Question

Consider a rectangle ABCD formed by the points A ≡ (0, 0), B ≡ (6, 0), C ≡ (6, 4) and D ≡ (0, 4). P(x, y) is a moving interior point of the rectangle, moving in such a way that d(P, AB) < mini. {d(P, BC), d(P, CD), d(P, AD)}. Here d(P, AB), d(P, BC), d(P, CD) and d(P, AD) represents the distances of the point P from the sides AB, BC, CD and AD respectively. Area of the region representing all possible positions of the point P is equal to

• A. 8 sq. units
• B. 4 sq. units
• C. 12 sq. units
• D. 6 sq. units

Answer 1

Answer: (A)

8 sq. units

Answer 2

d(P, AB) = y, d(P, BC) = 6 − x,

d(P, CD) =4 – y, d(P, AD) = x,

We must have y ≤ 6 – x, y ≤ 4 – y, y ≤ x

⇒ x + y ≤ 6, y ≤ 2, y ≤ x

Shaded region represents the required area. This area is equal to area of trapezium ABB₁A.

∆ABB₁A₁ = $\frac { 1 } { 2 }$(8 + 2).2 = 8 sq. units