is a diameter of a circle and is any point on the circumference of the circle. Then,
- A.the area of is maximum when it is isosceles
- B.the area of is minimum when it is isosceles
- C.the perimeter of is minimum when it is isosceles
- D.None of the above
Show answer and solution
Correct option: A
Solution
Let the radius be r, with AB = 2r as diameter. Since angle ACB = 90°, place C on the circle with the angle θ that AC makes. Then AC = 2r·cosθ and BC = 2r·sinθ.
Area = (1/2)·AC·BC = (1/2)·(2r cosθ)(2r sinθ) = r²·sin2θ.
This is maximum when sin2θ = 1, i.e., θ = 45°, meaning AC = BC = r√2, so the triangle is isosceles (right-angled isosceles triangle). Hence area is maximum when △ABC is isosceles.
Area = (1/2)·AC·BC = (1/2)·(2r cosθ)(2r sinθ) = r²·sin2θ.
This is maximum when sin2θ = 1, i.e., θ = 45°, meaning AC = BC = r√2, so the triangle is isosceles (right-angled isosceles triangle). Hence area is maximum when △ABC is isosceles.
[1983]