Question

Let $ A(4, -2), B(1, 1) $ and $ C(9, -3) $ be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.

Hint:

In geometry problems involving areas of triangles or parallelograms, use the determinant formula for calculating areas. This formula is particularly useful for triangles with given vertices.

Answer 1

Correct Answer: 3

The area of triangle \( \triangle ABC \) is: \[ \text{Area of } \triangle ABC = \frac{1}{2} \left| \begin{array}{ccc} 4 & -2 & 1 1 & 1 & 1 9 & -3 & 1 \end{array} \right| \] \[ = 6 \text{ square units} \] The maximum area of parallelogram \( AFDE \) is given by: \[ \text{Maximum area of } AFDE = \frac{1}{2} \times 6 = 3 \text{ square units} \]