AB is a diameter of a circle and C is any point on the circumference of the circle. Then the area of ` A B C` is maximum when it is isosceles the area of ` A B C` is minimum when it is isosceles the perimeter of ` A B C` is minimum when it is isosceles none of these

AB is a diameter of a circle and C is any point on the circumference of the circle. Then a) the area of A B C is maximum when it is isosceles b) the area of A B C is minimum when it is isosceles c) the perimeter of A B C is minimum when it is isosceles d) none of these

AB is a diameter of a circle and C is any point on the circumference of the circle. Then (a) the area of DeltaABC is maximum when it is isosceles (b) the area of DeltaABC is minimum when it is isosceles (c) the perimeter of DeltaABC is minimum when it is isosceles (d) none of these

A B is a diameter of a circle and C is any point on the circle. Show that the area of A B C is maximum, when it is isosceles.

If a chord AB of a circle subtends an angle theta(nepi//3) at a point C on the circumference such that the triangle ABC has maximum area , then

If A is the area and C is the circumference of a circle, then its radius is (a) A/C (b) (2A)/C (c) (3A)/C (d) (4A)/C

The circumference of Circle A is twice the circumference of circle B. {:("Quantity A","Quantity B"),("The area of Circle A","Twice the area of Circle B"):}

Prove that the area of right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Prove that the area of right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

A B C P is a quadrant of a circle of radius 14 cm. With A C as diameter, a semi-circle is drawn. Find the area of the shaded portion.

An isosceles triangle of vertical angle 2theta is inscribed in a circle of radius a . Show that the area of the triangle is maximum when theta=pi/6 .