Question

Let ABCD be convex cyclic quadrilateral suppose P is a point in the Plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least if
{PA, PB, PC, PD}={3, 4,6,8}
What is the maximum possible area of ABCD

Answer 1

Answer:

Maximum possible area of ABCD = 54

Step-by-step explanation:

Formula: If AB and AC are the sides of a triangle ABC and angle between them is θ, then the area of the traingle ABC is given by

$\boxed{\text{Area}=\frac{1}{2} \times AB\times AC\times \sin\theta}$

Since ABCD is covex cyclic quadrilateral

Therefore point P will be the point of intersection of the diagonals of the quadrilaterls

As shown in the figure

If ∠DPC = θ then

Area of ΔPDC = (1/2) × PD × PC × sinθ

                        = (1/2) × 8 × 6 × sinθ

                       = 24sinθ

Area of ΔPBC = (1/2) × PB × PC × sin(180° - θ)

                        = (1/2) × 4 × 6 × sinθ

                       = 12sinθ

Area of ΔPAB = (1/2) × PA × PB × sinθ

                        = (1/2) × 3 × 4 × sinθ

                       = 6sinθ

Area of ΔPAD = (1/2) × PA × PD × sin(180° - θ)

                        = (1/2) × 3 × 8 × sinθ

                       = 12sinθ

Therefore, area of the quadrilateral ABCD = (24 + 12 + 6 + 12)sinθ

                                                                      = 54sinθ

The area of the quadrilateral will be maximum when sinθ = 1

Therefore, the maximum area of the quadrilateral = 54