JEE Main 2025 Apr 2 Shift 2, Mathematics Q24: Maxima & Minima

JEE Main2025Apr 2, Shift 2Mathematics

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

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Solution

Area of $△ A B C = \frac{1}{2} ∣ x_{A} (y_{B} - y_{C}) + x_{B} (y_{C} - y_{A}) + x_{C} (y_{A} - y_{B}) ∣$

$= \frac{1}{2} ∣4 (1 - (- 3)) + 1 ((- 3) - (- 2)) + 9 ((- 2) - 1) ∣ = \frac{1}{2} ∣16 - 1 - 27∣ = 6$ .

Classical result: among all parallelograms inscribed in a triangle with one vertex at a fixed corner of the triangle and the opposite vertex on the opposite side, the maximum area equals half the area of the triangle (achieved when $F$ and $E$ are the midpoints of the adjacent sides, making $F D E$ the medial configuration).

$∴$ maximum area of $A F D E = \frac{1}{2} \times 6 = 3$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q24: Maxima & Minima

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