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  • AB is a diameter of a circle and C is any point on the circumference of the circle. ThenOption: 1 The area of triangle ABC is maximum when

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

Option: 1

The area of triangle ABC is maximum when it is isosceles. 


Option: 2

The area of triangle ABC is minimum when it is isosceles. 


Option: 3

The perimeter of triangle ABC  is minimum when it is isosceles. 


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\text { Clearly } \angle C=90^{\circ} \text {. }}

\mathrm{\text { Let } \angle A B C=\theta \text { and let } a \text { be the area of triangle } A B C \text {. }}

\mathrm{\begin{aligned} & A C=A B \sin \theta=2 r \sin \theta \\ & B C=A B \cos \theta=2 r \cos \theta \\ & a=\frac{1}{2} A C \cdot B C \\ & =\frac{1}{2} \times 2 r \sin \theta \times 2 r \cos \theta \\ & =2 r^2 \sin \theta \cos \theta=r^2 \sin 2 \theta \end{aligned}}

\mathrm{a \text { is maximum when } \sin 2 \theta=1, \theta=\pi / 4}

i.e. Triangle ABC is isosceles.

Posted by

Divya Prakash Singh

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