In a rhombus, one diagonal is 60% of the other (let the larger be D and the smaller be 0.6D). By what factor is the rhombus area compared to D^2?

Introduction / Context:The area of a rhombus equals half the product of its diagonals: A = (1/2) * d1 * d2. When one diagonal is a fixed percentage of the other, the area becomes a fixed fraction of the larger diagonal squared. This is a direct substitution problem.

Given Data / Assumptions:

• Larger diagonal = D.
• Smaller diagonal = 0.6D.
• Rhombus area A = (1/2) * D * (0.6D).

Concept / Approach:Substitute and simplify to find A as a multiple of D^2, then express that multiple as a simple fraction.

Step-by-Step Solution:A = (1/2) * D * (0.6D) = 0.3 D^20.3 = 3/10 ⇒ A = (3/10) * D^2

Verification / Alternative check:Pick D = 10: then the other diagonal is 6; area = (1/2)*10*6 = 30; and (3/10) * D^2 = (3/10)*100 = 30, matching.

Why Other Options Are Wrong:1/5 = 0.2 and 2/5 = 0.4 do not match 0.3; 6/7 is unrelated to diagonal products.

Common Pitfalls:Mistaking 60% as 0.06; forgetting the 1/2 factor in the rhombus area formula.

Final Answer:3/10