Torque balance in rotational equilibrium Steps:

• For rotational equilibrium about pivot P, net torque is zero: torque from F equals torque from 2.5F.
• Torque from F is F × 20 cm.
• Torque from 2.5F is 2.5F × d, where d is its moment arm; set equal: 2.5F × d = F × 20 cm, so d = 20 / 2.5 = 8 cm.
• Force F at corner Q is vertical (moment arm = length/2 = 20 cm); force 2.5F at adjacent corner R is horizontal (moment arm = width/2 = 8 cm).
• Thus, length = 40 cm, width = 16 cm; area = 40 × 16 = 640 cm². Why C is correct:
• Matches area calculation from torque balance using moment arms as half-dimensions for perpendicular forces at adjacent corners. Why the others are wrong:
• A: Underestimates dimensions (e.g., assumes equal arms, yielding 20 × 8 = 160 cm²).
• B: Half the correct area (ignores full dimensions).
• D: Overestimates (e.g., uses diagonals or full distances without halving). Final answer: C