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 A B of the triangle A B C respectively, is___________

Solution

Maths sir · 4 steps

Slide D along BC and watch two small similar triangles; the parallelogram is whatever they leave behind.

1. Area of triangle. With A(4,-2), B(1,1), C(9,-3): Δ = 1/2|4(1+3) + 1(-3+2) + 9(-2-1)| = 1/2|16 - 1 - 27| = 6.
2. Set up the shape. In parallelogram AFDE, DF ∥ AC and DE ∥ AB. Let BD = t · BC with 0 < t < 1. Then triangle BFD is similar to BAC with ratio t, so its area is t²Δ. Triangle DEC is similar with ratio 1-t, so its area is (1-t)²Δ.

   Area(AFDE) = Δ[1 - t² - (1-t)²] = 2Δ t(1-t)

3. Maximise. Differentiate t(1-t): 1 - 2t = 0 gives t = 1/2, and the second derivative -2 confirms a maximum. So D is the midpoint of BC, and the area is 2Δ × 1/4 = Δ/2.
4. So. Maximum area = 6/2 = 3.

The common slip: 6. Stopping at the area of triangle ABC gives 6. The parallelogram can cover at most half of the triangle, so halve it.

The idea to keep. A parallelogram inscribed in a triangle, sharing one of its angles, can cover at most half the triangle. It reaches that when its far corner is the midpoint of the opposite side.

Answer from the NTA final answer key. Solution by PrepWiser.