theta , \frac{AB}{2}sin\theta )
Then area A of triangle ABC will be
A = \frac{1}{2}.AB.\frac{ABsin\theta }{2}
A = \frac{AB^{2}}{4}sin\theta
A is maximum when sin value is maximum which is 1.
sin\theta =1
cos\theta =0
Then coordinates of C will be:
(0, \frac{AB}{2})
which proves that ABC is an isosceles triangle.
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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