Question
$$AB$$ is a diameter of a circle and $$C$$ is any point on the circumference of the circle. Then
A.
the area of $$\Delta ABC$$ is maximum when it is isosceles
B.
the area of $$\Delta ABC$$ is minimum when it is isosceles
C.
the perimeter of $$\Delta ABC$$ is minimum when it is isosceles
D.
none of these
Answer :
the area of $$\Delta ABC$$ is maximum when it is isosceles
Solution :
$${\text{Area}}\,{\text{of}}\,\Delta ABC,\,A = \frac{1}{2} \times d\cos \alpha \times d\sin \alpha = \frac{{{d^2}}}{4}\sin 2\alpha $$
$$\eqalign{ & {\text{which is max}}{\text{. when}}\,\sin 2\alpha = 1 \cr & {\text{i}}{\text{.e}}{\text{.}}\,\alpha = {45^ \circ } \cr & \therefore \Delta ABC\,{\text{is an isosceles triangle}}{\text{.}} \cr} $$
$${\text{Area}}\,{\text{of}}\,\Delta ABC,\,A = \frac{1}{2} \times d\cos \alpha \times d\sin \alpha = \frac{{{d^2}}}{4}\sin 2\alpha $$
$$\eqalign{ & {\text{which is max}}{\text{. when}}\,\sin 2\alpha = 1 \cr & {\text{i}}{\text{.e}}{\text{.}}\,\alpha = {45^ \circ } \cr & \therefore \Delta ABC\,{\text{is an isosceles triangle}}{\text{.}} \cr} $$