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]

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 __________.. [2025]

Smart Achievers · Accepted Answer

(3)

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

Now, Area of $△ A B C = \frac{1}{2}$ $| 4 (1 - (- 3)) + 1 (- 3 - (- 2)) + 9 (- 2 - 1) |$

$= \frac{1}{2}$ $| 16 - 1 - 27 |$ $= 6 sq . units$

$∴$ Area of parallelogram AFDE = 3 sq. units.

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 $△ A B C = \frac{1}{2}$ $| 4 (1 - (- 3)) + 1 (- 3 - (- 2)) + 9 (- 2 - 1) |$

$= \frac{1}{2}$ $| 16 - 1 - 27 |$ $= 6 sq . units$

$∴$ Area of parallelogram AFDE = 3 sq. units.

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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.

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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 $△ A B C = \frac{1}{2}$ $| 4 (1 - (- 3)) + 1 (- 3 - (- 2)) + 9 (- 2 - 1) |$

$= \frac{1}{2}$ $| 16 - 1 - 27 |$ $= 6 sq . units$

$∴$ Area of parallelogram AFDE = 3 sq. units.