Straight Lines
Parallelogram and Diagonals
Grade 11
Question:
<p>Let <em>ABCD</em> be a parallelogram, the equations of whose diagonals are \(AC: x + 2y - 3 = 0\) and \(BD: 2x + y - 3 = 0\). If the length of the diagonal \(AC = 4\) units and the area of the parallelogram \([ABCD] = 8\) square units. The length of side <em>BD</em> is:</p>
<p>\(\dfrac{20}{3}\)</p>
<p>5</p>
<p>\(\dfrac{10}{3}\)</p>
<p>2</p>
Step-by-Step Solution
Key Concept: In a parallelogram, diagonals bisect each other at their intersection point. The area formula relates to both diagonals and the angle between them: Area = (1/2)d₁·d₂·sin(θ), where θ is the angle between diagonals.
<p><strong>Step 1:</strong> Find the intersection point of diagonals by solving AC and BD simultaneously.</p><p>From AC: x + 2y - 3 = 0 and BD: 2x + y - 3 = 0</p><p>Solving: x = 1, y = 1. Center O = (1, 1)</p><p><strong>Step 2:</strong> Find the angle θ between the diagonals using their slopes.</p><p>Slope of AC: m₁ = -1/2; Slope of BD: m₂ = -2</p><p>tan(θ) = |m₁ - m₂|/(1 + m₁m₂) = |-1/2 + 2|/(1 + 1) = (3/2)/2 = 3/4</p><p>Therefore: sin(θ) = 3/5 (using sin²θ + cos²θ = 1)</p><p><strong>Step 3:</strong> Use the area formula for parallelogram in terms of diagonals.</p><p>Area = (1/2)·AC·BD·sin(θ)</p><p>8 = (1/2)·4·BD·(3/5)</p><p>8 = (6/5)·BD</p><p>BD = 40/6 = 20/3 units</p><p>∴ Answer: A</p>
Correct Answer: A