Straight Lines · Question 24

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 .

Answer

3

Solution

Area of △ABC = 12 |4(1 + 3) + 1(-3 + 2) + 9(-2 - 1) | = 12|16 - 1 - 27| = 6

In parallelogram AFDE, DE ∥ AB and DF ∥ AC.

Let BDBC = t (0 < t < 1).

△BFD ~ △BAC area(△BFD) = t2 6

△DEC ~ △BAC area(△DEC) = (1 - t)2 6

area(AFDE) = 6 [1 - t2 - (1 - t)2 ] = 12t(1 - t)

t(1 - t) is maximum at t = 12 (value 14)

Maximum area = 12 14 = 3 (half the area of the triangle)

Answer: 3