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Solution


Q.

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 __________..         [2025]

Ans.

(3)

The maximum area of a parallelogram inscribed in a triangle is half the area of the triangle.

Now, Area of △ABC=12|4(1–(–3))+1(–3–(–2))+9(–2–1)|

                                             =12|16–1–27|=6 sq. units

∴   Area of parallelogram AFDE = 3 sq. units.